Level 2 — RecallRedox & Electrochemistry (Intro)

Redox & Electrochemistry (Intro)

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Difficulty: Level 2 — Recall & standard problems Time limit: 30 minutes Total marks: 40

Use $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , $F = 96500\ \text{C mol}^{-1}$ , $T = 298\ \text{K}$ unless stated. At 298 K, $\tfrac{RT}{F}\ln = \tfrac{0.0591}{1}\log$ .

Q1. Define the following, giving the direction of electron flow where relevant: (a) anode, (b) cathode, in a galvanic cell. (3 marks)

Q2. State the Nernst equation for a general cell reaction and write its simplified form at 298 K using log base 10. (3 marks)

Q3. For the cell $\text{Zn} \mid \text{Zn}^{2+} \parallel \text{Cu}^{2+} \mid \text{Cu}$ , given $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$ and $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ : (a) Identify anode and cathode. (b) Calculate $E^\circ_{\text{cell}}$ . (4 marks)

Q4. For the cell in Q3, calculate $\Delta G^\circ$ in kJ (with $n = 2$ ), and state whether the reaction is spontaneous. (4 marks)

Q5. Why is the Standard Hydrogen Electrode (SHE) assigned a potential of exactly $0.00\ \text{V}$ , and what are its standard conditions? (3 marks)

Q6. For the reaction in Q3 ( $n=2$ , $E^\circ_{\text{cell}} = 1.10\ \text{V}$ ), calculate the equilibrium constant $K$ at 298 K using $\log K = \dfrac{nE^\circ}{0.0591}$ . (4 marks)

Q7. A concentration cell is built as $\text{Cu} \mid \text{Cu}^{2+}(0.01\ \text{M}) \parallel \text{Cu}^{2+}(0.10\ \text{M}) \mid \text{Cu}$ . (a) State $E^\circ_{\text{cell}}$ . (b) Calculate $E_{\text{cell}}$ ( $n=2$ ). (4 marks)

Q8. A current of $2\ \text{A}$ is passed through molten $\text{AlCl}_3$ for $30$ minutes. Calculate the mass of aluminium deposited. ( $M_{\text{Al}} = 27\ \text{g mol}^{-1}$ , $n=3$ ). (5 marks)

Q9. Explain the electrochemical mechanism of iron rusting, and state how cathodic protection prevents corrosion. (5 marks)

Q10. Distinguish between a primary and a secondary cell, giving one example of each. (3 marks)

END OF PAPER

Answer keyMark scheme & solutions

Q1. (3 marks)

Anode: electrode where oxidation occurs (loss of electrons); it is the negative terminal in a galvanic cell. (1.5)
Cathode: electrode where reduction occurs (gain of electrons); it is the positive terminal. (1)
Electrons flow from anode → cathode through the external wire. (0.5) Why: Oxidation releases electrons which must leave at the anode and travel to the cathode where a species accepts them.

Q2. (3 marks) $E = E^\circ - \frac{RT}{nF}\ln Q \quad \text{(2)}$ At 298 K: $E = E^\circ - \frac{0.0591}{n}\log Q \quad \text{(1)}$ Why: Substituting $R$ , $F$ , $T=298$ and converting $\ln$ → $2.303\log$ gives the numeric coefficient $0.0591$ .

Q3. (4 marks)

Anode = Zn (lower/more negative $E^\circ$ , oxidised); Cathode = Cu. (2)
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76)$ (1)
$= \mathbf{1.10\ V}$ (1)

Q4. (4 marks) $\Delta G^\circ = -nFE^\circ_{\text{cell}} = -(2)(96500)(1.10)$ (2) $= -212300\ \text{J} = \mathbf{-212.3\ kJ}$ (1) Since $\Delta G^\circ < 0$ (and $E^\circ > 0$ ), the reaction is spontaneous. (1)

Q5. (3 marks)

SHE is the arbitrary reference electrode assigned $0.00\ \text{V}$ so all other potentials can be measured relative to it. (1)
Standard conditions: $\text{H}_2$ gas at 1 bar (1 atm) pressure, $\text{H}^+$ at 1 M activity, 298 K, over a platinised Pt electrode. (2)

Q6. (4 marks) $\log K = \frac{nE^\circ}{0.0591} = \frac{2 \times 1.10}{0.0591} = 37.22$ (2) $K = 10^{37.22} \approx \mathbf{1.7 \times 10^{37}}$ (2) Why: Large $K$ confirms the reaction goes essentially to completion (consistent with strong spontaneity).

Q7. (4 marks)

(a) $E^\circ_{\text{cell}} = \mathbf{0\ V}$ (same electrode/species both sides). (1)
(b) Reaction moves ions from high→low concentration; cathode = higher conc. $E = -\frac{0.0591}{2}\log\frac{[\text{dilute}]}{[\text{conc}]} = -\frac{0.0591}{2}\log\frac{0.01}{0.10}$ (1.5) $= -\frac{0.0591}{2}(-1) = +0.02955 \approx \mathbf{0.0296\ V}$ (1.5)

Q8. (5 marks) $Q = It = 2 \times (30\times 60) = 3600\ \text{C}$ (1.5) $m = \frac{QM}{nF} = \frac{3600 \times 27}{3 \times 96500}$ (2) $= \frac{97200}{289500} = \mathbf{0.3358\ g}$ (1.5) Why: Faraday's law $m = \dfrac{ZIt}{1} = \dfrac{ItM}{nF}$ ; $\text{Al}^{3+}+3e^-\to\text{Al}$ needs 3 mol e⁻ per mol Al.

Q9. (5 marks) Rusting mechanism (electrochemical):

Anodic region: $\text{Fe} \to \text{Fe}^{2+} + 2e^-$ (oxidation). (1)
Cathodic region: $O_2 + 2H_2O + 4e^- \to 4OH^-$ (reduction, in presence of water & O₂). (1)
$\text{Fe}^{2+}$ is further oxidised to $\text{Fe}^{3+}$ ; combines with $O_2/H_2O$ to form hydrated ferric oxide rust ( $\text{Fe}_2O_3\cdot x\text{H}_2O$ ). (1)
Cathodic protection: connect iron to a more easily oxidised (more active) metal, e.g. Mg/Zn (sacrificial anode); this metal becomes the anode and corrodes instead, forcing iron to act as cathode so it does not oxidise. (2)

Q10. (3 marks)

Primary cell: non-rechargeable; reaction irreversible. Example: dry (Leclanché) cell. (1.5)
Secondary cell: rechargeable; reaction reversible by passing current. Example: lead-acid battery or Li-ion battery. (1.5)

[
  {"claim":"E_cell = 0.34 - (-0.76) = 1.10 V","code":"result = (0.34-(-0.76)) == 1.10"},
  {"claim":"dG = -nFE = -212.3 kJ","code":"dG=-2*96500*1.10/1000; result = abs(dG+212.3)<0.5"},
  {"claim":"log K = 2*1.10/0.0591 ~ 37.22","code":"lk=2*1.10/0.0591; result = abs(lk-37.22)<0.1"},
  {"claim":"Al mass = 3600*27/(3*96500) ~ 0.3358 g","code":"m=Rational(3600*27,3*96500); result = abs(float(m)-0.3358)<0.001"},
  {"claim":"Conc cell E = -(0.0591/2)*log10(0.01/0.10) ~ 0.02955 V","code":"import math; E=-(0.0591/2)*math.log10(0.01/0.10); result = abs(E-0.02955)<0.0005"}
]