CHAPTER 19: ELECTROCHEMISTRY
Part One: Introduction
A. Terminology.
1. Electrochemistry deals with:
a. Chemical reactions produced by electric current. (electrolysis)
b. Production of electric current by chemical reactions. (voltaic)
2. Involves the electrons transferred in redox reactions.
3. Very convenient way of studying chemical reactions.
4. Sites of oxidation and reduction are physically separated so that the e- must flow through an external circuit.
5. This introduces the concept of half-reactions, with an oxidation half-reaction occurring at the anode and a reduction half-reaction occurring at the cathode.
6. Setup is called an electrochemical cell. Two main types:
a. Electrolytic cell = electrical energy from external source causes nonspontaneous reactions to occur. (lysis = splitting)
b. Voltaic cell = spontaneous reaction produces electricity in external circuit.
B. Balancing Oxidation-Reduction Reactions. (Section 19.1)
Two methods:
a. change-in-ox# method
b. half-reaction method
C. Balancing Redox Rxns - Change-in-ox# Method. (optional)
1. Based on equal total increases and decreases in ox#’s.
2. Procedure:
a. Write as much of overall unbalanced eqn. as possible.
Chapter 19 Page 1 b. Assign ox#.
c. Draw brackets connecting oxidized and reduced species, show change in ox# per atom.
d. Insert coefficients to make net change of ox# = 0.
e. Balance the other atoms by inspection.
3. All balanced eqns. must satisfy two criteria:
a. Mass balance. (same # of atoms of each kind on both sides)
b. Charge balance.
4. Convenient to omit spectator ions also.
+ 2+ For example: H (aq) + Zn(s) → Zn (aq) + H2(g)
5. Example:
- + - 2+ Balance: MnO4 (aq) + H (aq) + Br (aq) → Mn (aq) + Br2(l) + H2O(l)
Answer:
+ - D. Adding H , OH , or H2O to balance O or H atoms.
1. Frequently must work with an incomplete given equation.
+ - O & H species balanced by adding H , OH , or H2O by rules:
Chapter 19 Page 2 2. Example:
2+ - 3+ 2+ Balance in acidic solution: Fe (aq) + MnO4 (aq) → Fe (aq) + Mn (aq)
E. The Half-Reaction Method.
1. Write as much of the overall unbalanced equation as possible, omitting spectator ions.
2. Construct unbalanced redox half-reactions (these are usually incomplete as well as unbalanced). Show complete formulas for polyatomic ions and molecules.
3. Balance all elements in each half-reaction, except H and O. Then use the previous chart to balance H and O in each half-reaction.
4. Balance the charge in each half-reaction by adding e- as “products” or “reactants.”
5. Balance the electron transfer by multiplying the balanced half-reactions by appropriate integers.
6. Add the resulting half-reactions and eliminate any common terms.
7. Example: Given (in acidic solution):
- 2- Br2(l) + SO2(g) → Br (aq) + SO4 (aq)
Answer:
- - 2e + Br2(l) → 2 Br (aq)
2- + - 2 H2O + SO2(g) → SO4 (aq) + 4 H (aq) + 2e
- 2- + 2 H2O + Br2 + SO2 → 2 Br + SO4 + 4 H
F. Electrical Conduction.
1. Electric current is transfer of charge, measured in Amperes (A).
Chapter 19 Page 3 2. Metallic conduction = only e- move, no atoms move.
3. Ionic or electrolytic conduction = motion of ions through a solution.
G. Electrodes.
1. Surfaces at which oxidation and reduction half-reactions occur.
2. Cathode = site of reduction (e- are gained by some species).
3. Anode = site of oxidation (e- are lost by some species).
4. Which is + and - depends on whether cell is electrolytic or voltaic type.
Part Two: Voltaic (Galvanic) Cells
A. Basic Construction. (Section 19.2)
1. These are electrochemical cells in which spontaneous redox reactions produce e- current.
Therefore, ΔGrxn < 0
2. Examples: flashlight batteries, etc.
3. Must isolate the 2 half-reactions in separate compartments connected by salt bridge.
4. Here is the Daniell cell (Cu/Zn cell), based on the reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Chapter 19 Page 4
5. Cell notation:
Zn|Zn2+(1.0M )||Cu2+(1.0 M)|Cu
6. Cell potential (voltage) of this cell is 1.10 volts.
ΔGrxn = -νFE
where ν = # of e- transferred (here = 2) E = cell potential (in volts)
B. Half-Reactions.
1. Any redox reaction may be expressed as the sum of two half-reactions, one involving electron loss by a species and the other electron gain.
