Chapter 18 Electrochemistry AP*. AP Learning Objectives LO 3.2 The student can translate an observed chemical change into a balanced chemical equation

Chapter 18

Electrochemistry

AP*

AP Learning Objectives

LO 3.2 The student can translate an observed chemical change into a balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. (Sec 18.1)

LO 3.8 The student is able to identify redox reactions and justify the identification in terms of electron transfer. (Sec 18.1)

LO 3.12 The student can make qualitative or quantitative predictions about galvanic or electrolytic reactions based on half-cell reactions and potentials and/or Faraday’s laws. (Sec 18.3-18.5, 18.8)

LO 3.13 The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying redox reactions. (Sec 18.3-18.5, 18.8)

LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. (Sec 18.8)

AP Learning Objectives

LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec 18.3-18.5, 18.7)

Section 18.1Balancing Oxidation-Reduction Equations

AP Learning Objectives, Margin Notes and References Learning Objectives LO 3.2 The student can translate an observed chemical change into a balanced chemical equation and justify the

choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. LO 3.8 The student is able to identify redox reactions and justify the identification in terms of electron transfer.


Copyright © Cengage Learning. All rights reserved 5

Review of Terms Electrochemistry – the study of the interchange of chemical

and electrical energy Primarily concerned with the generation of an electrical current fro a

spontaneous chemical reaction and, the opposite process, the use of current to produce a chemical reaction.

Oxidation–reduction (redox) reaction – involves a transfer of electrons from the reducing agent to the oxidizing agent

Oxidation – loss of electrons Reduction – gain of electrons Reducing agent – electron donor Oxidizing agent – electron acceptor


Synopsis of Assigning Oxidation Numbers (as a Reminder)

1. Elements = 02. Monatomic ion = charge3. F: –14. O: –2 (unless peroxide = –1)5. H: +1 (unless a metal hydride = –1)6. The sum of the oxidation numbers equals

the overall charge (0 in a compound).


Oxidation Numbers To keep track of what loses electrons and what gains them, we assign oxidation numbers.

If the oxidation number increases for an element, that element is oxidized.

If the oxidation number decreases for an element, that element is reduced.


Oxidation and Reduction

A species is oxidized when it loses electrons. Zinc loses two electrons, forming the Zn2+ ion.

A species is reduced when it gains electrons. H+ gains an electron, forming H2.

An oxidizing agent causes something else to be oxidized (H+); a reducing agent causes something else to be reduced (Zn).


Half-Reactions The oxidation and reduction are written and

balanced separately. We will use them to balance a redox reaction. For example, when Sn2+ and Fe3+ react,



1. Write separate equations for the oxidation and reduction half–reactions.

2. For each half–reaction:A. Balance all the elements except H and O.B. Balance O using H2O.

C. Balance H using H+.D. Balance the charge using electrons.

The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution



3. If necessary, multiply one or both balanced half–reactions by an integer to equalize the number of electrons transferred in the two half–reactions.

4. Add the half–reactions, and cancel identical species.

5. Check that the elements and charges are balanced.






The Half-Reaction Method

Consider the reaction between MnO4– and C2O4

2–:

MnO4–(aq) + C2O4

2–(aq) Mn2+(aq) + CO2(aq)

Assigning oxidation numbers shows that Mn is reduced (+7 +2) and C is oxidized (+3 +4).


Oxidation Half-Reaction

C2O42– CO2

To balance the carbon, we add a coefficient of 2:

C2O42– 2 CO2


Oxidation Half-Reaction

C2O42– 2 CO2

The oxygen is now balanced as well. To balance the charge, we must add two electrons to the right side:

C2O42– 2 CO2 + 2e–


Reduction Half-Reaction

MnO4– Mn2+

The manganese is balanced; to balance the oxygen, we must add four waters to the right side:

MnO4– Mn2+ + 4 H2O



MnO4– Mn2+ + 4 H2O

To balance the hydrogen, we add 8H+ to the left side:

