Chapter 18 1 EEEElectrochemistry - is the study of the relationships between chemical reactions and...

ELECTROCHEMISTRY

Chapter 18

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Definitions (Ch. 4.9/4.10 and Ch. 18.1)

Electrochemistry - is the study of the relationships between chemical reactions and electricity.

Oxidation Number – The charge on an ion for a monoatomic ion. Otherwise, it is the hypothetical charge based on a set of rules. (See Slides 6 - 7.)

If a substance has lost electrons, its oxidation number has become more positive and thus oxidized

If a substance has gained electrons, its oxidation number is more negative and thus has been reduced.

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Definitions

LEO:Lose Electrons Oxidation

GER:Gain Electrons Reduction

GER

LEO says GER

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Definitions

Redox – Shorthand name for an oxidation-reduction reaction. Oxidation states are really just used as ‘electron bookkeepping’ and don’t represent an actual charge (unless you are talking about monoatomic ions). It does represent, though, a movement of electrons from one atom(s) to another. Electrons are particles of matter, so the number that leave one substance must the same number that go to another substance.

Oxidation Number Rules

1)Oxidation number of an element is zero.2)For monoatomic ion, the oxidation number

is equal to its charge. (ex: Na+ is +1, F- is -1)

3)Oxygen usually has a -2 oxidation number. (except in peroxides where it’s -1)

4)Hydrogen’s oxidation number is +1 when its attached to a nonmetal (e.g. HCl) and -1 when attached to a metal (e.g. CaH2) as a hydride.

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Oxidation Number Rules

5) Fluorine is always -1 in compounds.6) The other halogens are mostly -1 in binary

compounds. When they combine w/oxygen, they can be positive (ClO4

-).

7) The sum of the oxidation numbers in a neutral compound is zero.

8) The sum of the oxidation numbers in a polyatomic ion is equal to its charge.

Example from #6 & #8: oxygen is always -2 and there are 4 of them, Cl has to be +7 in order for the ClO4

- ion to be -1.

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Practice

COMPOUND OXIDATION NUMBER OF RED BOLDFACE ELEMENT

P2O5

NaH

Cr2O72-

SnBr4

K2O2

Practice Oxidation #’s in Reactions

Determine the oxidation number of every element in the following reactions.

1. Sn(s) + 2H+(aq) Sn2+ (aq) + H2(g)

2. 2MnO4-(aq) + 5HSO3

-(aq) +H+

2Mn2+(aq) + 5SO42-(aq) +

3H2O

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• Zn added to HCl yields the spontaneous reaction

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g).

• The oxidation number of Zn has increased from 0 to 2+.

• The oxidation number of H has reduced from 1+ to 0.

• Zn is oxidized to Zn2+ while H+ is reduced to H2.

Oxidation-Reduction Reactions

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Oxidation-Reduction Reactions

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• Law of conservation of mass: the amount (moles) of each element present at the beginning of the reaction must be present at the end.

• Conservation of charge: electrons are not lost in a redox chemical reaction, just moved around

Half Reactions• Half-reactions are a convenient way of separating

oxidation and reduction reactions.

Balancing Oxidation-Reduction Reactions

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Half Reaction Method• The skeleton reaction is:

MnO4-(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(aq)

• We want to make this such that it is easy to balance the atoms and the charges.

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

So, how do we balance the previous reaction?

1. Write down the two half reactions.

2. Balance each half reaction:a. First with elements other than H and O.

b. Then balance O by adding water.

c. Then balance H by adding H+.

d. Balancing the charge by using electrons

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

3. Multiply each half reaction to make the number of electrons equal.

4. Add the reactions and simplify.

5. Check!

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

1. The two incomplete half reactions are

MnO4-(aq) Mn2+(aq)

Fe2+(aq) Fe3+(aq)

2. Adding water (to balance O’s) and H+ yields:

8H+ + MnO4-(aq) Mn2+(aq) + 4H2O

• There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left:

5e- + 8H+ + MnO4-(aq) Mn2+(aq) + 4H2O

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

• In the iron ion reaction, there is a 2+ charge on the left and a 3+ charge on the right, so we need to add an electron:

Fe2+(aq) Fe3+(aq) + 1e-

• Now there is the same overall +2 charge on both sides.