2. Example in Daniell Cell:
a. Oxidation of Zn: Zn(s) → Zn2+(aq) + 2e-
b. Reduction of Cu2+: Cu2+(aq) + 2e- → Cu(s)
3. Half-reactions are conceptual reactions showing the loss and gain of electrons.
4. A redox reaction is the sum of an oxidation and a reduction half-reaction.
Chapter 19 Page 5
5. Example: Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
6. However, it is a common practice to write all half-reactions as reductions. Then the overall reaction is the difference of the two reduction half-reactions.
7. Example: Cu2+(aq) + 2e- → Cu(s) -{Zn2+(aq) + 2e- → Zn(s)} Cu2+ + Zn(s) → Cu(s) + Zn2+(aq) net reactions
C. The Cell Potential. (Section 19.4)
1. If cell reaction is not at equilibrium the reaction drives e- through external circuit.
2. A potential difference exists between the two electrodes called the cell potential, E, measured in volts, V.
3. The larger the E, the more work a given number of electrons can do.
4. A cell in which the reaction is at equilibrium can do no work, and its voltage is zero.
5. The maximum work that a cell can do is equal to the ΔG of reaction.
6. The cell potential, E, is related to the reaction free energy, ΔG, at a particular composition of the cell compartments by:
ΔG = -νFE
a. ν is the moles of electrons transferred according to the balanced equation.
b. For example, in the Daniell cell, ν = 2 mol.
c. F = Faraday’s constant.
7. The equation ΔG = -νFE means that, if we know the reaction free energy, then we can state the cell potential, and vice versa. (We now have an electrical method for measuring a reaction free energy.)
Chapter 19 Page 6 8. Problem: A given reaction has a ΔG = -100 kJ at certain specified concentrations of reactant and products and ν = 1 moles are transferred according to the balanced equation. Calculate the voltage of the cell.
E = -(ΔG/νF ) = -{(-100 kJ)/[(1 mol)(96486 C mol-1)]} E = +1.036 V
D. The Standard Reduction Potential E° of Half-Reactions. (Section 19.5)
1. Each electrode in a galvanic cell makes a characteristic contribution to the overall cell potential.
2. It is not possible to measure the potential of a single electrode alone.
3. One electrode can be assigned a value zero and the others assigned values relative to that electrode. The specially selected electrode is the standard hydrogen electrode (SHE): + - 2 H (aq) + 2e → H2(g) E° = 0
4. The standard reduction potential of any half-reaction is then measured by constructing a cell in which the half-reaction of interest is the cathode and the SHE is the anode.
5. Example: The standard reduction potential of the Ag+ ion is the standard potential of the following cell:
+ + (anode) Pt H2(g) H (aq) Ag (aq) Ag(s)
(cathode) E° = +0.8 V
Chapter 19 Page 7 6. Standard Aqueous Electrode Potentials at 25° C - The Electromotive Series.
