Chemistry notes Important terms

*Mass of element in a sample Periods- horizontal rows on a periodic table Groups- vertical rows on a periodic table Ionic compound – electron transfer from a metal to a non-metal Covalent compound- electron sharing between two non-metals Hydrates- have a specific number of water molecules associated with each formula unit ( shown with a *#H2O) Aqueous solutions are solutions in water Combustion analysis- add O2 H2O +CO2

Limiting reactant- substance that stops the reactions from proceeding Molarity- concentration of a solution (mole/liters) Polar molecule : a molecule that is electronegativity charged Precipitate reaction- when two soluble ionic compounds combined to form an insoluble product Acid-base reaction- acid reacts with a base to form a neutral substance Titration- one solution of known concentration is used to determine the concentration of another solution through a monitored reaction End point- tiny excess of OH- ion changes Oxidation-reduction reaction(redox)- net movement of electrons from one reactant to the other, goes from less electronegative to more electronegative rules

1. For atoms in their elemental form, the oxidation number is 0

2. For ions, the oxidation number is equal to their charge

3. For single hydrogen, the number is usually +1 but in some cases it is -1

4. For oxygen, the number is usually -2

5. The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal

to its total charge.

Standard temperature and pressure(STP)- 273.15K and 1 atm Standard molar volume = 22.4 L Universal gas constant= .082058 atm*L Mol*K Mole fraction(X) the fraction of each element in a compound Kinetic-molecular theory- describes behavior of gas at macroscopic levels Root mean squared speed (rms) a molecule moving at this speed has the average kinetic energy Effusion- gas escaping through a tiny hole Diffusion- gas moving through another gas Equations

Mass of compound in sample * mass of element in compound Mass of compound

Molecular mass= sum of atomic masses Avogadro’s number= 6.022*1023= 1 mole Mass % of element x = moles of X in formula * molar mass of X (g/mole) Mass(g) of 1 mole of compound %Yield=actual yield *100 theoretical yield PV=nRT n= m=PV M RT Pa=Xa * Ptotal

Urms= √3RT/M R= 8.314 (J/k*mol) Rate of effusion Rate a =√Mb Rateb √Ma Vander Waals equation = (Pn2a) (V-nb) =nRT V2 a and b are Vander Waals constants A relates to the number of electrons B relates to molecular volume Laws

*law of multiple proportions if element A and B react to form tow compounds, the different masses of B than combine with a fixed mass of A can be expressed as a ratio of small whole numbers *law of mass conservation the total mass of a substance does not change during a chemical reaction. The number of substance may change. The total amount of substance is constant *law of definite composition no matter what its source a particular compound is composed of the same elements in the same parts by mass. *coulombs law the energy of attraction is directly proportional to the product of the charges and inversely proportionate to the distance between them *ideal gas law PV=nRT *Boyles law at constant temperature the volume occupied by a fixed amount of gas is inversely proportionate to the applied pressure *Charles law

at constant pressure, the volume occupied by a fixed amount of gas is directly proportionate to its absolute temperature *Dalton’s law of partial pressure in a mixture of unreacting gas the total pressure I the sum of the partial pressure of the individual gases Ptotal=P1+P2+P3+P4… *Grahams law of effusion the rate of effusion of a gas is inversely proportional to the square root of molar mass Chapter 16 kinetics Factors that influence reaction rate

1. Concentration – molecules must collide to react a. Reaction rate is proportional to concentration of reaction

2. physical state – molecules must mix to collide a. the more finely divided a solid or liquid reaction , the greater its

surface area per unit volume, the more contact it makes with the other reactant, the faster it occurs

3. Temperature- molecules must collide with enough energy to react a. At higher temperature more collisions occur in a given time. b. Raising the temperature raises the reaction rate

Express your reaction Rate of motion = Change in position = x2-x1 = Δx Change in time t2-t1 Δt Reaction rate- change in concentration; reactant decreases while product increase at set rate Rate = - Δ[A] [ ] represent concentration in moles per liter Δ t Average , instantaneous, and initial reaction rates

Rate itself varies with time as the reaction proceeds The instantaneous rate decreases as during the course of the reaction The slope of straight line from two points on an x/y graph is the average

Initial rate- the instantaneous rate when the reaction occurs aA+ bB cC+ dD Rate= - 1 Δ[A] = - 1 Δ[B] = 1 Δ[C] = 1 Δ[D] c Δt b Δt c Δt d Δt 16.3 Rate law and its components

rate law expresses rate as a function of reactant concentrations, product concentrations, and temperature. Rate depends only on reactant concentration and temperature Rate= k[A]m[B]n k is reaction constant M and n are the reaction orders aka how the rate is affected by reactant concentration The component of the rate law- rate, reaction, order and rate constant must be found through experiment Reaction order terms First order Rate = k[A] Second order Rate = k[A]2 Zero order- doesn‘t depend on A Rate = k{A}0 Example 2NO + 2H2 N2 + O2 Rate = k[NO]2[H2] Second order in NO and first order in H2 overall it‘s a third order Another example: CHCl3 + Cl2 CCl4+HCl

Rate= k[CHCl3][Cl2]1/2

Reaction orders experimentally Run a series of experiments staring each with a different set of reactant concentration and obtaining an initial rate for each case

Integrated rate law Rate -Δ{A} =k[A]m Δt First order is Ln[A]0 =kt [A]t second order is 1 -1 =kt [A]t [A]0 zero order is [A}t -[A]0 = -kt half-life of rate law half-life of first order is constant ln 2 k = half life second order half life t1/2= 1 . k[A]0

zero order half life t1/2= [A]0 2k Effects of temperature on reaction

K increases as temperature increase Use Arrhenius equation K=Ae=Ea/RT A is a constant Ea is activation energy- minimum energy needed for a reaction R is universal gas constant T in temperature in K Higher T larger k increase in rate Rewrite the equation Ln k2 = Ea(1 - 1 ) K1 R(T2 T1)

16.6 Explanation of effect of concentration and temperature

There are two theories that explain 1. collision theory-views the reaction rates as the result of parties colliding with certain frequencies and minimal energy. 2. transition state theory- offers close up view of how energy of collision converts from reactant to product collision theory measured in collisions per time explains why reactant concentrations are multiplied together in the rate law Why concentrations are multiplied Because the different paths that a particle can take to get to the end result Temperature affects the rate because collisions must have enough energy to overcome the minimum amount needed in a reaction called activation energy Two types of activation energy

1. Ea forward- energy difference between activated state and reactant. 2. Ea reverse- difference between activated energy and product

the smaller the Ea the larger the value of k, and the faster a reaction larger Ea (or lower T) smaller k decrease rate effective collision- the molecules must collide so that the reacting atoms make contact. Aka must have enough energy and particular molecular orientation TRANSITION STATE THEORY Explains why activation energy is needed and what the molecules look like.

- if the potential energy is less than the activation energy, molecules recoil, like billiard balls.

- The kinetic energy pushes them together with enough force to overcome repulsions and react

Transition state aka atiatied complex: a state that is not product or reactant; very frail; highest potential energy 16.7 Reaction mechanisms: steps in the overall reaction

reaction mechanism- a sequence of single reaction steps that sum to an overall reaction A+B E +D

Reaction intermediate- a substance that is formed and used up during the overall reaction Elementary steps- a single molecular event, molecules decompose or collide Unimolecular reaction A B +C Bimolecular reaction A +B C We use the equations coefficient as the reaction order in rate law for elementary steps Rate determining step: a step in a reaction slower than every other that affects overall rate Slow step determines rate We can‘t prose from just data that particular mechanism represents actual chemical change 16.8 Catalysis: speeding up a chemical reaction

catalysis: a substance that increases the rate without being consumed in the reaction catalysis lower activation energyrate constant higher homogeneous catalysis: exist in solution with reaction mixture heterogeneous catalysis: speed up reaction that occur in separate stages enzyme- a catalyst found in the human body, it is a protein active site- a small region whose shape results from those of side chains of amino acids, every enzyme has one 2 models of enzymes

1. lock and key – key(substance) lock is active site 2. induced fit – hand entering a glove

all enzymes function by binding to the reactions transition state and thus stabilizing it ozone reactions O2

uv 2O O + O2O3 (ozone) O + O3 2O2 (ozone breakdown)

Chapter 17 Equilibrium: the extent of chemical reactions

Focuses on 17.1 Equilibrium state & equilibrium constant summary:

kinetics and equilibrium are distinct aspects of a chemical reaction, thus the rate extent of a reaction are related.

When the forward and reverse reactions occur at the same rate, the system has reached dynamic equilibrium.

The equilibrium constant (K) is a number based on a particular ratio of products and reactant concentrations: K is small for reaction that reach equilibrium with a high concentration of reactant(s) and large for reactions that reach equilibrium with a low concentration of reaction

17.2 reaction quotient & equilibrium constant

17.3 expressing equilibrium w/ pressure

The reaction quotient and the equilibrium constant are most often expressed in terms of concentration(Qc & Kc). for gases, they can also be expressed in terms of partial pressure(Qp & Kp)

The value of Kp and Kc are related by the derivation that relies on the ideal gas law Kp=Kc(RT)Δngas

17.4 reaction direction: comparing Q and K

We compare the value of Q and K to determine the direction I which a reaction will proceed towards equilibrium

If Qc<Kc, more product forms

If Qc> Kc more reactant is formed

If Qc=Kc there is no net charge 17.5 how to solve equilibrium problem

In equilibrium problems, we typically use quantities( concentrations or pressure) of reactants and products to find K, or we use K to find quantities

Reaction tables summarize the initial quantities, how they change and the equilibrium quantities

To simplify calculation, we assume that if K is small and the initial quantity of reactant is large, the unknown change in reactant (x) can be neglected. If this assumption is not justified( that is, if the error that results is greater than 5%), we use the quadratic formula to find x

To determine reaction direction, we compare the values of Q and K

17.6 reaction conditions : Le Chatelier‘s principle

Le Chatelier principle states that if a system at equilibrium is disturbed, the system undergoes a net reaction that reduces the disturbance and allows equilibrium to be retained

Changes in concentration cause a net reaction away from the added component or toward the removed component

For a reaction that involves a change in the number of moles of gas, an increase in pressure(decrease in volume) causes a net reaction toward fewer moles of gas, an increase in pressure(decrease in volume) cause a net reaction towards fewer moles of gas and a decrease in pressure causes the opposite change.

