electrochemistry brown, lemay ch 20 ap chemistry

Electrochemistry

Brown, LeMay Ch 20AP Chemistry

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20.1: Oxidation-reduction reactions“Redox”: rxns with a change in oxidation number• At least one element’s oxidation number will

increase and one will decrease• Consists of two half-reactions:

oxidation: loss of e-; oxidation number increasesreduction: gain of e-; oxidation number decreases

(“reduces”)

Oxidizing agent or oxidant: causes another substance to be oxidized; gains e-

Reducing agent or reductant: causes another substance to be reduced; loses e-

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Ex: Determine the half-reactions.

Cr2O72- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2

(g)+6

-2 -1 +3

0

Reduction half-reaction:

Cr2O72- (aq) → Cr3+ (aq)

Oxidizing agent

Oxidation half-reaction:

Cl1- (aq) → Cl2 (g)

Reducing agent

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20.2: Balancing Redox Reactions“Method of Half-Reactions” steps:1. Assign oxidation numbers to all species; based on

this, split the half-reactions of reduction and oxidation.

2. For each, balance all elements except H and O.3. Balance oxygen by adding H2O to the opposite

side.4. Balance hydrogen by adding H+ to the opposite

side.5. Balance charges by adding e- to side with overall

positive charge.6. Multiply each half-reaction by an integer so there

are an equal number of e- in each.7. Add the half reactions; cancel any species; check

final balance.

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Ex: Balance the following reaction, which takes place in acidic solution:Cr2O7

2- (aq) + Cl1- (aq) → Cr3+ (aq) + Cl2 (g)

Reduction:

Cr2O72- (aq) → Cr3+ (aq)2 + 7 H2O (l)14 H+ (aq) +6 e- +

Oxidation:

Cl- (aq) → Cl2 (g)

6 Cl- (aq) → 3 Cl2 (g) + 6 e-

2 + 2 e-3 [ ]

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6 e- + 14 H+ (aq) + Cr2O72- (aq) → 2 Cr3+ (aq) + 7

H2O (l)

6 Cl- (aq) → 3 Cl2 (g) + 6 e-

14 H+ (aq) + Cr2O72- (aq) + 6 Cl- (aq) →

2 Cr3+ (aq) + 7 H2O (l) + 3 Cl2 (g)

•Since occurs in acidic solution, H+ appears as a reactant

•If reaction occurs in basic solution, balance using same method, but neutralize H+ with OH- as a last step.

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Ex: In basic solution, MnO4

- (aq) + Br- (aq) → MnO2 (s) + BrO3- (aq)

+7

-2 -1 +4

+5

-2 -2

Reduction: MnO4- (aq) → MnO2 (s)

MnO4- (aq) → MnO2 (s) + 2 H2O

(l)4 H+ (aq) + MnO4

- (aq) → MnO2 (s) + 2 H2O (l)3 e- + 4 H+ (aq) + MnO4

- (aq) → MnO2 (s) + 2 H2O (l)

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Oxidation: Br- (aq) → BrO3- (aq)

3 H2O (l) + Br- (aq) → BrO3- (aq)

3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq)

3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e-

2[3 e- + 4 H+ (aq) + MnO4- (aq) → MnO2 (s) + 2 H2O

(l)]

6 e- + 8 H+(aq) + 2 MnO4-(aq) → 2 MnO2(s) + 4 H2O(l)

3 H2O (l) + Br- (aq) → BrO3- (aq) + 6 H+ (aq) + 6 e-

2 H+(aq) + 2 MnO4- (aq) + Br- (aq) →

2 MnO2 (s) + BrO3- (aq) + H2O (l)

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Neutralize with OH-:2 H+(aq) + 2 MnO4

-(aq) + Br- (aq) + 2 OH- (aq) →2 MnO2 (s) + BrO3

- (aq) + H2O (l) + 2 OH- (aq)

2 H2O (l) + 2 MnO4-(aq) + Br- (aq) →

2 MnO2 (s) + BrO3- (aq) + H2O (l) + 2 OH- (aq)

Simplify:H2O (l) + 2 MnO4

-(aq) + Br- (aq) →

2 MnO2 (s) + BrO3- (aq) + 2 OH- (aq)

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20.3: Voltaic (or Galvanic) Cells

• Device that utilizes energy released in a spontaneous electrochemical reaction by directing e- transfer along an external pathway, rather than directly between reactants

Electrodes: metals (or graphite) connected by an external circuitAnode: half-cell where oxidation occursCathode: half-cell where reduction occurs

Luigi Galvani(1737–1798)

Alessandro Volta(1745–1827)

Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

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20.4: Electromotive force (emf or E)• Potential difference between cathode and

anode that causes a flow of e-.– The result of the difference in electric fields

produced at each electrode

• Also called cell potential or cell voltage, Ecell

• Units = Volt (1 V = 1 J / 1 C)

• Standard conditions (25°C, 1 atm, 1 M): Eºcell

• Eºcell is unique to each set of half-cells involved Charles-Augustin de Coulomb

(1736–1806)

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Standard Reduction Potentials• Assigned to each reduction reactionEºcell = Eºreduction(cathode) - Eº reduction(anode)

• All Eºred are referenced to standard hydrogen electrode (S.H.E.):

2 H+ (aq, 1 M) + 2 e- → H2 (g, 1 atm)

Eºreduction = 0 V

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Ex: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)• Anode: Zn (s) → Zn2+ (aq) + 2 e-

– This is oxidation, but Eºreduction is recorded as the reduction reaction:

Zn2+ (aq) + 2 e- → Zn (s), where Eºreduction = - 0.76 V

• Cathode:Cu2+ (aq) + 2 e- → Cu (s) Eºreduction = + 0.34 V

Eºcell = Eºreduction (cathode) - Eºreduction (anode)= 0.34 V – (- 0.76 V)

Eºcell = +1.10 V

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• Positive voltages = reaction is spontaneous (cell will produce voltage, G is -)

• Changing the stoichiometric coefficients of a half-cell does not affect the value of Eº, since potential is a measure of the energy per electrical charge.