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

3. To balance the 5 electrons for permanganate and 1 electron for iron, we need to have 5 electrons move from 1 species to the other. Multiplying the iron equation by 5 gives:

5e- + 8H+ + MnO4-(aq) Mn2+(aq) + 4H2O

5Fe2+(aq) 5Fe3+(aq) + 5e-

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

3. 5e- + 8H+ + MnO4-(aq) Mn2+(aq) + 4H2O

5Fe2+(aq) 5Fe3+(aq) + 5e-

• Notice now that the number of electrons is the same for both equations, appearing on opposite sides.• In reduction half-reaction, e- are reactants• In oxidation half-reaction, e- are products

Balancing Oxidation-Reduction Reactions

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Balancing Equations by the Method of Half Reactions

4. Adding gives:

8H+(aq) + MnO4-(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l)

+ 5Fe3+(aq)

5. Which is balanced!

Notice that atoms and charges are both balanced. There is a +17 charge on left (8-1+5(2)), and +17 charge on right (2+5(3)).

Balancing Oxidation-Reduction Reactions

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Fe2+(aq) + Al(s) Fe(s) + Al3+(aq)

Balancing Oxidation-Reduction Half-Reactions Practice

Mn2+(aq) + NaBiO3(s) Bi3+(aq) + Na+(aq) + MnO4-(aq)

Balancing Oxidation-Reduction Half-Reactions Practice

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Balancing Equations for Reactions Occurring in Basic Solution

• We can use OH- and H2O rather than H+ and H2O, respectively.

• However, it’s easier to balance in acid, and then ‘neutralize’ the acid with OH-. But if you add OH- to one side, you have to add it to the other side.

• In other words, balance in acidic solution and then convert it to a basic solution.

Balancing Oxidation-Reduction Reactions

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Balance the following reaction in a basic solution

Cr(OH)3(s) + ClO- CrO42- + Cl2(g)

1. Cr(OH)3 CrO42-

2. 2ClO- Cl2(g)

3. Multiply equation 1 by 2 and equation 2 by 3

2H2O + 2Cr(OH)3 2CrO42- + 10H+ + 6e-

6e-+ 12H++ 6ClO- 3Cl2(g) + 6H2O

Add to get:

2H+ + 2Cr(OH)3 + 6ClO- 2CrO42- + 3Cl2 + 4H2O

+ 5H+H2O +

Balancing Oxidation-Reduction Half-Reactions Example

+ 3e-

+ 2H2O2e- + 4H+ +

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Balancing Oxidation-Reduction Half-Reactions Example

2H+ + 2Cr(OH)3 + 6ClO- 2CrO42- + 3Cl2 + 4H2O

+2OH- +2OH -

“Neutralize” by adding OH- to both sides.2H2O +2Cr(OH)3 +6ClO-2CrO4

2- +3Cl2+ 4H2O+2OH-

Cross out common substances on both sides.

2Cr(OH)3 +6ClO-2CrO42- +3Cl2+ 2H2O+2OH-

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• The energy released in a spontaneous redox reaction is used to perform electrical work.

• Galvanic cells or voltaic cells are devices in which electron transfer occurs via an external circuit.

• Galvanic cells are spontaneous.• If a strip of Zn is placed in a solution of CuSO4, Cu is

deposited on the Zn and the Zn dissolves by forming Zn2+.

Ch. 18.2 - Galvanic Cells

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• Zn is spontaneously oxidized to Zn2+ by Cu2+.• The Cu2+ is spontaneously reduced to Cu0 by Zn.• The entire process is spontaneous.

Galvanic Cells

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Spontaneity and Potential EnergyRedox reactions occurring in voltaic cell are

spontaneousWhy do electrons flow spontaneously from

one electrode to other? Flow spontaneously due to difference in potential

energy between anode and cathode

Voltaic Cell: Cathode Reaction

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Voltaic Cell: Anode Reaction

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• Galvanic cells consist of– Anode: Zn(s) Zn2+(aq) + 2e2- (Oxidation)– Cathode: Cu2+(aq) + 2e- Cu(s) (Reduction)– Salt bridge (used to complete the electrical circuit): cations

move from anode to cathode, anions move from cathode to anode.

• The two solid metals are the electrodes (cathode and anode).

Galvanic Cells

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• The flow of electrons from anode to cathode is spontaneous.

• Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode.

• Potential difference: difference in electrical potential. Measured in volts.

• One volt is the potential difference required to impart one joule of energy to a charge of one coulomb:

Galvanic Cells

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Electron Flow and Potential Energy

Anode higher

potential energy

Cathode lower

potential energy

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High Voltage

Low Voltage

Height is an analogy for voltage

• As oxidation occurs, Zn is converted to Zn2+ and 2e-. The electrons flow towards the cathode where they are used in the reduction reaction.

• We expect the Zn electrode to lose mass and the Cu electrode to gain mass.