Although the equilibrium concentration of components changes as a result of concentration and volume, K does not change. A temperature change, however, does change K: higher T increase K for an endothermic reaction (positive ΔHrxn) and decreases K for an exothermic reaction ( negative ΔHrxn)

A catalyst causes a system to reach the equilibrium point more quickly

Ammonia production is favored by high pressure, low temperature, and continual removal product. To make the process economical, an intermediate temperature and a catalyst are used

17.1 equilibrium state and the equilibrium constant

equilibrium: reactant and product concentrations stop changing because the forward reverse rates have become equal ratefwd = raterev no further change is observed at equilibrium because changes in one direction are equal to changing in the opposite direction Kfwd[N2O4]eq=krev[NO2]

2eq

K =Kfwd = [NO2]

2eq = product

Krev = [N2O4]eq reactant K is the equilibrium constant- number equal to particular ratio of equilibrium concentration of product and reactant at a particular temperature

1. small K – reaction yields very little product ; it looks like there is ‗no reaction‘

2. Large K – reaction reaches equilibrium with very little reactant remaining; ― goes to completion‖

3. Intermediate K – significant amount of both reactant and product are present

17.2 reaction quotient and equilibrium constant

History: In 1864 two Norwegian chemist, Cato Goldberg & Peter Waage found that at a given temperature a chemical system reaches a state in which a

particular ratio of reactant to product concentration has a constant value Aka law of chemical equilibrium or law of mass action Reaction quotient (Q) –aka mass-action expression

Q= [product] [reactant]

@ equilibrium Q=K note

1. ratio of initial concentration varies widely but always gives the same ratio of equilibrium concentration

2. individual equilibrium concentrations are different in each case, but the ratio of these concentrations is constant

reaction quotient (Q) is a ratio made up from product concentration terms multiplied together and divided by reactant concentration terms multiplied together. aA +bB cC + dD a , b , c, d are stoichiometry coefficients Qc=[C]c[D]d [A]a[B]b To construct the reaction quotient you must write the balance equation first Q & K are unitless numbers because it is a ratio If the overall reaction is the sum of two or more reactions, the overall reaction quotient is the product of the reaction quotients Qoverall = Q1 *Q2 * Q3 *…. Koverall = K1 *K2 *K3 *… The form of the reaction depends on the direction the balanced equation is written. Qc(fwd] = 1/Qc(rev) Kc(fwd) = 1/ Kc(rev) If you multiply a reaction by a coefficient the Q must be raises to that power If all coefficients of the balanced equation are multiplied by some factor, that factor becomes the exponent for relating the reaction quotients and the equilibrium constant. N (aA +bB cC +dD) Q‘=Qn=([C]c[D]d)n K‘=Kn

([A]a[B]b)n

heterogeneous equilibrium- components are in different phases a pure solid and liquid always has the same concentration, same density We eliminate the terms for pure liquids and solids from reaction quotient… they do not change 17.3 expressing equilibria with pressure terms: relation between Kc & Kp

we start with the ideal gas law PV=nRT if T is constant then pressure is directly proportional to molar concentration reaction quotient based on partial pressures (Qp

)

2NO(g) + O2(g)2NO2(g) Qp = P2

NO2 . P2

NO*PO2

Kp= equilibrium constant based upon pressure Kp =/= Kc

But you can use change in moles of gasΔn to find Kp = Kc(RT)Δn 17.4 reaction direction: comparing Q &K

more products make Q larger more reactant makes Q smaller 3 types of relative size of Q and K

1. if Q< K reactant products 2. if Q>K, reactant product 3. if Q=K, reactantproducts

17.5 How to solve equilibrium problems

Review: -reactant and product concentrations are constant over time -the forward reaction rate equals the reverse reaction rate -the reaction quotient equals the equilibrium constant Q=K Principles

- we are given equilibrium quantities and solve for K - we are given K and initial quantities and some for equilibrium quantities

USE QUANTITES TO FIND EQUILBRUM CONSTANT Example: flask 1.5 Liters H2(g) + I2(g) 2HI(g) We have At equilibrium 1.8 mol H2 1.8 mol I2 and .520 mol HI. We find Kc through concentrations and substituting them into reaction quotient. Qc = [HI]2 . [H2][I2] We first convert amount (mol) to concentration (mol/Liters) using the 1.5 liter flask. [H2]=1.8 mol = 1.2 M 1.5 L We get [I2] =1.2 mol, & [HI] =.347M Substitute values for Qc Kc = (0.347)2 . = 8.36 *10-2 (1.2)(1.2) Using a reaction table CO2 + C(graphite) 2CO(g) Qp= P2CO PCO2 X atm CO22x atm CO Pressure at equilibrium PCO2(initial) –X =PCO2 Setup a table Pressure (atm) CO2 + C(graphite) 2CO(g)

Initial 0.458 ------- 0 Change -x -------- +2x

Equilibrium 0.458-x -------- 2x Use the quadratic formula to solve if in doubt If a reaction has a relatively small K and a relatively large initial reactant concentration, the concentration change (x) can often be neglected, x does not equal 0 [A]initial – [A]reacting = [A}eq =[A]initial

17.6 reaction conditions and the equilibrium state: Le Chatelier‘s principle

Le Chatelier Principle: when a chemical system at equilibrium is disturbed, it retains equilibrium by undergoing a net reaction that reduces the effects of the disturbance Q does not equal K ― Disturbance‖ Three common are change in

1. concentration of component a. if the concentration goes up, system reacts to consume some

i. equilibrium position moves right when reactant added b. if the concentration does down, systems reacts to reduce some

i. equilibrium position move to the left when reactant reduced The equilibrium system reacts to consume some of added substance or produce some of the removed substance New table Concentration(M) PCl3(g) + Cl2(g) PCl5(g)

Original equilibrium .2 .125 .6 Disturbance 0 +.075 0 New initial .2 .2 .6 Change -x -x +x New equilibrium .2-x .2-x .6+x Kc stays the same

2. change in pressure(volume) a. change in concentration of gas b. adding an inert gas (gas does not take part)

i. no effect c. changing the volume of vessel

i. decrease volume raises concentration ii. increase volume decreases concentration iii. does not alter Kc iv. more volume = more product v. more pressure leads to side with less moles gas

3. change in temperature

a. only way to alter K b. temperature rises, K rises and ΔHrxnx is positive c. temperature down, K down ΔHrxn negative

A catalyst shortens the time it takes to reach equilibrium but has no effect on the equilibrium position. Haber Process N2(g) + 3H2(g) 2NH3(g) ΔHrxn =-91.8Kj 1. Decrease concentration of ammonia. NH3 is the product, so removing it will

shift the equilibrium position towards producing more 2. Decreasing Volume(increase in pressure). Because 4 moles of gas react to

form two moles of gas, decrease volume will shift the equilibrium position toward fewer moles of gas.

3. Decrease temperature. Because the formation of ammonia is exothermic, decreasing the temperature will shift the equilibrium position towards formation of product, thereby increasing Kc

Chapter 18 Acid base equilibria

18.1 acid bases in water

In aqueous solution, water binds the proton released from an acid to form the hydrated species represented by H3O

+(aq)

In Arrhenius definition , acids contain H and yield H3O+ in water, bases

contain OH and yield OH- in water, and acid-base reaction(neutralization) is the reaction of H+ and OH- to form H2O

Acid strength depends on [H3O] related to [HA] in aqueous solution. Strong acids dissociate completely and weak acids slightly

The extent of dissociation is expressed by the acid-dissociation constant Ka. weak acids range from 10-1 to 10-12

Many acids and bases can be classified qualitatively as strong or weak based on their formulas

18.2 auto ionization of water and pH scale

Pure water has a low conductivity because it auto ionizes to a small extent. This process is described by an equilibrium reaction whose equilibrium constant is the ion-product for water, Kw(10-14). Thus [H3O] and [OH] are inversely related: in acidic solution, [H3O] is greater than[OH]; the reverse is true in basic solutions; and the two are equal in a neutral solution

To express small values of [H3O]more simply, we use the pH scale (pH = -log[H3O]). A high pH represents a low [H3O]. In acidic solution, pH <7.00; I basic solutions, pH>7.00; and in neutral solutions pH=7.00. Similarly pOH= -log[OH], and pK = -logK. The sum of pH and pOH equals pKw(14)

18.3 proton transfer and he Bronsted-Lowry acid- base definition

The bronsted-lowry acid-base definition does not require that bases contain OH or that acid-base reaction occur I aqueous solutions

An acid is a species that donates a proton and a base is one that accepts it

An acid and a base act together in proton transfer. When an acid donates a proton, it becomes the conjugate base; when a base accepts a proton it becomes the conjugate acid. I an acid-base reaction, acids and bases form their conjugates. A stronger acid has a weaker conjugate base and vice versa

An acid-base reaction proceeds in the net direction I which a stronger acid and base form a weaker base and acid

18.4 solving problems involving weak-acid equilibra

Two common types of weak- acid equilibrium problems involve Ka from a concentration and find a concentration from Ka

We summarize the information in a reaction table (ICE) and we simplify the arithmetic by assuming [H3O ]from H2O is so small relative to [H3O]from HA that it can be neglected and weak acids dissociate so little that [HA]init=[HA]equilibrium

The fraction of weak acid molecules that dissociates is greater in a more dilute solution, even though the total [H3O] is less

Polyprotic acids have more than one ionizable proton, but we assume that the first dissociation provides virtually all H3O

18.5 weak bases and their relation to weak acids

The extent a weak base accepts a proton from water to from OH is expressed by base dissociation constant Kb

Bronsted-lowry base include NH3 and amines and the anions of weak acids. All produce basic solution by accepting H+ from water, which yields OH and makes [H3O]<[OH}

A solution of HA is acidic because [HA] >>[A], so [H3O]>>[OH].

A solution of A is basic because [A]>>[HA] so [OH]>>[H3O]

By multiplying the expression for Ka of HA and Kb of A we obtain Kw. this relationship allows us to calculate either Ka of BH, the cation conjugate acid of weak base B, or Kb of A, the anion conjugate base for weak acid HA

18.6 molecular properties and acid strength

The strength of an acid depends on the ease with which the ionizable proton is released

For nonmetal hydrides, acid strength increases across the period, with the electronegativity of the nonmetal (E, and don a group, with the length of E—H bond

For oxoacids with the same number of O atoms, acid strength increases with electronegativity of E; for oxoacids with the same E, acid strength increases number of O

Small, highly charged metal ions are acidic in water because they withdraw electron density from the O-H bond of bound H2O molecules, releasing an H ion to the solution

18.7 acid-base properties of salt solutions

Salts that always yield a neutral solution consist of ions that do not react with water

Salt that always yield an acidic solution contain unreactive anion and a cation that releases a proton to water

Salts that always yield a basic solution contain an unreactive cation and an anion that accepts protons from water

If both cation and anion react with water, the ion that reacts to the greater extent (higher K) determines the acidity or basicity of the salt solution

If the anion is amphiprotic(first anion a polyprotic acid) the strength of the anion as an acid (Ka) or a base (Kb) determines the acidity of the salt solution

18.8 electron-pair donation and the Lewis acid-base definition

The Lewis acid base definition focuses on the donation or acceptance of an electron pair to form a new covalent bond in an adduct, the product of an acid-base reaction. Lewis bases donate the electron pair, the Lewis acids accept it. Many species that do not contain are Lewis acids.

Molecules with polar double bonds act as Lewis acids, as do those with electron deficient atoms.