E. Calculating Cell Potentials using Table of Reduction Potentials.
1. Predict standard potential of a cell E° formed from any two electrodes by taking the difference of their standard reduction potentials:
E° = E°cathode - E°anode
E°(whole cell) = {reduction pot of cathode rxn} - {reduction pot of anode rxn}
2. Example: One of the reactions important in corrosion in an acidic environment is:
+ 2+ Fe(s) + 2 H (aq) + 1/2 O2(g) → Fe (aq) + H2O(l)
The two reduction half-reactions are:
+ - a) 2 H (aq) + 1/2 O2(g) + 2e → H2O(l) E° = +1.23 V
b) Fe2+(aq) + 2e- → Fe(s) E° = -0.44 V
Chapter 19 Page 8 The difference (a) - (b) gives the desired net reaction, with
E° = +1.23 - (-0.44) = +1.67 V
3. Example: Calculate the cell potential for the reaction at 298 K:
Sn2+(aq) + Pb(s) → Sn(s) + Pb2+(aq)
Sn2+ + 2e- → Sn E° = -0.140 - {Pb2+ + 2e- → Pb} E° = -0.126
E° = -0.140 - (-0.126) = -0.014V
F. The Electrochemical Series.
1. Species with a high reduction potential tend to be reduced easily.
2. Species with a low reduction potential tend to be oxidized easily.
3. Important: low reduces high, high oxidizes low.
A species with a low reduction potential has a thermodynamic tendency to reduce a species with a high reduction potential.
4. For example:
E°(Zn2+, Zn) = -0.76 V
E°(Cu2+, Cu) = +0.34 V
Reaction will be spontaneous if Cu2+ + 2e- → Cu is the cathode reaction (reduced) and Zn → Zn2+ + 2e- is the anode reaction (oxidized).
Total E° = +0.34 - (-0.76) V = +1.10 V
G. Effect of concentration on cell potentials. (Section 19.7)
1. As electrochemical cell wears down its operating voltage E drops.
2. Why? E depends on concentrations of cell components.
Chapter 19 Page 9 3. Standard electrode potentials E° in tables are found at standard conditions:
1.00 M solution concentration 1.00 atm pressure for gases
4. Cell potentials E at other conditions will be different than E°.
5. E concentrations described by Nernst Equation.
H. The Nernst Equation.
2.303RT 1. E = E° - logQ nF where:
E = potential under conditions of interest € E° = potential under standard conditions R = gas constant T = Kelvin temperature n = # of e- transferred in chemical equation F = Faraday’s constant Q = reaction quotient
2. Collection of constants at 25°C:
2.303RT = 0.0592 Volts F
0.0592 E = E° - log Q n € 3. Work problem: Calculate the potential of a voltaic cell that consists of an Fe3+/Fe2+ electrode with [Fe3+] = 1.0 x 10-2M and [Fe2+] = 0.1M connected to a Sn4+/Sn2+ electrode€ with [Sn4+] = 1.0M and [Sn2+] = 0.10M.
a. Step 1: Calculate the E° of the cell by usual procedure using E° table. E°
Reduction 2(Fe3+ + e- ---> Fe2+) 0.771 V Oxidation 1(Sn2+ ---> Sn4+ + 2e-) -(0.15) V
Cell reaction 2 Fe3+ + Sn2+ ---> 2 Fe2+ + Sn4+
E°cell = +0.62 V
Chapter 19 Page 10 b. Step 2: Use Nernst Equation – substitute the ion concentrations into Q to
calculate Ecell.
0.0592 E = E° - log Q cell cell 2 2+ 4+ 0.0592 (Fe )(Sn ) Ecell = 0.62V − log 2 2 (Fe3+ ) (Sn 2+ ) €
2 0.0592 (0.10 )(1.0) Ecell = 0.62V − log 2 € 2 (1.0 ×10−2 ) (0.10) 2 0.0592 (0.10 )(1.0) Ecell = 0.62V − log 2 2 (1.0 ×10−2 ) (0.10) € 0.0592 Ecell = 0.62V − (3.00) 2
€ Ecell = 0.62 – 0.09V = 0.53V
€ I. Relationship of standard cell potential E° to ΔG° and equilibrium constant K. (Section 19.6)
1. Remember ΔGrxn = -nFE and ΔG°rxn = -nFE°.
2. But we also saw before that: ΔG°rxn = -RT ln Keq
3. Therefore: -nFE° = -RT ln Keq
nFE° 4. Re-arrange to: ln K = RT
nE° 5. At 25°C: log K = 10 0.0592V €
€
Chapter 19 Page 11
Part Three: Commercially Important Cells
A. The Dry Cell. (Leclanche, 1866) (Section 19.8)
1. Outer container made of Zn metal. (anode)
2. Inner electrode is carbon rod at center (cathode) surrounded by moist mixture of
NH4Cl, MnO2 and ZnCl2.