• “Rules” of galvanic cells:1. At the anode electrons are products. (Oxidation)

2. At the cathode electrons are reactants. (Reduction)

3. Electrons cannot swim.

Galvanic Cells

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• Electrons flow from the anode to the cathode.• Therefore, the anode is negative and the cathode is

positive. (Think anion is negative, cation is positive.)• Electrons cannot flow through the solution, they have to

be transported through an external wire. (Rule 3.)

Galvanic Cells

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Galvanic Cells - 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)

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• Anions and cations move through a porous barrier or salt bridge.

• Cations move into the cathodic compartment to balance the excess negatively charged ions since the Cu2+ ions are being reduced. (Cathode: Cu2+ + 2e- Cu)

• Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by the oxidation of Zn metal.

Galvanic Cells

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A Molecular View of Electrode Processes

During the spontaneous redox reaction between Zn(s) and Cu2+(aq):

• Zn(s) is oxidized to Zn2+(aq) and Cu2+(aq) is reduced to Cu(s).

• On the atomic level, a Cu2+(aq) ion comes into contact with a Cu(s) atom on the surface of the electrode that has extra electrons.

• Electrons form from Zn(s), forming Zn2+(aq) , travel through the circuit to the Cu electrode where Cu2+(aq) gain those electrons and form more Cu(s).

Galvanic Cells

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Standard Reduction (Half-Cell) Potentials• Convenient tabulation of electrochemical data.• Standard reduction potentials, Ered are measured relative

to the standard hydrogen electrode (SHE).

Ch. 18.3 – Standard Reduction Potentials

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Standard Reduction (Half-Cell) Potentials

Standard Reduction Potentials

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Standard Reduction (Half-Cell) Potentials

• The SHE is the cathode. It consists of a Pt electrode in a tube placed in 1 M H+ solution. H2(g) is bubbled through the tube.

• For the SHE, we assign

2H+(aq, 1M) + 2e- H2(g, 1 atm)

• Ered of SHE is zero by definition.

• The emf of a cell can be calculated from standard reduction potentials:

Standard Reduction Potentials

anodecathode redredcell EEE1

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Standard Reduction (Half-Cell) Potentials

• Consider Zn(s) Zn2+(aq) + 2e-. We measure Ecell relative to the SHE (cathode):

Ecell = Ered(cathode) - Ered(anode)

0.76 V = 0 V - Ered(anode).

• Therefore, Ered(anode) = -0.76 V.

• Standard reduction potentials must be written as reduction reactions:

Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V.

Standard Reduction Potentials

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Standard Reduction (Half-Cell) Potentials• Since Ered = -0.76 V we conclude that the reduction of

Zn2+ in the presence of the SHE is not spontaneous.• The oxidation of Zn with the SHE is spontaneous.• Changing the stoichiometric coefficient does not affect

Ered. (Measures potential energy/unit charge (Volt = J/C) Intensive property.)

• Therefore,

2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V.

• Reactions with Ered > 0 are spontaneous reductions relative to the SHE.

Standard Reduction Potentials

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Standard Reduction (Half-Cell) Potentials

• Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.

• The larger the difference between Ered values, the larger Ecell.

• In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode).

• Recall

Standard Reduction Potentials

anodecathode redredcell EEE

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Ease of Oxidation and Reduction

• The more positive Ered the more easily reduced the reactant on the left.

• The more negative Ered the more difficult the reactant is reduced. Conversely, the more easily the product is oxidized.

• A species higher and to the left of the table of standard reduction potentials will spontaneously oxidize a species that is lower and to the right (product) in the table. (Called the Northwest-Southeast Rule)

• That is, F2 will oxidize H2 or Li; Ni2+ will oxidize Al(s).

Standard Reduction Potentials

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Most easily reduced

Incr

easi

ng e

ase

of r

educ

tion

Incr

easi

ng e

ase

of o

xida

tion

Most easily

oxidized1

Standard Reduction PotentialsStandard Reduction (Half-Cell) Potentials

Rank the halogens in order of ease of being reduced.

Which of the halogens is capable of oxidizing Ag to Ag+.

Which of these metals is easiest to oxidize? Fe, Ag, Zn, Mg, Au

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Cell EMF

Calculate the Eocell for the following reaction:

2Al(s) + 3I2(s) 2Al3+ + 6I-(aq)

1st: Determine anode & cathode 2nd: Write the reduction half-reactions.