8 H+ + MnO4– Mn2+ + 4 H2O



8 H+ + MnO4– Mn2+ + 4 H2O

To balance the charge, we add 5e– to the left side:

5e– + 8 H+ + MnO4– Mn2+ + 4 H2O


Combining the Half-Reactions

Now we combine the two half-reactions together:

C2O42– 2 CO2 + 2e–

5e– + 8 H+ + MnO4– Mn2+ + 4 H2O

To make the number of electrons equal on each side, we will multiply the first reaction by 5 and the second by 2:



5 C2O42– 10 CO2 + 10e–

10e– + 16 H+ + 2 MnO4– 2 Mn2+ + 8 H2O

When we add these together, we get

10e– + 16 H+ + 2 MnO4– + 5 C2O4

2–

2 Mn2+ + 8 H2O + 10 CO2 +10e–



10e– + 16 H+ + 2 MnO4– + 5 C2O4

2– 2 Mn2+ + 8 H2O + 10 CO2 +10e–

The only thing that appears on both sides is the electrons. Subtracting them, we are left with

16 H+ + 2 MnO4– + 5 C2O4

2– 2 Mn2+ + 8 H2O + 10 CO2

(Verify that the equation is balanced by counting atoms and charges on each side of the equation.)



Balance the following oxidation–reduction reaction that occurs in acidic solution.

Br–(aq) + MnO4–(aq) Br2(l)+ Mn2+(aq)

10Br–(aq) + 16H+(aq) + 2MnO4–(aq) 5Br2(l)+ 2Mn2+(aq) + 8H2O(l)

EXERCISE!EXERCISE!



1. Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present.

2. To both sides of the equation obtained above, add a number of OH– ions that is equal to the number of H+ ions. (We want to eliminate H+ by forming H2O.)

The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution



3. Form H2O on the side containing both H+ and OH– ions, and eliminate the number of H2O molecules that appear on both sides of the equation.

4. Check that elements and charges are balanced.






Complete and balance this equation for a redox reaction that takes place in basic solution:

CN–(aq) + MnO4–(aq) → CNO– (aq) + MnO2(s)

(basic solution)

Balancing Redox Equations in Acidic Solution

Section 18.2Galvanic Cells

Galvanic Cell/ Voltaic Cell

Device in which chemical energy is changed to electrical energy.

Uses a spontaneous redox reaction to produce a current that can be used to do work.



A Galvanic Cell



Galvanic Cell

Oxidation occurs at the anode. Reduction occurs at the cathode. Salt bridge or porous disk – devices that allow ions to

flow to neutralize the charge imbalance without extensive mixing of the solutions. Salt bridge – contains a strong electrolyte held in a

Jello–like matrix. Porous disk – contains tiny passages that allow

hindered flow of ions.Copyright © Cengage Learning. All rights reserved 29



Cell Potential

A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment.

The “pull”, or driving force, on the electrons is called the cell potential ( ), or the electromotive force (emf) of the cell. The emf is the potential difference between the anode and cathode. Unit of electrical potential is the volt (V).

1 joule of work per coulomb of charge transferred. (1 V = 1 J/C).


cellE


Electromotive Force (emf) Water flows

spontaneously one way in a waterfall.

Comparably, electrons flow spontaneously one way in a redox reaction, from high to low potential energy.


The oxidation–reduction reaction

Cr2O72– (aq) + 14 H+ (aq) + 6 I– (aq) → 2 Cr3+ (aq) + 3 I2(s) + 7 H2O(l)

is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes.

Describing a Voltaic Cell-draw it


The following two half-reactions occur in a voltaic cell:

Ni(s) → Ni2+ (aq) + 2 e– (electrode = Ni)Cu2+(aq) + 2 e– → Cu(s) (electrode = Cu)

Which one of the following descriptions most accurately describes what is occurring in the half-cell containing the Cu electrode and Cu2+ (aq) solution?