• However, having more moles means the reaction will continue for a greater time.

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Using the Standard Reduction Potential table

• When Eºred is very negative, reduction is difficult (but oxidation is easy):Li1+ (aq) + e- → Li (s) Eºred = - 3.05 V– Li (s) oxidizes easily, so it is the strongest

reducing agent; lower on the Standard Reduction Potential chart signifies the anode.

• When Eºred is very positive, reduction is easy:

F2 (g) + 2 e- → 2 F1- Eºred = + 2.87 V– F2 reduces easily, so it is the strongest oxidizing

agent; higher on the chart signifies the cathode.

• Predicting metal “activity series”: most reactive metals are near bottom

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20.5: Spontaneity of Redox Reactions• Gibbs free energy and emf:

Go = - n F Eo

n = # of moles of e- transferred

Faraday’s constant, F= 96,500 C/mol of e-

• 1F = 1 mole of e-, which has a charge of 96,500 C• Units = (mol e-)(C/mol e-)(V) = (mol e-)(C/mol e-)

(J/C) = J

• As Eºcell increases, Gº decreases (becomes more likely to be spontaneous)

Michael Faraday(1791-1867)

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20.6: Effect of Concentration on Ecell

The Nernst Equation:G = Go + RT ln Q and Go = - n

F Eo

(R = 8.314 J/(mol•K))

- n F E = - n F Eº + RT ln Q Q

n

RTEE ln0

F

Walther Nernst

(1864-1941)

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• At equilibrium, G = 0

Go K Eocell Reaction

Negative > 1 Positive Spontaneous

0 1 0 Equilibrium

Positive < 1 Negative

Non-spontaneous

Kn

RTE ln0

F

Kn

E log0591.00 since T º = 298 K

Equilibrium composition measurements

Calorimetry and entropy: S0 and S0

Electrical measurements: E0 and E0

cell

3 Thermodynamic Quantities:

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* E0cell of cells

• Model of a portion of a cell membrane shows a K+

channel (blue) and a Na+ channel (red). The area outside the cell (top) is rich in Na+ and low in K+. Inside the cell, the fluids are relatively rich in K+ and low in Na+.

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20.8-9: Electrolytic cells• Electrical energy is used to cause a

non-spontaneous reaction to occur.• Current (I): a measure of the charge

(Q) per unit time (t)

• Units = amps or amperes (A)1 A = 1 C / 1 s

t

QI

André-Marie Ampère(1775 – 1836)

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Ex: Na+ (l) + Cl- (l) → Na (l) + Cl2 (g)Oxidation half-cell: 2 Cl- (l) → Cl2 (g) + 2 e-

Reduction half-cell: Na+ (l) + e- → Na (l)

anode

Pt Pt(-)(+)

cathode

e- flow

NaCl (l)

Cl2 (g) Na (l)

• In a voltaic cell: anode is (-), cathode is (+).

• In an electrolytic cell: anode is (+), cathode is (-).

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20.9: Quantitative ElectrolysisEx: A current of 0.452 A is passed

through an electrolytic cell containing molten NaCl for 1.5 hours.

• Write the electrode reactions.

• Calculate the mass of products formed at each electrodes

Oxidation: 2 Cl- (l) → Cl2 (g) + 2 e-

Reduction: Na+ (l) + e- → Na (l)

Strategy: calculate the total charge of e- transferred, and use F to convert to moles.

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C 2441sec) A)(5400 (0.452Q

Na mol 1

g 22.99

1

Na mol 1

C 96,500

1C) (2441Na Mass

FF

g 0.58

2

22 Cl mol 1

g 70.90

2

Cl mol 1

C 96,500

1C) (2441Cl Mass

FF

g 0.90

tIQor t

QI

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Electrolysis of Aqueous Mixtures• If the voltage on an aqueous mixture of CuCl2 and

ZnCl2 in an electrolytic cell is slowly increased, what products will form at each electrode?

o What exists in solution that could be reduced?Cu2+, Zn2+, and H2O (not Cl-)

Cu2+ + 2 e- → Cu Eºred = 0.34 V

Zn2+ + 2 e- → Zn Eºred = - 0.76 V

2 H2O + 2 e- → H2 + 2 OH- Eºred = - 0.83 V

o Since Eºred (Cu) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the cathode):Cu (s) > Zn (s) > H2 (g)

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• If the voltage on an aqueous mixture of CuCl2 and ZnCl2 in an electrolytic cell is slowly increased, what products will form at each electrode?

o What exists in solution that could be oxidized?Cl- and H2O (not Cu2+ and Zn2+)

2 H2O → O2 + 4 H+ + 4 e- Eºox = - 1.23 V2 Cl- → Cl2 + 2 e- Eºox = - 1.36 V

o Since Eºox (H2O) is the most positive, it is the most probable reaction, and therefore will occur most easily (at the anode):O2 (g) > Cl2 (g)

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Electroplating• The anode is a

silver bar, and the cathode is an iron spoon.

Anode:Ag (s) → Ag+ (aq)

Cathode:Ag+ (aq) → Ag (s)

electrochemistry brown, lemay ch 20 ap chemistry

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