2Al3+ + 6e- 2Al(s) Eored = -1.66 V

3I2(s) + 6e- 6I-(aq) Eored = +0.54 V

3rd: Calculate Eocell =Eo

cathode – Eoanode

Eocell = 0.54 – (-1.66) = +2.20 V

Standard Reduction Potentials

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A voltaic cell is based on a Co2+/Co half cell and a AgCl/Ag half cell. What is the reaction at the anode? What is the standard cell potential?

Co2+ + 2e- Co(s) Eored = -0.277

AgCl + 1e- Ag(s) + Cl-(aq) Eored = 0.22 V

Anode will be oxidation, more neg reduction potential (Co(s)Co2+ + 2e-)

Eocell = 0.22 – (-0.277) = 0.497 V

Cell Practice

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The Galvanic cell below has a cell potential of 1.19 V.Tl3+(aq) + 2Cr2+(aq) Tl+(aq) + 2Cr3+(aq)

What are the half reactions? Determine the Ered for Tl3+ to Tl+. Write the shorthand notation for the cell.

Tl3+ + 2e- Tl+(aq) Eored = ??

2Cr3+ + 2e- 2Cr2+(aq) Eored = -0.41 V

The cell potential is positive, so see what was oxidized (Cr2+) Eo

cell = Eocathode – Eo

anode then

Eocathode = Eo

cell+Eoanode

=1.19 + (-0.41) = +0.78 Cr2+(aq)|Cr3+(aq)||Tl3+(aq)|Tl+(aq)

Cell Practice

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• Electromotive force (emf) is the force required to push electrons through the external circuit.

• Cell potential: Ecell is the emf of a cell.

• For 1M solutions at 25 C (standard conditions), the standard emf (standard cell potential) is called Ecell.

• Electrical work is viewed from the point of view of the system.

C 1J 1

V 1

Ch. 18.4 - Cell Potential, Electrical Work and Free Energy

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Potential, Work & 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.

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Potential, Work & Energy

Maximum Cell Potential Directly related to the free energy difference

between the reactants and the products in the cell. ΔG° = –nFE°

F = 96,485 C/mol e–

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EMF and Free-Energy Change• The Free Energy of a Cell (ΔG) is related to it’s EMF:

• G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell.

• We define

• Since n and F are positive, if G > 0 then E < 0.• And if E>0, then ΔG <0 and cell is spontaneous.

Spontaneity of a Cell Reaction

nFEG

1 96,500 C/mol 96,500 J/V·molF

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• In a galvanic (voltaic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode) since

• More generally, for any electrochemical process

• A positive E indicates a spontaneous process (galvanic cell).

• A negative E indicates a nonspontaneous process.

Spontaneity of a Cell Reaction

anodecathode redredcell EEE

cell red redreduction process oxidation processE E E

Spontaneity of a Cell Reaction

What is n for the following reaction? What is the ∆G0 for the reaction?

Cu(s) + Ba2+(aq) Cu2+(aq) + Ba(s) n = 2 E0 = -2.90V (E0

cathode) – 0.337(E0anode) = -3.24 V

∆G0 = -nFE0

=-(2 mol)(96,500J/V mol)(-3.24V) =625 kJ

=> Reaction is not spontaneous.1

The Nernst Equation• A voltaic cell is functional until E = 0 at which point

equilibrium has been reached.• The point at which E = 0 is determined by the

concentrations of the species involved in the redox reaction.

• The Nernst equation relates emf to concentration using

and noting that

Ch. 18.5 - Dependence of Cell Potential on Concentration

QRTGG ln

QRTnFEnFE ln1

The Nernst Equation• This rearranges to give the Nernst equation:

• The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K:

• (Note that change from natural logarithm to base-10 log.)• Remember that n is number of moles of electrons.

Concentration and Cell Potential

QnFRT

EE ln

0.0592logE E Q

n

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Concentration and Cell Potential

Example• Recall earlier we had a voltaic cell reaction of

2Al(s) + 3I2(s) 2Al3+ + 6I-(aq) E0cell = +2.20 V

• What is the Ecell when [Al3+] is 4.0x10-3M and [I-] = 0.010 M?

• Ecell = E0cell – 0.0592/n log [Al3+]2[I-]6

• = 2.20 V – 0.0592/6 log [4.0x10-3]2[0.010]6

• = 2.20 V – 0.0099*log 1.6x10-5(1x10-12)• = 2.20 V – 0.0099*log 1.6x10-17

• = 2.20 V – 0.0099*(-16.796)• = 2.20 V + 0.17 = 2.37 V1

Concentration Cells• We can use the Nernst equation to generate a cell that has

an emf based solely on difference in concentration.• One compartment will consist of a concentrated solution,

while the other has a dilute solution.• Example: 1.00 M Ni2+(aq) and 1.00 10-3 M Ni2+(aq).• The cell tends to equalize the concentrations of Ni2+(aq)

in each compartment.• The concentrated solution has to reduce the amount of

Ni2+(aq) (to Ni(s)), so must be the cathode.

