Electrochemistry (2)

Electrochemistry (2)

🟢 Lite — Quick Review (1h–1d)

Electrochemistry studies how electrical energy and chemical energy interconvert in electrochemical cells. Two cell types exist: a galvanic (voltaic) cell converts chemical energy to electrical energy via a spontaneous redox reaction, while an electrolytic cell uses electrical energy to drive a nonspontaneous reaction.

Key formulas to memorise:

• Ecell = Ecathode − Eanode (cell potential = difference between reduction potentials)
• E°cell = E°reduction + E°oxidation
• ΔG° = −nFE°cell
• Nernst equation: E = E° − (RT/nF)ln Q
• Faraday’s law: m = (Q × M) / (n × F) where m = mass deposited, Q = charge in coulombs, M = molar mass, n = electrons per mole, F = 96,485 C/mol
• Q = I × t

Exam tips: In galvanic cells the anode is negative (oxidation) and the cathode is positive (reduction) — the signs flip in electrolytic cells. E° values are measured under standard conditions (1 M, 1 atm, 25 °C). E° > 0 indicates a spontaneous reaction. Paper 2 at Makerere (3 % weightage) frequently tests Nernst equation calculations and identification of half-reactions at each electrode. Never multiply E° by stoichiometric coefficients when adding half-reactions — but always multiply ΔG° values by n.

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🟡 Standard — Regular Study (2d–2mo)

Galvanic Cell — The Daniell Cell

A classic example is the Daniell cell: a zinc electrode dipped in ZnSO₄ solution (anode, oxidation: Zn → Zn²⁺ + 2e⁻) joined by a salt bridge to a copper electrode dipped in CuSO₄ solution (cathode, reduction: Cu²⁺ + 2e⁻ → Cu). Electrons flow through the external wire from Zn to Cu. The salt bridge (usually KCl or KNO₃ in a gel) maintains electrical neutrality by allowing counter-diffusion of ions.

Calculating cell potential under standard conditions:

Zn²⁺ + 2e⁻ → Zn    E° = −0.76 V
Cu²⁺ + 2e⁻ → Cu    E° = +0.34 V

E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V

Since E°cell > 0, the reaction is spontaneous. For a non-standard concentration, apply the Nernst equation. With [Zn²⁺] = 0.1 M and [Cu²⁺] = 1.0 M:

Q = [Zn²⁺] / [Cu²⁺] = 0.1 / 1.0 = 0.1
n = 2,  F = 96485 C/mol,  RT/F ≈ 0.0257 V at 298 K
E = 1.10 − (0.0257/2) ln(0.1) = 1.10 + 0.0296 ≈ 1.13 V

Concentration shifts raise E when Q < 1.

Electrolytic Cells

In electrolysis, an external power source forces oxidation at the positive anode and reduction at the negative cathode. The electrolysis of aqueous NaCl produces Cl₂ at the anode (2Cl⁻ → Cl₂ + 2e⁻) and H₂ at the cathode (2H₂O + 2e⁻ → H₂ + 2OH⁻) because their reduction potentials favour these outcomes over Na⁺ and NaOH formation. This distinction between theoretical prediction and actual discharge at electrodes is a common exam trap. Ion concentration, electrode material, and overpotential all influence which species actually discharges.

Faraday’s Laws in Electrolysis

Law 1: Mass depositied (m) is proportional to charge passed: m ∝ Q. Law 2: For the same Q, m is proportional to molar mass M divided by n (valence change): m = (Q × M) / (n × F).

Worked numeric: Calculate the mass of copper deposited when 2.0 A flows through CuSO₄ for 30 minutes.

• Q = I × t = 2.0 × 1800 = 3600 C
• Cu²⁺ + 2e⁻ → Cu, so n = 2, M = 63.5 g/mol
• m = (3600 × 63.5) / (2 × 96485) = 228600 / 192970 ≈ 1.18 g

Common mistake to avoid: Confusing the anode and cathode polarities. In galvanic cells, E° values themselves determine cathode/anode assignment — but in electrolytic cells the external power source imposes the opposite polarity.

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🔴 Extended — Deep Study (3mo+)

Connecting Electrochemistry to Thermodynamics

The link between cell potential and Gibbs free energy is ΔG = −nFEcell. At standard conditions: ΔG° = −nFE°cell. Because E° is an intensive property (does not scale with reaction quantity), it directly predicts spontaneity without needing the balanced equation coefficients. This contrasts with ΔG°, which is extensive.

The standard cell potential also connects to the equilibrium constant via:

nFE°cell = RT ln K    or    log K = nE°cell / 0.0592 V (at 298 K)

For the Daniell cell (E° = 1.10 V, n = 2): log K = 2(1.10)/0.0592 ≈ 37.2, giving K ≈ 1.6 × 10³⁷. An astronomically large K confirms the reaction proceeds essentially to completion.

Concentration Cells and pH Determination

A concentration cell has identical electrodes but different ion concentrations. The EMF depends only on the concentration ratio via:

E = (0.0592/n) log ([anode]/[cathode])  (at 298 K)

This principle underlies pH meters: a hydrogen electrode immersed in solutions of different [H⁺] generates a measurable potential proportional to the pH difference. The glass electrode (a hydrogen ion-selective membrane) exploits the same Nernstian response: E = E° − (0.0592) pH.

Concentration Polarisation and Overpotential

In real electrolytic cells, overpotential (η) is the extra voltage beyond the theoretical value required to sustain a given current density. It arises from activation energy barriers at the electrode–electrolyte interface and concentration gradients near the electrodes. The actual decomposition voltage is Edecomp = E°cell + ηanode + ηcathode. This is particularly significant in industrial electrolyses (chloro-alkali process, aluminium extraction via Hall-Héroult) and explains why the discharge of competing ions (e.g., H₂ vs. Na⁺ during NaCl electrolysis) does not always follow reduction-potential rankings alone.

Common mistakes examined at Makerere:

1. Sign reversal on Q: In the Nernst equation, Q uses the reaction quotient with products/ reactants in the same order as the cell reaction — inserting Q inverted flips the result.
2. Using E° for concentration-dependent predictions: E° applies only at 1 M; once concentrations change, recalculate with Nernst — never use E° as a shortcut for non-standard conditions.
3. Assuming E° = 0 for identical half-cells: This is only true for concentration cells under the same conditions; even small concentration differences generate measurable potentials.

Practice Prompts

1. A galvanic cell uses the reaction: Fe²⁺(aq) + Ce⁴⁺(aq) → Fe³⁺(aq) + Ce³⁺(aq). Given E°(Ce⁴⁺/Ce³⁺) = +1.44 V and E°(Fe³⁺/Fe²⁺) = +0.77 V, calculate E°cell and determine whether the reaction is spontaneous under standard conditions.
2. During electrolysis of 1.0 M CuSO₄ using a platinum cathode, a current of 0.50 A flows for 1 hour. Calculate the mass of copper deposited and a the volume of O₂ gas liberated at the anode (at STP: 1 mol gas ≈ 22.4 L). The anode reaction is 2H₂O → O₂ + 4H⁺ + 4e⁻.

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