Metal ions act as Lewis Acids when they dissolve in water, which acts as a Lewis base to from an adduct, a hydrated cation

Many metal ions function as Lewis acids in biomolecules 18.1 acids and bases in water

Acids and bases react, each cancels the properties of the other in a process called neutralization. Water is a product of all reactions between strong acids and strong bases: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) An acid dissociates in water to make a hydronium molecule HA + H2O A- + H3O Arrhenius acid-bade definition- acids and bases are classified in terms of their formula and their behavior in water

- an acid is a substance that has H in its formula and dissociates into water to yield H3O

+ - a Base is a substance that has OH in its formula and dissociates into

water to yield OH-

Acids contain covalently bonded H atoms that ionize when their molecules dissolve in water. The H+ ion and OH- ion form the base combine h2O

The heat of reaction is abut -57Kj per mole of water formed Strong acids dissociate completely into water. Weak acids dissociate slightly into water Specific equilibrium (Ka)- weak acid in water Kc[H2O]= Ka=[H3O

+][A-] [HA] stronger acid higher[H3O

+]larger Ka smaller Ka lower dissociation of HA weaker acid Acids Strong acids Hydrohalic acids HCl, HBr, HI Oxoacids- 2 more O‘s than H‘s HNO3, H2SO4, HClO4 Weak acids- a lot Hydrohalic acid HF H is not bonded to O or Halogens HCN, H2S Oxoacids- number of O‘s exceeds by 1 over H‘s HClO, HNO2, H3PO4 Carboxylic acids- (RCOOH) CH3COOH, C6H5COOH Bases Strong bases- water soluble compound containing O2- or OH- Group 1A(Li, Na, K, Rb, Cs) M2O, MOH Group 2A(Ca, Sr, Ba) MO, M(OH)2 Weak bases- nitrogen with lone pair Ammonia NH3 Amines (RNH2, R2NH, R3N) CH3CH2NH2, (CH3)2NH, (C3H7)3N

18.2 autoionization of water and the pH Scale

Water is an extremely weak electrolyte. Water dissociates in equilibrium process called autoionization (self- ionizing) Ion-product constant for water (Kwx) Kc[H2O]2 = Kw = [H3O

+][OH-] = 1.0*10-14 (at 25degrees C) Pure waters concentration 55.5M One H3O and one OH ion appear for each H2O molecule that dissociates 1. A change in{H3O] causes an inverse change in [OH] Higher [H3O] lower [OH] 2.Both ions are present in all aqueous systems

In an acidic solution [H3O]>[OH] In a basic solution [H3O]<[OH] In a neutral system [H3O] = [OH] [H3O]=Kw . [OH]

pH scale pH = -log[H3O

+] [H3O] can varies from 10M to 10-15M [H3O] = 5.4*10-4M find pH pH=-log[H3O

+]= (-1) (log5.4 +log10-4)=3.27 the number of the significant figures in the concentration equals the number of digits to the right of the decimal point in the pH value The higher the pH, the lower the [H3O] pH of an acidic solution <7.00 pH of a neutral solution = 7.00 pH of a basic solution >7.00 hydroxide ion concentration can be expressed as pOH pOH = -log[OH] acidic solutions have a higher pOH than basic solution pK= -logK a lower pK corresponds to a higher K relationship between pH, pOH, pKw

Kw=[H3O][OH]=1.0*10-14 -logKw = (-log[H3O]) + (-log [OH]) = 1*10-14 pKw = pH + pOH= 14 @25degrees C

table 18.1 Relationship between Ka and pKa

Acid Name (formula) Ka@25C pKa Hydrogen sulfate ion (HSO4

-) 1.0*10-2 1.99 Nitrous acid (HNO2) 7.1*10-4 3.15 Acetic acid (CH3COOH) 1.8*10-5 4.74 Hypobromous acid (HBrO) 2.3*10-9 8.64 Phenol (C6H5OH) 1.0*10-10 10.00 In the lab we measure acidity with a pH indicator or a pH meter. 18.3 proton transfer and the Bronsted-lowry acid base definition

The Bronsted- Lowry acid-base definition took out the limitations of Arrhenius definition

1. an acid is a proton donor, and species that donates a H+ ion a. there must be a H in the formula

2. a base is proton acceptor, any species that accepts a H+ ion a. a base must contains a lone pair of electron to bind the H b.

the only requirement for an acid-base reaction is that one species donates a proton another species accepts it; an acid-base reaction is a proton-transfer process. Typical acidic and basic solutions

1. acid donates a proton to water- a. HCl + H2O Cl- +H3O

+ 2. base accept a proton from water

a. NH3 + H2O NH4+ + OH-

H2S+NH3 HS + NH4 Conjugate acid-base pair- HS is conjugate base of acid H2S NH4 is the conjugate acid of the base NH3 Every acid has a conjugate base, every base has a conjugate acid

Table 18.2 the conjugate pairs in some acid-base reactions

_____________________conjugate pair__ | | acid + Base Base + Acid

| | ------------------conjugate pair--------

reaction1 HF + H2O F- + H3O

+ reaction 2 HCOOH+ CN- HCOO- + HCN+ reaction 3 NH4

+ + CO32- NH3 + HCO-

reaction 4 H2PO4- + OH- HPO4

-2 + H2O reaction 5 H2SO4 + N2H5

+ HSO4- + N2H6

+2 reaction 6 HPO4

2- + SO32- PO3

3- + HSO3-

A reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base forma weaker acid and weaker base Strong weak Kc>1 if strong is on left A weaker acid has a stronger conjugate base An acid-base reaction proceeds to the right if the acid reacts with a base that is lower on the list

18.4 solving problems involving weak-acid equilibria

2 general problems 1. given equilibrium concentration, find Ka 2. given Ka and some concentration information, find the other equilibrium

constants - problem solving approach

o start with what is given in the problem, then apply steps 1. write the balanced equation an Ka expression; these will tell you

what to find 2. define x as the unknown change in concentration during reaction.

Through the use of certain assumptions, also equals [H3O+] and [A-]

at equilibrium 3. construct a reaction table that incorporates the unknown 4. make any assumptions that simplify calculation

a. [H3O] from autoionization is so much smaller [H3O] of dissociation that it can be ignored

b. weak acid has a small Ka 5. substitute the value of Ka and solve for x 6. check the assumption are justified, test it

percent HA dissociated =[HA]dissoc *100 [HA]init

as the original acid concentration decreases, the percent dissociation of the acid increases. do not confuse the concentration of HA dissociated with eh percent of HA dissociated polyprotic acid- acids with more than one ionizable proton. Each time a proton is taken away it gets a new Ka. Ka decreases as protons go Ka1>Ka2>Ka3 18.5 Weak bases and their relation to weak acids

a base must have a lone pair base-dissociation constant (Kb)-no base is dissociated during the process

B(aq) + H2O BH (aq) +OH(aq) Kb = [BH+][OH-] [B]

ammonia is the simplest nitrogen-containing compound that acts as a weak base in water NH3(aq) + H2O NH4(aq) +OH(aq) kb=1.76*10-5 Kb=[NH4][OH] [NH3] Amine group: was a NH3 but H atom gets replaced by N group

RNH2, R2NH, R3N Has a lone pair of electrons to bind proton Rules

1. the acidity of HA(aq) a. the equilibrium position moves to the left.

2. basicity of A-(aq) a. the relative concentration of HA and A determine the acidity or

basicity of the solution 3. In a HA solution, [HA]>>[A-] & [H3O]from HA>>[OH]from H2O, so the solution is

acidic 4. In a A solution, [A]>>[HA]&[OH]from A>> [H3O]from the H2O, so the solution is

basic HA + H2O H3O +A A + H2O Ha +OH

2H2O H3O + OH the sum of the two dissociated reactions is the autoionization of water.

Ka * Kb = Kw

18.6 Molecular properties and acid strength

The strength of an acid depends on its ability to donate a proton. Factors for nonmetal hydrides

1. cross a period, nonmetal hydride acid strength increase, because electronegativity increases.

2. Down a group, nonmetal hydride acid strength increases, as E becomes longer the E-H bond increases, decreasing the bong strength

a. H2O<H2S<H2Se<H2Te

Trends in acid strength in oxoacids -All oxoacids have the acidic H atom bonded to an O atom -2 factors determine acid strength a. Electronegativity of central non metal b. the number of O atoms 1. for oxoacids with the same number of O around the e, acid strength increases across with electronegativity of E

2.for oxoacids with different number of oxygen around a given E, acid strength increase with the number of O atoms.

a. HNO3> HNO2 acids of hydrated metal ions hydrated ions transfer H+ to water. M(NO3)n(s) +xH2O(l) M(H2O)x

n+(aq) +nNO3(aq)

If M is small and highly charged

18.7 Acid-base properties of salt solutions

Salts that yield neutral solutions

A salt consist of the anion of a strong acid and the cation of a strong base yields a neutral solution because the ions do not react with water HNO3(l) +H2O NO3

- (aq) + H3O(aq) The anion of a strong acid is a weak base in water, a strong acid anion is hydrated but nothing more NaOH(s)h2o Na + Oh All strong bases behave this way Salt containing ions such as NaCl or Ba(NO3)2 yield neutral solutions because no reaction takes place between the ions and water. Salts the yield acidic solutions -A salt consisting of the anion of a strong acid and the cation of a weak base yield an acidic solution because the cation acts as a weak acid and the anion does not react. NH4Cl(s)H2O NH4 +Cl {dissolution and hydration} NH4 (aq) + H2O(l) NH3 + H3O { dissociation of weak acid} Small highly charged metal ions make up another group of cations that yield H3O in solution Fe(NO3)3 + 6H2O H2O Fe(H2O)6(aq) +3NO3 (dissociates and hydrates) FE(H2O)6(AQ) + H2O Fe(H2O)5OH2++H3O (dissociate of weak acid) Salts that yield basic solutions

-a salt consist of the anion of a weak acid and the cation of a strong base yields a basic solution in water because the anion acts as a weak base and the cation does not react. Salts of weak acid and weak base ions Acidity depends on Ka and Kb NH4 +H2O NH3 + H3O CN+H2O HCN +OH Ka of NH4 = Kw . = 1.0*10-14 = 5.7*10-10 Kbof HCN 1.76*10-5 Kb of CN = Kw . = 1.0*10-14 = 1.6*10-5 Ka of HCN 6.2*10-10 Solution is basics because Kb>Ka Table 18.3 behavior of salts in water

Salt solution examples

pH Nature of ions Ion that react with water ex:

Neutral: Nacl,Kbr,Ba(NO3)2

7.0 Cation of strong acid Anion of strong base

None

Acidic: NH4Cl, NH4NO3, CH3NH3Br

<7.0 Cation of weak base Anion of strong acid

Cation: NH4 + H2O NH3 + H3O

Acidic: AL(NO3)3, CrBr3, FeCl3

<7.0 Small, highly charged cation Anion of strong acid

Cation: Al(H2O)6 +H2O Al(H2O)5OH

Acidic/ basic NH4ClO2, NH4CN, Pb(CH3COO)2

<7.0 if Ka(cation>Kb(anion)

