11.4 Electrochemical Cells & Redox Applications

Electrochemical Cells and Redox in Practice

Quick Answer: Galvanic (voltaic) cells convert spontaneous redox reactions into electrical energy; electrolytic cells use external electrical energy to drive nonspontaneous redox. On the PA-CAT, calculate E°cell = E°cathode − E°anode (both written as reductions), use the Nernst equation for concentration effects, and apply Faraday's law for electrolysis quantities.

Chemistry is 16% (~38 items) of the PA-CAT (Bulletin of Information, rev. 20240815). Electrochemistry items are typically Application-level: you must compute a voltage or a mass of metal plated.

Galvanic Cell Anatomy

A Zn/Cu galvanic cell:

• Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻. E°anode (as reduction) = −0.76 V.
• Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s). E°cathode = +0.34 V.
• E°cell = 0.34 − (−0.76) = +1.10 V (spontaneous).
• Salt bridge maintains charge balance; electrons flow anode → cathode through the wire.

Standard Reduction Potentials (selected, 25 °C)

A species with a higher (more positive) E° is a stronger oxidizing agent and will be reduced; a species with a more negative E° is a stronger reducing agent in its oxidized form.

Free Energy and Equilibrium

ΔG° = −nFE°cell. If E°cell = +1.10 V and n = 2, ΔG° = −2 × 96,485 × 1.10 = −212,267 J ≈ −212 kJ/mol. The equilibrium constant follows ΔG° = −RT ln K, so ln K = nFE°/RT. For the Zn/Cu cell at 25 °C, log K = nE°/0.0592 = 2(1.10)/0.0592 = 37.2 → K ≈ 1.6 × 10³⁷.

Nernst Equation — Concentration Matters

E = E° − (RT/nF) ln Q. At 25 °C: E = E° − (0.0592/n) log Q.

Worked: a Zn/Cu cell with [Zn²⁺] = 0.010 M and [Cu²⁺] = 1.0 M. Q = [Zn²⁺]/[Cu²⁺] = 0.010. E = 1.10 − (0.0592/2) log(0.010) = 1.10 − 0.0296 × (−2) = 1.10 + 0.059 = 1.159 V. The cell voltage rises when product ion is dilute or reactant ion is concentrated—Le Châtelier's principle in electrochemical form.

Electrolysis and Faraday's Law

moles = (I × t) / (nF), where I is current (A), t is seconds, n is electrons per ion, F = 96,485 C/mol.

Worked: How many grams of Cu (MW 63.55) plate out when 2.50 A flows through Cu²⁺ solution for 45.0 min?

Total charge Q = I × t = 2.50 × 45.0 × 60 = 6,750 C. Moles e⁻ = 6,750 / 96,485 = 0.06996 mol e⁻. Cu²⁻ + 2e⁻ → Cu, so moles Cu = 0.06996 / 2 = 0.03498 mol. Mass = 0.03498 × 63.55 = 2.22 g Cu.

Applications

• Lead-acid battery: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O; E° ≈ 2.04 V per cell; rechargeable because products adhere to plates.
• Lithium-ion battery: Li⁺ intercalates between graphite anode and metal-oxide cathode; high energy density from Li's very negative E° (−3.04 V).
• Corrosion of iron: Anode: Fe → Fe²⁺ + 2e⁻; Cathode: O2 + 2H2O + 4e⁻ → 4OH⁻. Water and electrolyte catalyze the cell. Sacrificial anodes (Zn, Mg) oxidize preferentially because their E° is more negative than Fe's.
• Redox titration: MnO4⁻ (purple) + 5Fe²⁺ + 8H⁺ → Mn²⁻ (colorless) + 5Fe³⁺ + 4H2O. The first persistent pink signals equivalence; moles Fe²⁺ = 5 × moles MnO4⁻ added.

The Nernst Equation and Concentration Cells

Cell potential depends on concentration as well as standard potentials. The Nernst equation at 25 °C is Ecell = E°cell − (0.0592 / n) log Q, where n is the electrons transferred and Q is the reaction quotient. When all species are at unit activity, Q = 1, log Q = 0, and Ecell = E°cell. A concentration cell uses two half-cells of the same chemistry at different concentrations; E°cell is 0 but Ecell is not, because Q ≠ 1, driving spontaneous dilution.

Worked example: Consider a Cu | Cu²⁺ (0.10 M) || Cu²⁺ (1.0 M) | Cu cell. E°cell = 0, n = 2, and Q = 0.10 / 1.0 = 0.10. Ecell = 0 − (0.0592 / 2) × log(0.10) = −0.0296 × (−1) = +0.0296 V, spontaneous, driving the two Cu²⁺ concentrations toward equality. The PA-CAT may ask which direction electrons flow or require an Ecell calculation from given concentrations — remember the more dilute half-cell is the anode, because oxidation there raises its ion concentration.