Concentration and Cell Potential

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Concentration Cells

Concentration and Cell Potential

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Cell Potential and Chemical Equilibrium• A system is at equilibrium when G = 0.• From the Nernst equation, at equilibrium and 298 K (E =

0 V and Q = Keq):

Concentration and Cell Potential

0.05920 log

log0.0592

eq

eq

E KnnE

K

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Spontaneity of Redox Reactions Calculate the Keq for the following reaction

Fe(s) + Ni2+ Fe2+(aq) + Ni(s) Fe2+ + 2e- Fe(s) E0= -0.440 V Ni2+ + 2e- Ni(s) E0 = -0.280 V E0

cell = -0.280 – (-0.440) = +0.16V

Log Keq = nE0/0.0592 = 2(0.16)/0.0592 = 5.405 Keq = 2.54x105

Concentration and Cell Potential

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Electrolysis of Aqueous Solutions• Nonspontaneous reactions require an external current in

order to force the reaction to proceed.• Electrolysis reactions are nonspontaneous.• In voltaic and electrolytic cells:

– reduction occurs at the cathode, and– oxidation occurs at the anode.– However, in electrolytic cells, electrons are forced to flow

from the anode to cathode. (Electrons are forced to flow from low potential to high potential. So it’s a measure of the work you put into the system.)

Ch. 18.8 - Electrolysis

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Electrolysis of Aqueous Solutions

Electrolysis

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Electrolysis of Aqueous Solutions• Example, decomposition of molten NaCl.• Cathode: 2Na+(l) + 2e- 2Na(l)• Anode: 2Cl-(l) Cl2(g) + 2e-.

• Industrially, electrolysis is used to produce metals like Al.

Electrolysis

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Electroplating• Active electrodes: electrodes that take part in electrolysis.• Example: electrolytic plating.

Electrolysis

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Electroplating• Consider an active Ni electrode and another metallic

electrode placed in an aqueous solution of NiSO4:

• Anode: Ni(s) Ni2+(aq) + 2e-

• Cathode: Ni2+(aq) + 2e- Ni(s).• Ni plates on the inert electrode.• Electroplating is important in protecting objects from

corrosion.

Electrolysis

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Quantitative Aspects of Electrolysis• We want to know how much material we obtain with

electrolysis.• Consider the reduction of Cu2+ to Cu.

– Cu2+(aq) + 2e- Cu(s).– 2 mol of electrons will plate 1 mol of Cu.– The charge of 1 mol of electrons is 96,500 C (1 F).– Since q = It, the amount of Cu can be calculated from the

current (I) and time (t) taken to plate. (C = Ampere*second)

Electrolysis

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Electrolysis In an electrolysis cell, Ni(s) is deposited on the

cathode in a solution of Ni2+ ions. If the current is 0.150 A for 12.2 min, how much Ni is plated?

1st: Calculate how many Coulombs of e- are needed.

Q=0.150 A * 12.2 min * 60 sec/min=109.8 Coulombs

2nd: Calculate the number of moles of electrons (Faradays) that is by…

109.8 C = 0.0011 mol e- (F) 96500C/mol

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Electrolysis 3rd: Determine the stoichiometric ratio of

electrons to Ni2+

Ni2++ 2e- Ni(s) 0.0011 mol e-(1 mol Ni2+)(58.7g Ni/mol Ni) 2 mol e-

=0.0334 g or 33.4 mg of Ni

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Electrolysis Metallic magnesium can be made by the

electrolysis of molten MgCl2. How many minutes are needed to form 10.00 g of Mg from molten MgCl2 using a 3.50 A current?

1st: Calculate how many moles of Mg you need to make. 10.00g Mg*1mol/24.3 g = 0.4115 mol Mg

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Electrolysis 2nd: Determine stoichiometric ratio of e- to Mg

Mg2+ + 2e- Mg(s) 0.4115 mol Mg* 2 mol e-/mol Mg =0.8230 mol e-

3rd: Determine the charge on 0.8230 mol e-

0.8230 mol e- * 96,500 C/mol = 79423 C

4th: Calculate the time using t=q/I or q * 1/It = 79423C * 1 sec/3.5C * 1min/60sec

= 378.2 min1

Electrolysis Practice 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?

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AP Exam Practice

2011 #3(d-g) (Fuel Cell question)2010 #6(c-g)2009B #62008 #3

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