>7.0 If Kb(anion)> Ka(cation)

Cation of weak base Anion of weak acid

NH4 +H2O NH3 +H3O CN+ H2O HCN + OH

Acidic/ basic: NaH2PO4, KHCO3, NaHSO3

<7.0 if Ka(anion)>Kb(cation) >7.0 if Kb(cation)>Ka(anion)

Cation of strong base Anion of polyprotic acid

Anion: HSO3 +H2O SO3 + H3O HSO3+ H2O H2SO3+OH

18.8 electron-pair donation and the Lewis acid-base definition

definition: - a base is any species that donates an electron pair - an acid is any species that accepts an electron pair

compared to Bronsted- Lowry Lewis expands on classes of acid the product of a Lewis acid-base reaction is called Adduct – a single species that contains a new covalent bond B: + H+

B-H+ \ covalent bond Lewis acid has electron deficiency

- one surrounded by less than 8 electrons

Chapter 19 ionic equilibria in aqueous systems 19.1 equilibria of acid- base buffer systems

a buffered solution exhibits a much smaller change in pH when H3O or OH is added that does an unbuffered solution

a buffer consists of relatively high concentrations of the components of a conjugate weak acid-base pair. The buffer component concentration ratio determines the pH and the ratio and pH are related by the Henderson- hasselbalch equation. As H3O or OH is added, one buffer component concentration ratio, and consequently the free[H3O] (and pH), change only slightly

a concentration buffer undergoes small changes in pH than a dilute buffer. When the buffer pH equals the pKa of the acid component, the buffer has its highest capacity

a buffer has an effective range of pKa+- pH unit.

To prepare a buffer choose the conjugate acid-base pair, calculate the ratio of buffer components, determine buffer concentration, and adjust the final solution to the desires pH

19.2 acid-base titration curves

An acid-base (pH) indicator is a weak acid that has differently colored acidic and basic forms and changes color over about 2pH unit

In a strong acid-strong base titration, the pH starts out low, rises slowly, then shoots up near the equivalence point (pH=7)

In a weak acid- strong base titration, the pH starts out higher, than the strong acid-strong base titration, rises slowly in the buffer region

(pH=pKa at the midpoint), then rises more quickly at the equivalence point (pH>7)

A weak base-strong acid titration curve has a shape that is the inverse of the weak acid-strong base curve, with the pH decreasing to the equivalence point(pH<7)

19.3 equilibria of slightly soluble ionic compounds

As an approximation, the dissolved portion of a slightly soluble salt dissociates completely into ions

In a saturated solution, the ions are in equilibrium with the solid and the product of the ion concentrations, each raise to the power of its subscript in the compound‘s formula, has a constant value (Qsp=Ksp)

The value of Ksp can be obtained from the solubility , and vice versa

Adding a common ion lowers an ionic compound‘s solubility

Adding H3O (lowering pH) increases a compounds solubility if the anion of the compound is that of a weak acid

If Qsp> Ksp for an ionic compound, a precipitate forms when two solution, each contains one of the compounds ions, are mixed

Lakes bounded by limestone-rich soil form buffer systems that prevent harmful acidification by acid rain

19.4 equilibria involving complex ions

A complex ion consists of a central metal ion covalently bonded to two or more negatively charges or neutral ligands. Its formation is described by a formation constant( Kf)

A hydrated metal ion is a complex ion with water molecules as ligands, other ligands can displace the water in a stepwise process. In most cases, the Kf value of each step is large, so the fully substituted complex ion forms almost completely in the presence of excess ligand

Adding a solution containing a ligand increases the solubility of an ionic precipitate if the cation forms a complex ion with the ligand

19.1 equilibria of acid base buffer systems

buffer- something that lessens the impact of an external force. Acid-base buffer- a solution that lessens the impact of pH from the addiction of acid or base. Example: a shelter when it rains The components of a buffer are the conjugate acid-base pair of a weak acid (or base). How a buffer works

Common-ion effect- it‘s how buffers work

Example : CH3COOH(aq) +H2O H3O(aq) + CH3COO(aq) From Le Chatelier principle equations shifts left if CH3COO is added CH3COO is common ion in this case, it occurs when a given ion is added to an equilibrium mixture that already contains that ion, and shifts the position away from forming more of it. Common ion suppresses the dissociation of CH3COOH, making the solution less acidic. %Dissociation = dissoc/init *100 [CH3COOH] a Buffer consist of high concentrations of acidic (HA) and basic (A-) components when H3O or OH are added small amounts of one buffer component to convert into the other. As long as the amount is small the added ions have little effect on the pH because they are consumed by one or the other buffer components. Ka= [CH3COO][H3O] [H3O] = Ka *[CHCOOH] [CH3COOH] [ CHCOO] [H3O] of solution depends directly on the buffer-component ratio [CH3COOH] [CH3COO] The conversion of one component into the other produces a small change in the buffer- component concentration ratio and consequently a small change in [H3O] and in pH. Henderson equation Henderson-Hasselbalch equation pH= pKa + log[base]/[acid] can find [H3O] very quick allows us to prepare a buffer of desired pH Buffer capacity- is a measure of the ability to resist pH change and depends on both the absolute and relative component concentration The more concentrated the components of a buffer, the greater the buffer capacity. The pH of a buffer is distinct from its buffer capacity. For the given addition of acid or bases, the buffer-component concentration ratio changes less when the concentrations are similar than when they are different. *a buffer has the highest capacity when the component concentrations are equal

[A]/[HA] =1 A buffer whose pH is equal to or near the pKa of its acid component has the highest buffer capacity

buffer range- the pH range over which the buffer acts effectively, and its related to the relative component concentration. Buffers have a usable range with +- 1 pH unit of pKa of acid component Preparing a buffer Step 1. Decide on the conjugate acid-base pair. Based off desired pH. pH is close to pKa step 2. Find the ratio [A-]/[HA] that gives the desired pH, using the Henderson Hasselbalch equation pH = pKa +log[A]/[HA] step 3. Choose the buffer concentration and calculate the amount to mix .

find the amount of other component using the buffer- component concentration ratio

step 4. Mix the amounts together and adjust the buffer pH to the desired value. Add small amounts of strong acids or strong base while monitoring the solution with a pH meter 19.2 acid-base titration curves

we track the pH of titration with an acid –base titration curve a plot of pH vs. volume of titrant added monitoring pH with an indicator what is an acid base indicator? A weak organic acid (Denoted as Hln) that has different color than its conjugate base (IN) HIn + H2O H3O + In

Ka= [H3O][In]/[HIn] [H3O]/Ka = [HIn]/[In]

An indicator changes color in units of 2pH Strong acids strong base titration curve

Features of the curve

1. pH starts out low, shows high [H3O], goes up gradually as base is added 2. sudden, steep pH rise , dues to moles of OH equaling mole H3O 3. after steep there is an increase but slowly until equilibrium

equivalence point- number of moles OH is equal to number of mole H3O at this point the solution consist of the anion of a strong acid and the cation of a strong base ions do not react with water at this point so the solution is neutral. Before the titration begins , we add an indicator to acid to signal when equivalence point. The end point occurs when the indicator changes color An indicator with an end point close to the equilibrium point **** The visible change in color of indicator(end point) indicates point where moles of base equal mole of acid originally.****** calculating the pH of a system

1. original solution of strong HA. pH=-log[H3O] 2. before the equivalence point. Initial amount of H3O – change= which equals

amount of OH added. Concentration [H3O] 3. at the equilibrium point.

Weak acid- strong base titration

features of the curve 1.the initial pH is higher , because weak acid dissociates slightly 2. a gradual rise portion of curve called the buffer region, appears before the steep rise to the equivalence point, the pH equal the pKa at half the original HPr reacted 3. the pH at the equivalence point is greater than 7.00 4. beyond the equivalent point the pH increases slowly as excess OH is added 19.3 equilibria of slightly soluble ionic compounds

ion-product expression(Qsp) and the solubility-product constant (Ksp) equilibrium exists between solid solute and aqueous ions. PbSO4 (s) Pb2+ (aq)+SO4

2-(aq) Qc= [Pb2+][SO4

2-]/[PbSO4]solid Ion-product expression Qsp= Qc[PbSO4] =[Pb2+][SO4

2-] When [PbSO4] reaches equilibrium then we get Ksp solubility equilibrium constant Depends on the temperature not the concentration Also equals the subscript of each ion in the compounds formula

Cu(OH)2 Cu2+ + 2OH- Ksp= [Cu][OH]2

Exceptions include insoluble metal sulfides . Sulfur ion is so basic it is not soluble in water

MnS Mn +S

S + H2O HS + OH MnS + H2O Mn + HS + OH

Ksp=[Mn][HS][OH] The value of Ksp indicates how far to the right the dissolution proceeds at

equilibrium (saturation) Calculations with Ksp Most are in grams of solute dissolved in 100 grams H2O =100ml of solution Then we convert grams per 100ml solution into molar solubility: The amount of mol of solute dissolved per liter of solution. Using Ksp values to compare solubilities If compounds have the same total number of ions. Then the higher the Ksp the greater the solubility Effects of a common ion on solubility The presence of a common ion decreases the solubility of a slightly soluble ionic compound….. using le Chateliers principle Effects of pH on solubility If the compound contains the anion of a weak acid or CO3, addition of H3O(from

a strong acid) increases its solubility. CaCO3(s) Ca(aq) +CO3(aq)

Add some strong acid , reacts with CO3 to form weak HCO3 CO3(aq) + H3O(aq) HCO (aq) + H2O(l)

More CaCO3 dissolves

CaCO3 Ca + CO3-H3O->HCO3-H3O-> H2CO3 CO2+H2O+Ca Predicting the formation of a precipitate: Qsp vs. Ksp

Qsp=Ksp when the system is saturated Qsp>Ksp precipitate forms until solution is saturated( need more water) Qsp<Ksp Solution is unsaturated and no precipitate forms(can add more solid) ACID RAIN Substances involved

1. sulfurous acid(H2SO3)- coal burns to form sulfurous acid a. H2O2+H2SO3 H2SO4 +H2O

2. sulfuric acid(H2SO4)- forms through atmosphere oxidation of SO2 3. nitric acid- from car engines

a. CO2+ 2H2O H3O +HCO3

Lakes bounded by limestone-rich soils form buffer system that prevent harmful

acidification by acid rain

19.4 equilibria involving complex

complex ion consist of central metal ion covalently bonded o two or more anions or molecules called ligands OH, Cl, CN, all complex ions are Lewis adducts the metal ion acts as a Lewis acid formation of complex ion

M(H2O)4(aq) + 4NH3(aq) M(NH3)4(aq) + 4H2O(l) Kc=[M(NH3)4][H2O]4/[M(H2O)4][NH3]4

formation constant (kf)= Kc/[H2O]=[M(NH4)]/[M(H2O)4][NH]4 Kf=Kf1*Kf2*Kf3….. Complex ion and the solubility of precipitates Ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation

Chapter 20 Thermodynamics: entropy, free energy and the direction of chemical reactions 20.1 the second law of thermodynamics: predicting spontaneous change

A change in spontaneous under specified conditions if it occurs in a

Neither the first law of thermodynamics nor the sign ΔH predicts the direction of spontaneous change

All spontaneous processes involve an increase in the dispersion of energy

Entropy is a state function that measures the extant of energy dispersal into the number of microstates possible for a system, which is related to the freedom of motion of its particles

The second law of thermodynamics states that, in a spontaneous process, the entropy of the universe( system plus surroundings) increases

Absolute entropy values can be found because perfect crystals have zero entropy at 0K(third law)

Standard molar entropy S°(j/mol*K) is affected by temperature, phase changes, dissolution and atomic size or molecular complexity

20.2 calculating the change in entropy of a reaction

The standard entropy of reaction, ΔSrxn, is calculated from S values.