3. Overall reaction:
+ 2+ Zn(s) + 2 NH4 → Zn + 2 NH3(g) + H2(g); E = 1.6 V
+ - a. reduction at cathode 2 NH4 + 2e → 2 NH3(g) + H2(g)
b. oxidation at anode Zn → Zn2+ + 2e-
4. H2 gas evolved is absorbed by reaction:
H2 + 2 MnO2 → 2 MnO(OH)
5. NH3 gas evolved is absorbed by reaction:
2+ 2+ 4 NH3 + Zn → [Zn(NH3)4]
Chapter 19 Page 12 6. See Figure 19-9.
B. Alkaline Cells.
1. Similar to Leclanche cell except:
a. KOH electrolyte is used, making it alkaline.
b. Inner Zn surface is roughened to increase surface area.
2. Rxn:
- - Zn(s) + 2 OH (aq) → Zn(OH)2(s) + 2e (anode)
- - 2 MnO2(s) + 2 H2O + 2e → 2 MnO(OH)(s) + 2 OH (cathode)
Zn(s) + 2 MnO2(s) + 2 H2O → Zn(OH)2(s) + 2 MnO(OH) (overall)
3. Longer shelf life.
4. E = +1.5 V
C. Lead Storage Battery.
1. “Battery” means whole row or array of cells in series.
Chapter 19 Page 13 2. Rows of Pb plates are anodes.
3. Alternating rows of PbO2 plates are cathodes.
4. Immersed in 40% H2SO4(aq).
5. Reaction:
2- +0.356V - SO4 (aq) + Pb → PbSO4(s) + 2e (anode - oxid.)
+ 2- - +1.685V PbO2 + 4 H (aq) + SO4 + 2e → PbSO4(s) + 2 H2O (cathode - red.)
Pb + PbO2 + 2 H2SO4(aq) → 2 PbSO4(s) + 2 H2O (overall)
Overall E° = +0.356 + 1.658 = +2.041 V
6. Six cells in series creates 6 × 2.041V ≈ 12 Volts. (auto battery)
7. This reaction is reversible, so application of external potential can recharge the battery.
D. Nickel-Cadmium (Nicad) Cell.
1. Rechargeable and sealed.
2. Reaction:
- - Cd(s) + 2 OH (aq) → Cd(OH)2(s) + 2e (anode)
- - NiO2(s) + 2 H2O + 2e → Ni(OH)2(s) + 2 OH (cathode)
Cd(s) + NiO2(s) + 2 H2O → Cd(OH)2(s) + Ni(OH)2(s) (overall)
E° = 1.4 V
3. No gases produced, can be sealed.
4. Wristwatches, calculators, cameras, etc.
Chapter 19 Page 14 E. H2/O2 Fuel Cell.
1. Fuel cells = reactants continually fed in during operation, products continually removed.
2. Reaction:
- - H2 + 2 OH (aq) → 2 H2O + 2e (anode)
- - O2 + 2 H2O + 4e → 4 OH (cathode)
2 H2 + O2 → 2 H2O (overall)
3. Used aboard spacecraft.
4. Nonpolluting since product is H2O.
5. Twice efficiency of burning H2 to product heat and then converting it to electricity.
Part Four: Electrolytic Cells
A. Electrolysis of molten NaCl (Downs Cell). (Section 19.9)
1. NaCl melts at 801° C to produce a clear, colorless liquid of freely moving Na+ and Cl- ions. Can conduct.
2. Setup:
Chapter 19 Page 15 3. Used commercially to produce pure Na metal and Cl2 gas.