(a) The electrode is losing mass and cations from the salt bridge are flowing into the half-cell.(b) The electrode is gaining mass and cations from the salt bridge are flowing into the half-cell.(c) The electrode is losing mass and anions from the salt bridge are flowing into the half-cell.(d) The electrode is gaining mass and anions from the salt bridge are flowing into the half-cell.


Voltaic Cell: Cathode Reaction


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Voltaic Cell: Anode Reaction


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Section 18.3Standard Reduction Potentials

AP Learning Objectives, Margin Notes and References Learning Objectives LO 3.12 The student can make qualitative or quantitative predictions about galvanic or electrolytic reactions based

on half-cell reactions and potentials and/or Faraday’s laws. LO 3.13 The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying

redox reactions. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or

environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.


Galvanic Cell

All half-reactions are given as reduction processes in standard tables. These values are compared to the reduction of hydrogen as a standard. Table 18.1 1 M, 1atm, 25°C

When a half-reaction is reversed, the sign of E° is reversed. When a half-reaction is multiplied by an integer, E° remains the

same. A galvanic cell runs spontaneously in the direction that gives a

positive value for


cellE


Standard Hydrogen Electrode

Their reference is called the standard hydrogen electrode (SHE).

By definition as the standard, the reduction potential for hydrogen is 0 V:

2 H+(aq, 1M) + 2e– H2(g, 1 atm)


Figure 18-5 p843



Oxidizing and Reducing Agents The more positive the

value of E°red, the greater the tendency for reduction under standard conditions.

The strongest oxidizers have the most positive reduction potentials.

The strongest reducers have the most negative reduction potentials.


Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell° = Ered (cathode) – Ered (anode)° °

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

Section 18.3Standard Reduction PotentialsCell Potentials

For the anode in this cell, E°red = –0.76 V

For the cathode, E°red = +0.34 V

So, for the cell, E°cell = E°red (cathode) – E°red (anode) = +0.34 V – (–0.76 V) = +1.10 V


Example: Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)

Half-Reactions: Fe3+ + e– → Fe2+ E° = 0.77 V Cu2+ + 2e– → Cu E° = 0.34 V

To balance the cell reaction and calculate the cell potential, we must reverse reaction 2. Cu → Cu2+ + 2e– – E° = – 0.34 V

Each Cu atom produces two electrons but each Fe3+ ion accepts only one electron, therefore reaction 1 must be multiplied by 2. 2Fe3+ + 2e– → 2Fe2+ E° = 0.77 V

Copyright © Cengage Learning. All rights reserved45


Overall Balanced Cell Reaction

2Fe3+ + 2e– → 2Fe2+ E° = 0.77 V (cathode)

Cu → Cu2+ + 2e– – E° = – 0.34 V (anode) Balanced Cell Reaction:

Cu + 2Fe3+ → Cu2+ + 2Fe2+

Cell Potential: = E°(cathode) – E°(anode)

= 0.77 V – 0.34 V = 0.43 V


cellE

cellE


Line Notation

Used to describe electrochemical cells. Anode components are listed on the left. Cathode components are listed on the right. Separated by double vertical lines which indicated salt bridge

or porous disk. The concentration of aqueous solutions should be specified

in the notation when known. Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s)

Mg → Mg2+ + 2e– (anode) Al3+ + 3e– → Al (cathode)



Description of a Galvanic Cell

The cell potential (always positive for a galvanic cell where E°cell = E°(cathode) – E°(anode)) and the balanced cell reaction.

The direction of electron flow, obtained by inspecting the half–reactions and using the direction that gives a positive E°cell.



Description of a Galvanic Cell

Designation of the anode and cathode. The nature of each electrode and the ions present in

each compartment. A chemically inert conductor is required if none of the substances participating in the half–reaction is a conducting solid.



Order the following from strongest to weakest oxidizing agent and justify. Of those you cannot order, explain why.

Fe Na F- Na+ Cl2

CONCEPT CHECK!CONCEPT CHECK!