When the amount (mol) of gas (Δngas) increases in a reaction, usually ΔSrxn>0

The value of ΔSsurr is related directly to ΔHsys and inversely to T at which the change occurs

In a spontaneous change, the entropy of the system can decrease only if the entropy of the surroundings increases even more

For a system at equilibrium, ΔSuniv=0

20.3 entropy, free energy and work

The sign of the free energy change ΔG = ΔH – TΔS, is directly related to reaction spontaneity: a negative ΔG corresponds to a positive ΔSuniv

We use the standard free energy of formation ΔGf to calculate ΔGrxn at 25C

The maximum work a system can do is never obtained from a real (irreversible) process because some free energy is always converted to heat

The magnitude of T influences the spontaneity of temperature-dependent reactions( same signs of ΔH &ΔS) by affecting the size of TΔS. For such reactions, the T at which the reaction becomes spontaneous can be estimated by setting ΔG=0

A nonspontaneous reaction (ΔG >0) can be coupled to a more spontaneous one (ΔG<<0) to make it occur. For example, in organisms, the hydrolysis of ATP drives many reactions with a positive ΔG

20.4 free energy, equilibrium, and reaction direction

Two ways of predicting reaction direction are from the value of ΔG and from the relation of Q o K. these variables represent different aspects of the same phenomenon and are related to each other by ΔG = RT ln Q/K. When Q=K the system can release no more free energy

Beginning with Q at the standard state, the free energy change is ΔGo, and it is related to the equilibrium constant ΔGo = -RT ln K. For nonstandard conditions, ΔG has two components: ΔGo & RT ln K

And non-equilibrium mixture of reactants and products move spontaneously (ΔG<0) toward the equilibrium mixture. A product-favored reaction goes predominantly towards products and has K>1 and ΔGo <0

20.1 the second law of thermodynamics: predicting spontaneous change

spontaneous change- occurs by itself under specified conditions, without a continuous input of energy from outside the system,; can be physical, chemical or change in position example: burning something- just need a spark then the system takes over pushing a book off a desk, a force to start but then system takes over A non-spontaneous change occurs when the surrounding supply the system with constant energy. Example: picking a book up , constant supply of energy Putting a fire out, constant supply of energy *********A system can only be spontaneous in one direction, it cannot be spontaneous in both**********

spontaneous is not instantaneous , it implies the system acting by itself example: Ripening, rust A chemical reaction proceeding towards equilibrium is an example of spontaneous change The directions the equation moves it’s the direction it is spontaneous Limitations of first law of thermodynamics First law (law of conservation of energy)- states that internal energy (E), sum of kinetic and potential energy, changes when heat or work are gained or lost

ΔE=q+w Eunivers=Esystem+Esurrounding

(q+w)system= - (q+w)surround

because total energy of the universe is conserved=0 the first law does not help tell what direction the system is going (no spontaneous) the sign of ΔH cannot predict spontaneous change freedom of particle motion and dispersal of particle energy disperse- spread over more quantized energy levels

less freedom of particle motion more freedom of particle motion localized energy of motion dispersed energy of motion

phase change: solid liquid gas dissolving of salt: crystalline solid + liquid ions in solution chemical change: crystalline solids gases +ions in solution ************* in thermodynamic terms a change in freedom of motion of particles in a system, that is, in the dispersal of their energy of motion, is a key factor determining the direction of a spontaneous process********** entropy and the number of microstates what is a microstate? All the quantized states of the whole system molecules At any instance the energy is dispersed about a microstate A second later the energy is dispersed about a different microstate Number of microstate for a system of 1 mole is 10^10^23 Microstates with relevance to thermal energy is the number of ways it can disperse its thermal energy amongst various modes of motion of all its molecules Boltzmann equation relating microstate (W) with entropy(s) of a system

S=k ln(W) K is Boltzmann constant= R/Na=1.38*10-23 J/K

A system with fewer microstates (smaller W) has a lower Entropy(lower S)

A system with more microstates (larger W) has a larger Entropy (larger S) Lower entropy (fewer microstates) higher entropy (more microstates) Change in entropy ΔSsystem = ΔSfinal - ΔSinitial Quantities meaning of entropy change

1 mole neon (initial:1L and 298K) 1 mole neon(final:2 L and 298K) two approaches

1. approach based on number of microstates a. use S= k ln(W) b. W in this case is Wfinal/Winitial = 2NA NA is number of particles c. ΔSsys =Sf – Si = k lnWf – k lnWi = k ln (Wf - Wi)

2. approach based on change in heat a. ΔSsys = qrev/T b. T = temperature q = heat absorbed c. Rev refers to reversible process,

Entropy and the second law of thermodynamics The second law of thermodynamics- considering both the change in system and its surroundings, all real processes occur spontaneously in the direction that increases the entropy of the universe( system plus surroundings)

ΔSuniverse = ΔSsystem + ΔSsurrounding > 0

Standard molar entropies and the third law -We can measure change in enthalpy but not absolute enthalpy -We can measure the absolute entropy of a substance using the third law Third law of thermodynamics- A perfect crystal has zero entropy at a temperature of absolute 0 Ssys = 0 @ 0K Standard molar entropy (S°) is at 1 molar Predicting relative S° values of a system

1. temperature change (S° increases as temperature increases) T(K): 273 295 298 S°: 31.0 32.9 33.2

2. physical state and phase change

S° increases as we go Solid liquid gas Na H2O C(graphite)

S°(s or l) 51.4(s) 69.9(l) 5.7(s) S° (q) 153.6(g) 188.7(g) 158

3. dissolving a solid or liquid S° increases from ions to solution NaCl AlCl3 CH3OH

S°(s or l) 72.1 (s) 167(s) 127(s) S° (aq) 115.1 -148 132

4. dissolving a gas

gasliquid/solid S° goes down

5. atomic size or molecular complexity bigger the atom more microstates more entropy Li Na K Rb Cs

Atomic radius 152 186 227 248 265 Molar mass 6.941 22.99 39.1 85.47 132.9 S° (s) 29.1 51.4 64.7 69.5 85.2 ***** trends hold true for substances in the same state*** ****states dominate over trends****** chapter 20.2 calculating the change in entropy of a reaction

standard entropy of reaction ΔSrxn - the entropy change that occurs when all reactants and products are in their standard states. When the number of moles of gas increases, ΔSrxn is usually positive When the number of moles of gas decreases, ΔSrxn is usually negative * we cannot predict the sign of entropy change unless the reaction involves a change in the number of moles of gas***

ΔSrxn = ΣmSproducts – ΣnSreactants

Chang in entropy in surrounding -For a spontaneous process, decreases in the entropy of the system can occur only increases in the entropy of the surroundings outweigh them. ( second law) - the surrounding’s role is to either ass het to the system or remove heat from it. 1. exothermic change heat loss by system ending in entropy of surrounding increasing. Qsys <0 Qsurr>0 ΔSsurr >0

2. endothermic change heat gained by system ending with entropy of surrounding decreasing Qsys >0 Qsurr <0 ΔSsurr <0 Analogy if you are a light weight in drinking and you take a shot to start. Then you take another shot now you have drunk 100% more. If you are a heavy weight and are about 8 deep and you take another shot, it will not affect you as much because you already have a lot in your system The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to an opposite change in heat of the system and inversely related to the temperature at which the heat is transferred. ΔSsurr – ΔΗsys T The entropy change of forward reaction is equal in magnitude but opposite in sign to the entropy change of the reverse reaction. When a system is in equilibrium, neither the forward nor the reverse reaction is spontaneous, so no net reaction in either direction Summary of spontaneous exothermic and endothermic reaction

1. for an exothermic reaction ΔHsys < 0. He is released increasing freedom of movement thus ΔSsurr >0

a. if the reaching system yields whose entropy is greater than reactants ΔSsys >0 then total entropy is positive ΔSsys +ΔSsurr

b. if the entropy of system decreases ΔSsys<0 the entropy of surrounding must increase even more ΔSsurr>> 0 making total ΔS positive

2. for an endothermic reaction ΔHsys >0 heat lost by the surroundings decreases movement resulting in decreases of entropy of surroundings. ΔSsurr <0. If ΔSsys>>0 and can outweigh ΔSsurr then it is positive 20.3 entropy, free energy, and work

the Gibbs free energy aka free energy (G) is a function that combines the system’s enthalpy and entropy free energy change ΔG is a measure of the spontaneity of a process and of the useful energy available from it. ΔGsys = ΔHsys – TΔSsys The sign of ΔG tells if a reaction is spontaneous

ΔSuniv for a spontaneous process ΔSuniv <0 for non-spontaneous process ΔSuniv = 0 for a process at equilibrium

Standard free energy change (ΔG) : occurs when all components of the system are in their standard states. Adapting the Gibbs equation we have ΔGo =ΔHosys – TΔSosys Standard free energy of formation ΔGof occurs when 1 mole of compound is made from its elements ΔGof vales of reactant and product to calculate ΔGorxn no matter how reaction occurs ΔGorxn = ΣmΔGof(products) –ΣnΔGof(rectants)

* ΔGof of an element in its standard state are zero * an equation coefficient (m or n above) multiplies ΔGof by that number * reversing a reaction changes the sign of ΔGof a spontaneous process (ΔG <0 ) at constant T, P, and ΔG is the maximum useful work obtainable from the system as a process takes place ΔG= wmax A nonspontaneous process (ΔG>0) at constant T, P, and ΔG is the minimum work done to the system to make it process take place ΔG =wmin The maximum work is done by a spontaneous process only if it is carried out reversibly. we can never obtain the max work No real process uses all the available free energy to do work because some is always changed to heat Table 20.1 reactions spontaneity and the signs of ΔH, ΔS, and ΔG