4. Other Group IA and IIA metals are gathered this way from salts.
B. Electrolysis of Aqueous NaCl Solutions. (Section 19.10)
1. Here, H2(g) is produced at cathode, not Na.
2. Na and other active metals have greater tendency to remain cationic than H+ does.
- - 3. So at cathode: 2 H2O + 2e → 2 OH + H2(g)
- - 4. Anode is still: 2 Cl → Cl2(g) + 2e
5. What’s left behind in the solution? NaOH(aq)
6. This is commercial source of H2(g), Cl2(g) and NaOH(s).
C. Stoichiometry of Electrolysis - Faraday’s Law. (1832) (Section 19.11)
1. Established relationship between amount of electric charge and amount of oxid. and red. substances.
amt. of substance ∝ amt. of electricity oxid. or red. passed through cell
2. 1 faraday = amt. of electric charge that delivers one mole of electrons at cathode. € 3. 1 faraday = charge of one mole of electrons
= 6.022 × 1023 × e Here e = | charge of electron |
1 faraday = 96485 C (Coulombs)
Faraday’s # = F = 96485 C/mol
4. Relation to other electrical units:
Current of 1 ampere (A) = 1 Coulomb/second
Q(C) I(A) = or Q = It t(s)
Chapter 19 Page 16 € 5. A 100 watt bulb uses a current of about 0.8 amps. How many electrons flow through the bulb in 1 minute?
0.8 A × 60 s = 48 Coulombs of charge flow 1 mole e- 48 C × = 4.975 × 10-4 moles of e- 96485 C = 3.0 × 1020 electrons
Q € ne = F charge in Coul moles of e- = OR: Coul Faraday's # ( mol ) € Q = neF
charge€ passed through = moles of e- passed × Faraday’s #
6. Used in electroplating. (electrical deposition)
7. Example: How many grams of silver could be electroplated on the cathode of an
electrolytic cell by the passage of 2.0 Amp of current through AgNO3 solution for 1.0 hour?
Reaction is: Ag+ + 1e- → Ag(s)
total charge Q = I t = 2.0 Amp × 3600 sec = 7200 C
neF = Q Q 7200 C moles of e- n = = = 7.46 × 10-2 mol e C F 96485 mol
1 mole of e- deposits 1 mole Ag
€ So: 7.46 ×€ 10 -2 mol Ag deposited
7.46 × 10-2 mol Ag × 107.868 g/mol = 8.05 grams Ag
neF = It
- where ne = moles of e , I = current in Amps, t = time in seconds
Chapter 19 Page 17 8. Suppose in the previous example we were electroplating copper out of CuSO4(aq) solution. How many grams?
Cu2+(aq) + 2e- → Cu(s)
It 2.0 A × 3600 s n = = = 7.46 × 10-2 mol e- e C F 96485 mol
2e- required for every Cu € 1 mol Cu 63.54 g Cu 7.46 × 10-2 mol e- × × = 2.37 g Cu 2 mol e- mol
9. Notice that number of e- in balanced equation affect amount that can be deposited. Thus electrolysis can be used to determine ox# of a species in solution. €
Example: Iron can exist in solution as Fe2+ or Fe3+. Suppose a 3.00 amp current applied for 1.00 hour produced 2.08 g of Fe(s). What oxidation state is the iron in solution?
Fe+? + ? e- → Fe(s)
neF = It
It 3.00 A × 3600 s n = = = 0.112 mol of e- e C F 96485 mol
2.08 g Fe = 0.0373 mol Fe 55.85 g € mol
0.112 mol e- = 3.00 e-/Fe .0373 mol Fe € Fe3+ in solution
€
Chapter 19 Page 18 10. Summary:
m νF = It M
where m = mass ox. or red. M = molar mass of electrolyzed species € ν = # of e- transferred in balanced eqn. = 1, 2, 3, ...
11. Electroplating Commercially. (p. 805)
• refining of impure metals • jewelry • tableware • chromium bumpers • tin cans (tin or steel)
Chapter 19 Page 19
NOTES:
Chapter 19 Page 20