A voltaic cell is based on the two standard half-reactions

Cd2+(aq) + 2 e– → Cd(s)Sn2+(aq) + 2 e– → Sn(s)

Use data in Appendix A5.5 to determine (a) which half-reaction occurs at the cathode and which occurs at the anode and (b) the standard cell potential.

Determining Half-Reactions at Electrodes and Calculating Cell Potentials


Sketch a cell using the following solutions and electrodes. Include:

The potential of the cell The direction of electron flow Labels on the anode and the cathode

a) Ag electrode in 1.0 M Ag+(aq) and Cu electrode in 1.0 M Cu2+(aq)




Consider the cell from part a.

What would happen to the potential if you increase the [Ag+]?

Explain.

The cell potential should increase.



Section 18.5Dependence of Cell Potential on Concentration

You make a galvanic cell at 25°C containing: A nickel electrode in 1.0 M Ni2+(aq) A silver electrode in 1.0 M Ag+(aq)

Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential.

1.03 V




Section 18.4Cell Potential, Electrical Work, and Free Energy






Work

Work is never the maximum possible if any current is flowing.

In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.



Work

The work done when e- a transferred through a wire depends on the thermodynamic driving force (the push) behind the electrons. This “push” or emf is defined in terms of a potential difference (in volts) between 2 points in the circuit.

(1 V = 1 J/C) Thus 1 joule of work is produced or required (depending on the direction of flow) when 1 coulomb of charge is transferred between 2 points in the circuit that differ by 1 volt.



Free Energy and Redox Spontaneous redox reactions produce a positive cell

potential, or emf. E° = E°red (reduction) – E°red (oxidation) Note that this is true for ALL redox reactions, not only for

voltaic cells. Dropped cell notation (cathode and anode) Since Gibbs free energy is the measure of spontaneity,

positive emf corresponds to negative ΔG.


Free Energy and Redox (Yet a 5th way to calculate G!!)

How do they relate? ΔG = –nFE

n = moles of electronsF = is the Faraday constant, 96,485 C/mol e-

(charge of 1 mole of electrons)E = maximum energy potential

ΔG° = –nFE° Max energy potential is directly related to the energy difference between the reactants and products in a

cell.


Free Energy, Redox, and K

How is everything related? ΔG° = –nFE° = –RT ln K


(a) Use the standard reduction potentials in Table 18.1 to calculate the standard free-energy change, ∆G°, and the equilibrium constant, K, at 298 K for the reaction

4 Ag(s) + O2(g) + 4 H+(aq) → 4 Ag+(aq) + 2 H2O(l)

(b) Suppose the reaction in part (a) is written

2 Ag(s) + O2(g) + 2 H+(aq) → 2 Ag+(aq) + H2O(l)

What are the values of E°, ∆G°, and K when the reaction is written in this way?

Using Standard Reduction Potentials to Calculate ∆G° and K







Nernst Equation Remember, ΔG = ΔG° + RT ln Q So, –nFE = nFE° + RT ln Q Dividing both sides by –nF, we get the

Nernst equation:E = E° – (RT/nF) ln Q

OR E = E° – (2.303 RT/nF) log Q Using standard thermodynamic

temperature and the constants R and F,E = E° – (0.0592V/n) log Q


Nernst Equation

The relationship between cell potential and concentrations of cell components

At 25°C:

or

(at equilibrium)Copyright © Cengage Learning. All rights reserved 65

0.0591 = log E E Q

n

0.0591 = logE K

n


Explain the difference between E and E°.When is E equal to zero?When the cell is in equilibrium ("dead" battery).

When is E° equal to zero? E is equal to zero for a concentration cell.







A Concentration Cell


A cell in which both compartments have the same components at different concentrations.

The electrons will flow in the direction in which to equalize the Ag+ concentration in the two compartments. (Flows to lower the more concentrated ion)


Concentration Cells

The difference in concentration is the only factor that produces cell potential.

E° = 0• For such a cell, would be 0, but Q would not.Ecell°• Once the concentrations are the same on

both sides, Q= 1 and E = 0 (equilibrium).