ΔH ΔS -TΔS ΔG description - + - - spontaneous at all T + - + + non spontaneous at all T + + - + or- spontaneous at higher T; nonspontaneous at lower T - - + + or- spontaneous at lower T; nonspontaneous at higher T explanation most exothermic reactions are spontaneous the temperature of a reaction influences the magnitude of the TΔS term, so, the overall spontaneity depends on the temperature -temperature-independent cases (ΔH & ΔS have opposite signs) the reactions occurs spontaneously at all temperatures or none 1. reaction is spontaneous at all temperatures: ΔH<0, ΔS>0. So -TΔS is negative; thus ΔG is always negative 2. reactions are nonspontaneous at all temperatures: ΔH>0; ΔS<0 so –TΔS is positive; thus ΔG is always positive -temperature-dependent cases (ΔH&ΔS have the same signs)

3. reactions are spontaneous at high temperature and non-spontaneous at lower temperature. ΔH >0 &ΔS >0. –TΔS I negative and ΔG is either positive or negative depending on the temperature 4. reactions are spontaneous at lower temperature and nonspontaneous at higher temperature. ΔH<0 &ΔS<0, o –TΔS is going to be positive. Making ΔG either positive or negative depending on the temperature. ΔH=TΔS & T=ΔH ΔS A non-spontaneous step is driven by a spontaneous step when coupling of reactions. One step supplies enough free energy for the other to occur Example Cu2O(s) 2Cu(s) +1/2O2(g) ΔG375=140kJ Oxidation of carbon C(s) +1/2O2 CO(g) ΔG375=-143.8kJ Coupling these steps to spontaneous ones is a life-sustaining strategy that is common to all organisms. A key spontaneous biochemical reaction is the hydrolysis of a high-energy molecule called :adenosine triphosphate (ATP) to adenosine diphosphate(ADP) ATP4- + H2O ADP3- +HPO42- +H+ ΔGo` =-30.5 kJ

20.4 free energy, equilibrium, and reaction directions

Recall from chapter 17 If Q<K (Q/K <1) the reaction as written proceeds to the right If Q>K (Q/K >1) the reaction as written proceeds to the left If Q=K (Q/K =1) the reaction is at equilibrium, no net change ΔG and Q/K are related

If Q/K<1, then ln Q/K <0; reaction proceeds to the right (ΔG <0)

If Q/K >1, then ln Q/K >0; reaction proceeds to the left (ΔG>0) If Q/K =1, then ln Q/K =0; reaction is at equilibrium (ΔG =0)

ΔG = RT ln Q/K = RT ln Q – RT ln K

When all constants are at 1 mol or 1 atm then ΔG equals ΔGo ΔGo = -RT ln K

Table 20.2 The relationship between ΔGo and K @ 298K

ΔGo (kJ) K significant 200 9 *10-36 essentially no forward reactions; 100 3 *10-18 reverse reaction goes to completion 50 2*10-9 10 2 *10-2 1 7 *10-1

0 1 forward reaction and reverse reaction proceed -1 1.5 to the same extent -10 50 -50 6 *108

-100 3 *1017 forward reaction goes to completion; -200 1*1035 essentially no reverse reaction

ΔG = ΔGo + RT ln Q Case Q. K so ΔG<0 The free energy decreases as the reaction proceeds, until it reaches a minimum at the equilibrium mixture: Q=K and ΔG=0 So for A B ΔGoB is smaller than ΔGoA so ΔGo is negative so K>1 aka product favored

Chapter 21 Electrochemistry: chemical change and electrical work 21.1 Redox reactions and electrochemical cells

An oxidation-reduction (redox) reaction involves transfer of electrons from reducing agent to an oxidizing agent

The half-reaction methods of balancing and dividing the overall reaction into half-reactions that are balanced separately and then recombined

The two types of electrochemical cells are based on redox reactions. In a voltaic cell, a spontaneous reaction generates electricity and does work to drive a nonspontaneous reaction. In both types, two electrodes dip into electrolyte solutions; oxidation occurs at the anode, and the reduction occurs at the cathode.

21.2 voltaic cells: using spontaneous reactions to generate electrical energy

A voltaic cell consist of oxidation (anode) and reduction (cathode) half-cells, connected by a wire to conduct electrons and a salt bridge to maintain charge neutrality as the cell operates

Electrons move from anode (left) to cathode (right), while cations move from the salt bridge into the cathode half-cell and anions from the salt bride into the anode half-cell

The cell notation shows the species and their phases in each half-cell, as well as the direction of current flow

21.3 Cell potential: output of a voltaic cell The output of a cell is called cell potential (Ecell) and is measured in volts

(1V= 1J/C) When all substances are in their standard states, the output is the standard

cell potential (Eocell). Eocell >0 for a spontaneous reaction at standard-state conditions

By convention, a standard electrode potential (Eohalf-cell) refers to the reduction half-reaction. Eocell equals Eohalf-cell of the cathodes minus Eohalf-cell of anode

Using a standard hydrogen electrode, other Eohalf-cell values can be measured and used to rank oxidizing (or reducing) agent

Spontaneous redox reactions combine stronger oxidizing and reducing agents to form weaker ones

A metal can reduce another species (H+, H2O or even ions of other metal) if Eocell for the reaction is positive

21.4 free energy and electrical work

A spontaneous process is indicated by a negative ΔG or a positive Ecell, which are related:ΔG =-nFEcell. The ΔG of the cell reaction represents the maximum about of electrical work the cell can do.

The standard free energy change, ΔGo is related to Eocell and to K, we can use Eocell to determine K

At nonstandard conditions the first?equation shows the Ecell depends on Eocell and a correction term based on Q. Ecell is high when Q is small (high [reactant]), and it decreases as the cell operates. At equilibrium, ΔG and Ecell are zero, which means that Q=K

Concentration cells have identical half-reactions, but solutions of differing concentration; they generate electrical energy as the concentrations become equal. Ion-specific electrodes, such as pH electrode, measure the concentration of one species

21.5 electrochemical processes in batteries

Batteries contain several voltaic cells in series and are classified as primary(eg. Alkaline, mercury, and silver), secondary ( eg. Lead-acid, nickel-metal hydride, and lithium-ion, or fuel cell

Supplying electricity to a rechargeable battery reserves the redox reaction, forming more reactant

Fuel cells generate a current through the controlled oxidation of a fuel such as H2

21.6 corrosion: the case of environmental electrochemistry Corrosion damages metal structures through a natural electrochemical

change Iron corrosion occurs in the presence of oxygen and moisture and is

increased by a high [H+], high [ion], or contact with a less active metal, such

as Cu. Fe is oxidized and O2 is reducing in one redox reaction, while rust (hydrated from Fe2O3) is formed in another reaction that often takes place at different location

Because Fe functions as both anode and cathode in the process. An iron or steel object can be protected by physically covering its surface or joining it to a more active metal (such as Zn, Mg, or Al) which act as the anode in place of Fe

21.7 electrolytic cells: using electrical energy to drive nonspontaneous reactions

An electrolytic cell uses electrical energy to drive a nonspontaneous reactions

Oxidation occurs at the anode and reduction at the cathode, but the direction of electron flow and the charges of the electrodes are opposite those in voltaic cells

When two products can form at each electrode, the more easily oxidized substance reacts at the anode and the more easily reduced at the cathode.

The reduction or oxidation of water takes place at nonstandard conditions Overvoltage causes the actual voltage to be unexpectedly high and can affect

electrode product that forms The industrial trial production of many elements, such as sodium, chlorine,

copper, and aluminum utilize electrolytic cells The amount of product that forms depends on the quantity of charge flowing

through the cell, which is related to the magnitude of the current and the time it flows 21.1 redox reaction and electrochemical cells

electrochemistry is the study of the relationship between chemical change and electrical work. It is typically investigated in electrochemical cells a system that incorporates redox reactions -electrochemical process always involves movement of electrons from one chemical species to another through a redox reaction. - oxidation is the loss of electrons - reducing is the gain of electrons -oxidizing agent is the species that reduces - reducing agent is the species that oxidizes After the reaction, the oxidized substances have a higher oxidation number O.N., and the reducing substance has a lower one.

half-reaction method: divides the overall redox reaction into oxidation and reduction half-reactions reasons why to study electrochemistry

it separates the oxidation and reduction steps, which reflects their actual physical separation in electrochemical cells

it is readily applied to redox reactions that take place in acidic or basic solution, which is common in these cells

it does not require assigning O.N.s. ( in some cases when half-reactions are not obvious, we assign O.N.s to determine which atoms undergo a change)

if the oxidizing form of a species is on the left side of the skeleton reaction, the reducing form of that species must be on the right side, and vice versa steps step 1: divide skeleton structure into half-reaction, each contains oxidizing and reduced forms of one species. Step 2: balance the atoms and charges in each half-reaction

atoms are balanced in order : atoms other than O and H, then O, and then H charge is balanced by adding electrons (e-). They are added to the left in the

reducing half-reaction because they are gains; they are added to the right in the oxidation half-reactions because they are lost

step 3: if necessary, multiply one or both half-reactions by an integer to make the number e- gained in the reduction equal the number lost in oxidation step 4: add the balanced half-reactions, and include states of matter step 5: check that the atoms and charges are balanced

balancing redox reaction in acidic solution let’s balance the equation

Cr2O72- (aq) + I-(aq) Cr3+ (aq) + I2(s) [acidic solution] Step 1: divide the reaction into half-reactions

Cr2O72- Cr3+ I- I2

Step 2: balance atoms and charges in each half-reaction Cr2O7/2Cr 3+ half reaction

a. balance atoms other than O and H. Cr2O7 2Cr

b. balancing the O atoms by adding H2O molecules Cr2O7—2Cr +7H2O

c. balance H atoms by adding H ions. 14H + Cr2O7 2Cr + 7 H2O

d. balance charges by adding electrons 6e- +14H+ + Cr2O72- 2Cr3+ + 7H2O

we can see the reaction is reducing because the electrons appear on the left balancing the I-/I2 half reaction

a. balancing atoms other than O and H 2I- I2

b. balance O atoms with H2O : not needed c. balance H atoms with H: not needed d. balance charge with e-

2I- I2 +2e- step 3: multiply each half-reaction, if necessary, by an integer

3(2I- I2 + 2e-) 6I- 3I2 + 6e-

step 4: adding the half-reaction together 6e- + 14 H+ + Cr2O72- 2Cr3+ +7H2O

6I- 3I2 + 6e-

6I(aq) + 14H+ + Cr2O7 2Cr3+ +7H2O + 3I2 step 5: check that atoms and charges balance

balancing redox reactions in basic solutions both half reactions have been balanced as if they took place in acidic solution we add one OH ion to both sides of the equation for every H+ ion present Excess H2O molecules are cancelled Half-reaction reveals a great deal and is essential to understanding electrochemistry

any redox reaction can be treated as the sum of a reduction and an oxidation half-reaction

atoms and charge are conserved in each half-reaction electrons lost in one half-reaction are gained in the order

although the half-reactions are treated separately, electron los and electron gain occur simultaneously

overview of electrochemical cells

1. voltaic cells( galvanic cell) – spontaneous reaction (ΔG <0) to generate electricity high level of energy to lower level of energy I converted to electric energy System does work on surrounding

2. electrolytic cell – uses an electric energy to drive a nonspontaneous reaction

the surroundings do work on the system. Two electrodes, which conduct the electricity between cell and surroundings, are dipped into an electrolyte An electrode is identified as either anode or cathode depending on the half-reaction that takes place

the oxidation half-reaction occurs at the anode. Electrons are lost by the substance being oxidized( reducing agent) and leave the cell at the anode

The reduction half-reaction occurs at the cathode. Electrons are gained by the substance being reduced ( oxidizing agent) and enter the cell at the cathode.

The relative charges of the electrodes are opposite in the two types of cells Analogy:

1. the words anode and oxidation start with vowels; the words cathode and reduction start with consonant

2. Alphabetically, the A in anode comes before the C in cathode, and the O in oxidation comes before the R in reduction

3. Look at the first syllable and use your imagination 21.2 Voltaic cells: using spontaneous reactions to generate electrical energy

Cu2+ (aq)+ 2e- Cu (s) [reduction] Zn(s) Zn2+(aq) + 2e- [oxidation]

Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s) [overall reaction] Construction and operation of a voltaic cell In the situation above no electrical energy is generated because the oxidation agent and the reduction agent are in the same beaker If we split into half equations then electricity would be generates

The components of each half-reaction are placed in a separate container, or half-reaction, which consist of one electrode dipped into an electrolyte solution. The two half cells are joined by a circuit which consist of a wire and a salt bridge The oxidation half-cell(anode compartment) is shown on the left and the reduction half-cell (cathode compartment) on the right. Here are the key points about the Zn/Cu2+ voltaic cell:

1. the oxidation half-cell. 2. The reduction half-cell 3. Relative charges on the electrodes. The electrode charges are determines by

the source of electrons and the direction of electron flow through a circuit. The electrons flow left to right through the wire to the cathode. In any Voltaic cell the anode is negative and the cathode is positive

4. The purpose of a salt bridge. To enable the cells to operate , the two-half-cells are joined by a salt-bridge, which acts as a “liquid wire” The circuit is completed as electrons move left to right through the wire, while anions move right to left and cations move left to right through a salt bridge

5. Active vs. inactive electrodes. The electrodes in Zn/Cu2+ cell are active because the metal bars themselves are components of the half-reactions . as the cell operates, the mass of the zinc electrode gradually decreases and the[Zn2+ ] in the anode half-cell increases.

Inactive electrons are commonly rods of graphite or platinum: they conduct electrons into or out of the half-cells but cannot take part in the half-reaction. Notation for a voltaic cell Short hand notation for Zn(s) + Cu2+(aq) Zn2+ (aq)+ Cu(s) Zn(s) | Zn2+ (aq)|| Cu 2+ (aq)| Cu(s) Key parts of the notation --The components of the anode compartment( oxidation half-cell) are written to the left of the components of the cathode compartment (reduction half-cell) -- a single vertical line represents a phase boundary. For example Zn(s) | Zn 2+ (aq) indicates that the solid Zn is a different phase from the aqueous Zn2+ . A comma represents half-call components in the same phase. Example Graphite| I-(aq) | I2(s) || H+ (aq), MnO4- (aq) , Mn2+ (aq) | graphite --half-cell components usually appear in the same order as in the half-reaction, and electrodes appear at the far left and far right of the notation -- a double vertical line indicated the separated half-cells and represents the phase boundary on either side of the salt bridge ( the ion in the salt bridge is omitted because they are not part of the reaction)

21.3 Cell potential: output of a voltaic cell

The purpose of a voltaic cell is to convert the free energy charge of a spontaneous reaction into kinetic energy of the electrons moving through an external circuit. This electrical energy is proportionate to the difference in electrical potential between the two electrodes, called Cell potential (Ecell) also the voltage or electromotive force(emf)

Ecell > 0 for a spontaneous process Ecell <0 for a nonspontaneous process

Ecell = 0 when at equilibrium Volt(v) electrical potential Coulomb (C) electrical charge When 2 electrodes differ by 1 volt of electrical potential, 1 joule of energy is released 1 V = 1 J/C standard cell potentials in order to compare the output of different cells we obtain standard cell potential (Eocell) – the potential measured at a specified temperature with no current flow and all components in their standard state Zn (s) + Cu2+(aq; 1M) Zn2+ (aq,1M) +Cu(s) Eocell = 1.1V Standard electrode potential (Eohalf-cell)- potential associated with a given half-reaction when all the components are in their standard states A standard electrode potential always refers to the half-reaction written as a reduction (Eozinc, anode compartment) the standard cell potential is the difference between the standard electrode potential of the cathode( reduction) half-cell and the standard electrode potential of the anode ( oxidation ) half-cell

Eocell = Eocathode(reduction) – Eoanode(oxidation)

Half-cell potentials, such as Eozinc and Eocopper are not absolute values but they are referenced to the change . the standard reference half-cell has its standard electrode potential defined as zero (Eoreference = 0.00V) Standard reference half-cell is a standard hydrogen electrode consist of specially prepared platinum electrode immersed in a 1M aqueous solution of strong acid, H+ (aq) Reaction : 2H+(aq;1M) +2e- H2(g; 1atm) Eoreference = 0.00V Relative strength of oxidation and reduction agents And their oxidation strengths by writing each half-reaction as a gain of electrons(reduction with corresponding standard electrical potential

Cu2+ (aq) +2e- Cu(s) Eo = 0.34V 2H+ +2e- H2 Eo = 0.00V

Zn2+(aq) + 2e- Zn(S) Eo = -0.76V The more positive Eo value, the more readily the reaction occurs Note**** the list of half-reaction is in order of decreasing half-strength

all the values are related to the standard hydrogen (reference electrode: by convention, the half-reaction are written as reductions, which means that

only reactants re oxidizing agents and only products are reducing agents the more positive the Eohalf-cell the more readily the half-reaction occurs half-reactions are shown with an equilibrium arrow because each can occur

as a reduction or an oxidation depending on Eohalf-cell of the other half-reaction

F2 is the strongest oxidizing agent but the weakest reducing agent, and Li+ = is the strongest reducing agent but the weakest oxidizing agent

Writing spontaneous redox reaction Every redox reaction is the sum of two half-reactions, so there is a reducing agent and an oxidizing agent on each side.

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Stronger reducing stronger oxidizing weaker oxidizing weaker reducing agent agent agent agent *** the stronger reducing and oxidizing agent react spontaneously to for the weaker ones**** it’s like acids and bases the stronger goes to the weaker The stronger oxidizing agent has a half-reaction with a larger Eo value, and the stronger reducing agent has a half-reaction with a smaller Eo value For a spontaneous reaction to occur, the half-reaction higher in the list at the cathode as written and the half-reaction lower in the list proceeds at the anode in reverse Ag+ (aq) +2e- Ag(s) Eosilver = 0.80V Sn2+ (aq) + 2e- Sn Eotin = -0.14V Steps

1. reverse one of the half reactions into an oxidation step such that the difference of the electrode potential gives a positive Eocell ***************when we reverse the half-reaction we need not reverse the sign of Eohalf-cell will be settled by Eocell= Eocathode – Eoanode)*******

2. Add the rearranged half-reactions to obtain a balanced overall equation. Don’t forget to multiply by coefficients so that e- lost equals e- attained

Half-reactions written in the correct direction must make sure the electrons lost are equal to the electrons gained . Very important******When we double the coefficient of the silver half-reaction to balance the number of electrons, we do not double the Eo value it remains 0.80V. It is an intensive property like density Relative reactivity of metals

1. metals that can displace H2 from acid Fe(s) Fe2+ (aq) +2e- Eo = -0.44V [anode] 2H+ (aq) +2e- H2 (g) Eo = 0.00V [cathode]

Fe(s) + 2H+(aq)Fe2+ (aq) + H2(g) Eocell= 0.00V- (-0.44V)=.44V Explanation If Eocell for the reduction of H+ is more positive for metal A than it is for metal B, metal A is a stronger reducing agent than metal B and a more active metal.

2. metals that cannot displace the H2 from acid

if a metal is above the standard hydrogen (reference) half-reaction then it cannot reduce H+ from acid.

The Eocell would be negative making it nonspontaneous

3. metals that can displace H2 from water these metal lie below the half-reaction of H2 so V < -0.42V example:

2Na(s) 2Na+ (aq) + 2e- Eo = -2.71 V (anode) 2H2O (l) + 2e- 2H+ (g)+ 2OH-(aq) Eo =-0.42V (cathode)

2Na(s) + 2 H2O(l) 2Na+(aq) + 2H- (g) + 2OH- (aq) Eo= -0.42-(-2.71)=2.29V

group 1a and larger alkine earth metals can reduce water

4. metals that can displace other metals from solution a metal that is lower than another metal on the list example

Zn(s) Zn2+ (aq)+ 2e- Eo =-0.76V [anode; oxidize] Fe2+ (aq)+ 2e- Fe (s) Eo = -0.44V [cathode]

Zn(s) + Fe2+ (aq) Zn2+(aq) + Fe(s) Eo= -0.44 –(-0.76) = .32V 21.4 Free energy and electrical work

explanation we examine the relationship between useful work, free energy, and equilibrium constants in terms of electrochemistry and see the effects standard cell potential and equilibrium constant free energy change ΔG < 0 for a spontaneous reactions and electrochemical reaction Ecell >0 for a spontaneous reaction

the sign of ΔG and Ecell are opposite for spontaneous reaction* ΔG α -Ecell

Ecell is the maximum voltage possible for the cell, the work is the maximum work possible (wmax) * for work done on the surroundings, this quantity is negative * Wmax = -Ecell X charge The change in 1 mole of electrons is the faraday constant (F)

F = 9.65 X 104 J / V*mol e- Substitute for charge

ΔG = -nFEcell

When all are in standard state ΔGo = -nFEocell

Substitute for ΔGo Eocell = RT ln(K)

nF

substitute the know values of *.314 J/(mol rxnK) for constant R substitute the known value of 9.65 104 J/ ( Vmol e-) for constant F substitute standard temp of 298.15 K for T, but remember cell can be at diff

temp multiply by 2.303 to convert from ln to log

we get new equations Eocell = 0.0592V log (K) or log(K) = nEocell at 298.15K

n 0.0592 effects of concentration on cell potential

relation between cell potential and concentration based on free energy and concentration

by dividing both sides by –nF we get Nerest equation Ecell = Eocell – RT ln(Q)

nF * when Q< 1 and thus [Reactants] > [Products], lnQ <0 , so Ecell > Eocell * when Q=1 and thus [Reactants] = [products] lnQ = 0, so Ecell = Eocell