A concentration cell is constructed using two nickel electrodes with Ni2+ concentrations of 1.0 M and 1.00 × 10-4 M in the two half-cells.

Calculate the potential of this cell at 25°C.

0.118 V


EXERCISE!EXERCISE!



EXERCISE!EXERCISE!





Section 18.6BatteriesOne of the Six Cells in a 12–V Lead Storage Battery


Section 18.6Batteries

A Common Dry Cell Battery


A Mercury Battery



Schematic of the Hydrogen-Oxygen Fuel Cell

Fuel cells are NOT batteries; the source of energy must be continuously provided.

Section 18.7Corrosion

AP Learning Objectives, Margin Notes and References Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or



Process of returning metals to their natural state – the ores from which they were originally obtained.

Involves oxidation of the metal.


Section 18.7CorrosionCorrosion

Corrosion is oxidation. Its common name is rusting.


Corrosion Prevention

Application of a coating (like paint or metal plating) to protect the metal from oxygen and moisture. Galvanizing (coating steel with Zn to form a mixed

oxide-carbonate coating) Alloying- mix with a metal with a lower reduction

potential. Cathodic Protection

Protects steel in buried fuel tanks and pipelines.



Preventing Corrosion

Corrosion is prevented by coating iron with a metal that is more readily oxidized.

Zinc is more easily oxidized, so that metal is sacrificed to keep the iron from rusting.


Cathodic Protection-Protects steel in buried fuel tanks and pipelines.


Section 18.8Electrolysis



redox reactions. LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable

with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.

Additional AP References LO 5.15 (see Appendix 7.11, “Non-Spontaneous Reactions”)


Forcing a current through a cell to produce a chemical change for which the cell potential is negative.




Note:1. External powers source greater than 1.10V needed

2. Electrons flow in the opposite direction.3. Anode and cathode are reversed.4. Ions in the salt bridge flow in the opposite direction.

Section 18.9Commercial Electrolytic Processes

Production of aluminum Purification of metals Metal plating Electrolysis of sodium chloride Production of chlorine and sodium hydroxide


Section 18.8ElectrolysisElectrolysis

Nonspontaneous reactions can occur in electrochemistry IF outside electricity is used to drive the reaction.

Use of electrical energy to create chemical reactions is called electrolysis.

(+) terminal connected to the anode and the (-) one to the cathode-forces electrons from anode to the cathode.

Molten NaCl


Electroplating-use active electrodes during electrolysis to deposit a thin layer of metal onto another metal to improve beauty or resistance to corrosion.

Since using a solution, you must consider the reduction and oxidation of water.


Ni2+ E◦red = -.28V and water E◦red = -.83V so Ni2+ is preferentially reduced and plated on the steel cathode.

At the anode Ni2+ E◦red = -.28V and water E◦red = 1.23V so Ni2+ is preferentially oxidized (more negative E◦red means more easily oxidized).

Section 18.9Commercial Electrolytic Processes

Electroplating a Spoon


Section 18.8ElectrolysisElectrolysis and “Stoichiometry”

1 ampere = 1 coulomb 1 Second

I = q t q = charge (C) I = current (A) t = time (s)

Section 18.8ElectrolysisElectrolysis and “Stoichiometry” 1 coulomb = 1 ampere × 1 second q = It = nF q = charge (C) I = current (A) t = time (s) n = moles of electrons that

travel through the wire inthe given time

F = Faraday’s constantNOTE: n is different than that

for the Nernst equation!


Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.

Relating Electrical Charge and Quantity of Electrolysis


An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO3)3.

What is the metal?

gold (Au)




Consider a solution containing 0.10 M of each of the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+.

Predict the order in which the metals plate out as the voltage is turned up from zero.

Cu2+, Pb2+, Sn2+, Ni2+, Zn2+

Do the metals form on the cathode or the anode? Explain.



Documents

Chapter 18 Electrochemistry AP*. AP Learning Objectives LO 3.2 The student can translate an observed chemical change into a balanced chemical equation