* when Q >1 and thus [reactants]<[products] lnQ >0, so Ecell < Eocell

Ecell = Eocell – 0.0592V log(Q) N

Note that Q contains only those species with concentration or pressure that can vary Cs(s) + 2Ag+(aq) Cd2+ (aq) +2 Ag(s) Q = [Cd2+] / [Ag+]2 Change in potential during operation We go through stages Stage 1. Ecell > Eocell when Q<1: so [reactants] > [products] , now more products are going to form so Ecell decreases Stage 2. Ecell = Eocell when Q=1 so [reactants] = [products] Q = 1 Stage 3. Ecell < Eocell when Q>1 so [reactants] < [products] Stage 4. Ecell = 0 when Q= K term becomes large then equals Eocell

This occurs when system reaches equilibrium and no more free energy is released, so the cell can do no more work, the battery is “dead” Change in Ecell and concentration Stage in cell operation

Q Relative [P] and [R]

0.0592V log(Q) n

1. E>Eo <1 [P] <[R] <0 2. E=Eo =1 [P] = [R] =0 3. E < Eo >1 [P]> [R] >0 E=0 =K [P]>>[R] =Eo Concentration cells When mixing concentrations of salt with dilute solutions of salt you get an intermediate value . A concentration cell does the same thing but with electrical energy The two half-cells are separated but they become equal In the concentration cell the half-reactions are the same but the concentrations are different Eocell is equal to 0 , but Ecell is not equal to 0 because it is depend on ratio of concentrations Cu(s) Cu2+ (aq;.10M) +2e- [anode] Cu2+ (aq; 1M)+2e- Cu (s) [cathode]

Cu2+ (aq; 1M) Cu2+(aq;0.1M) Ecell =? n=2 Ecell=Eocell -0.0592V log[Cu2+]dil = 0V – (0.0592V log0.1M N [Cu2+]con 2 1.0M =.0296V applications of concentration cells we can find pH using Ecell example H2(g;1atm) 2H+ (aq; unknown) +2e- anode 2H+(aq;1M) +2e- H2(g;1atm) cathode

2H+(aq;1M) 2H+(aq; unknown) Ecell=? Ecell is not 0 but E0cell is 0 Using Nernst equation Ecell = Eocell -0.0592V log(H+)2unknown

N [H+]2stand Ecell = 0v- 0.0592V log (H+]2unknown = -[0.0592V 2 log H+unknown

2 1 2 Ecell = -0.0592 log[H+]unknown

-log[H] = pH so Ecell = 0.0592V pH

21.5 electrochemical processes in batteries battery – a self-contained group of voltaic cells arranged in a series (plus minus plus minus etc.) so the individual voltages are added together 3 types of batteries ( primary, secondary, fuel cells) Primary (non- rechargeable) batteries It is discarded when the components have reached equilibrium Alkaline batteries Has a zinc anode case that houses a MnO2 mixture and a paste of KOH and H2O

mercury and silver(button) batteries found in watches, cameras, heart pacemakers, and hearing aids have a zinc container as cathode and HgO as anode, the silver uses Ag2O can uses steel can around cathode.

Secondary (rechargeable) batteries Recharges by supplying electrical energy to reverse the cell reaction, in simpler terms the cells flip when recharged Example: car batteries Lead acid battery Overall reaction during discharge PbO2(s) + Pb + 2H2SO4 (aq) 2PbSO4 (aq) + 2H2O (l) Ecell =2.1 V Overall reaction of recharge 2PbSO4 + 2H2O PbO2 + Pb + 2H2SO4

Fuel cells Sometimes called a flow battery, is not self-contained Uses combustion to produce electricity Does not burn because half-reactions are separated and electrons transfer through an external circuit. Most common is a proton exchange membrane (PEM) cell Example: 2H2(g) 4H+ (aq) + 4 e- anode O2 + 4H+(aq) +4e- 2H2O (g) cathode

2H2 (g)+ O2(g) 2H2O (g) Ecell = 1.2V lower rates than in other batteries electrocataylist decreases the activation energy seen in plane , the water and electricity

21.6 corrosion: a case of environmental electrochemistry

corrosion, the natural redox process that oxidizes metals to their oxides and sulfides iron 25% of all steel is made to replace corroded steel once irons loses electrons damage is done and pit starts to form reaction Fe(s) Fe2+ (aq) +2e-- anode O2(g) + 4H+(aq) + 2e- 2 H2O(l) cathode

2Fe(s) + O2(g) + 4H+(aq) 2Fe2+(aq) + 2H2O (l)

how to prevent corrosion wash road salt off bodies paint over iron so prevent oxidation make iron act as a cathode because it only corrodes as an anode 21.7 electrolytic cells: using electrical energy to drive nonspontaneous reactions

voltaic cells are the exact opposite of electrolytic cells electrolytic cells use electric energy from an external source to drive a non-spontaneous reaction oxidation takes place at the anode and reduction takes place at the cathode, but the direction of electron flow and the signs of electrodes are reversed

* in a voltaic cell, electrons are generated at the anode, so it is negative and electrons are consumed at the cathode, so it’s positive * in an electrolytic cell, electrons come from the external power source, which supplies them to the cathode, so it is negative, and removes them from the anode, so it’s positive comparison of voltaic cells and electrolytic cells

Cell type ΔG Ecell Name Process Sign Voltaic <0 >0 Anode Oxidation - Voltaic <0 >0 Cathode Reduction + Electrolytic >0 <0 Anode Oxidation + Electrolytic >0 <0 Cathode Reduction - Electrolysis – the splitting of a substance by the input of electrical energy, often used to decompose a compound into elements Electrolysis of molten salts and the industrial production of sodium Predicting the product at each electrode is simple if the salt is pure because the cation will be reduced and the anion oxidized Reaction 2Cl- (l) Cl2(g) +2e- Ca2+(l) +2e- Ca(s) The dry solid i crushed and fused (melted) in an electrolytic apparatus called downs cells Electrolysis of water Extremely pure water is difficult to electrolyze because very few ions are present to conduct a current Equation 2H2O(l) O2(g) + 4H+(aq) +4e- Anode E=0.82V 2H2O(l)+2e- H2(g) + 2OH-(aq) cathode E=-0.42V

2H20(l) 2H2(g) +O2(g) total equation Ecell = -1.24V * these electrode potentials are not written with a degree sign because they are not standard electrode potentials* Electrolysis of aqueous salt solutions; overvoltage and the chlori-alkali process Two half-reactions are possible * the reduction with the less negative (more positive) electrode potential occurs * the oxidation with the less positive( more negative) electrode potential occurs electrolyzing KI solution case 1 K+ (aq) +e-K(s) Eo = -2.93V 2H2O(l) + 2e- H2(g) + 2OH-(aq) E=-0.42 V [reduction] case 2 2I-(aq) I2(s) +2e- Eo = 0.53V

2H2O(l) O2 (g)+ 4H+ +4e- E=0.82V [reduction] the increment above the expected required voltage is called overvoltage overvoltage plays a key role in chlor-alkli process for industrial production of chlorine electrolytic oxidation of Cl- ion from concentrated aqueous NaCl solution the asbestos diagram separates the anode and cathode compartments

Must go through the polymeric membrane Membrane allows cations to move through and only from anode to cathode compartment Rules from aqueous solutions

1. cations of less active metals are reduced to the metal, including gold, silver, copper, chromium, platinum, and cadmium

2. cations of more active metals are not reduced, including those of group 1A and 2A and Al. Water is reduced to H2 and OH- instead

3. anions that are oxidized, because of overvoltage from O2 formation, include the halides([Cl] must be high], except F

4. anions are not oxidized including F- and common oxoanions, such as SO42-, CO32-,NO3-, PO43-, because the central nonmetal in these oxoanions is already in its highest oxidation state. Water is oxidized into O2 and H+ instead

purification is accomplished by electrorefining, which involves the oxidation of Cu to form Cu2+ ions in solution

1. copper and the more active impurities are oxidized to their cations, while the less active ones are not.

2. Because Cu is much less active than Fe Ni impurities, Cu ions are reduced at the cathode, but Fe and Ni ions remain to solution:

Cu2+(aq) +2e- cu(s) Eo =0.34 V Ni2+ (aq)+ 2e- Ni(s) Eo= -0.25

Fe2+ (aq)+2e- Fe (s) Eo=-0.44V Copper obtained by electrorefining is over 99.99% pure

How to isolate aluminum Bauxite, a mixed oxide-hydroxide that is the major ore of aluminum is used Step 1. The mineral oxide, Al2O3 is separated from bauxite Step 2. Oxide is converted to metal Al2O3 electrolyzed at 2030 oC so we have to lower than by using molten cryolite (Na3AlF6) to electrolyte around 1000oC This process is called the hall-heroult process, and takes place in a graphite furnace

Example 2Al2O3 + 1AlF6 – 3Al2O2F42- (l)

cathode reduction AlF62-(l) +3e- Al(l) + 6F-(l) [cathode reduction]

Anode oxidation Al2O2F42-(l) + 8F-(l) +C (graphite) 2AlF63- (l)+ CO2 (g)+ 4e- [anode; oxidation]

Anodes are consumed in half reactions fast and must be replaced quickly Aluminum production accounts for more than 5% of all electricity used in America The stoichiometry of electrolysis: the relationship between amounts and charge of product Faraday’s law of electrolysis: the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell

Questions asked How much material will form a result of given quantity of charge? How much charge us needed to produce a given amount of material? Use faradays law

1. balance the half-reactions to find the number of moles of electrons needed per mole of product

2. use the faraday constant (F = 9.65 *104 C/mol e-) to find the corresponding charge

3. use the molar mass to find the charge needed for a given mass of product

we measure current, the charge of flowing per unit time the Si unit is ampere(A) = 1 coulomb flowing through a conductor in 1 second

1 Ampere = 1 coulomb/second or 1A= 1C/s we find the charge by measuring the current and the time during which the current flows

M(g/mol) balanced Faraday Half- constant Reaction (C/mol e-)

time(s)

here is an example problem How long does it take to produce 3grams of Cl(g) by electrolysis of aqueous NaCl using a power supply with a current of 12A?

Mass(g) Of substance oxidized or reduced

Amount(mol) of substance oxidized or reduced

Amount (mol) of electrons transferred

Charge (C)

Current (A)

We know the mass of Cl2 produced, so we can find the amount of mol of Cl2 The half-reaction tells us the loss of 2 moles of electrons produced 1 mole of chlorine gas

2Cl- (aq) Cl2(g) + 2e- we use this as a conversion factor in the faraday equation

Charge (C) = 3.0 g Cl2 * 1 mole Cl2 * 2 mole e- * 9.65*104C = 8.2*103C 70.90g Cl2 1 mol Cl2 1 mole e-

we use the relationship between charge and current Time (s) = Charge (C) =8.2 *103 C * 1s = 6.8 *102s(11min)

Current (A, or C/s) 12C Grams Cl2 moles of Cl2 moles of e- coulombs seconds