Level 5 — MasteryRedox & Electrochemistry (Intro)

Redox & Electrochemistry (Intro)

⏱ 75 minutes60 marksprintable — key stays hidden on paper

Level 5 — Mastery (cross-domain: chemistry + math + physics + coding) Time limit: 75 minutes Total marks: 60

Constants: $F = 96485\ \text{C mol}^{-1}$ , $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T = 298.15\ \text{K}$ unless stated. Use $\ln$ where indicated; $2.303RT/F = 0.05916\ \text{V}$ at $298.15$ K.

Question 1 — The Daniell cell, thermodynamics & equilibrium (22 marks)

Consider the galvanic cell

$\text{Zn(s)} \mid \text{Zn}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu(s)}$

with $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$ and $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$ .

(a) Identify the anode and cathode, write the two half-reactions and the balanced overall cell reaction. State $n$ . (4)

(b) Compute $E^\circ_{\text{cell}}$ , then $\Delta G^\circ$ (kJ mol⁻¹). Comment on spontaneity. (4)

(c) Derive the relation $\ln K = \dfrac{nFE^\circ_{\text{cell}}}{RT}$ starting from $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ and $\Delta G^\circ = -RT\ln K$ . Evaluate $K$ numerically. (5)

(d) Using the Nernst equation, find $E_{\text{cell}}$ when $[\text{Zn}^{2+}] = 0.100\ \text{M}$ and $[\text{Cu}^{2+}] = 0.00100\ \text{M}$ . (4)

(e) Coding. Write a Python function cell_emf(Zn, Cu) returning $E_{\text{cell}}$ in volts (using the Nernst equation at 298.15 K). Then state, with justification, the concentration ratio $[\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ at which the cell reaches equilibrium ( $E_{\text{cell}} = 0$ ). (5)

Question 2 — Concentration cell + electrolysis coupling (20 marks)

Part A — Concentration cell. A cell is built from two copper electrodes in $\text{Cu}^{2+}$ solutions of concentrations $c_1$ (dilute) and $c_2$ (concentrated).

(a) Explain qualitatively why an EMF arises and identify which electrode (dilute or concentrated) is the cathode. (3)

(b) Derive the EMF expression and compute $E_{\text{cell}}$ for $c_1 = 0.0100\ \text{M}$ , $c_2 = 1.00\ \text{M}$ . (4)

Part B — Electrolysis (Faraday). The concentrated half-cell above is now electrolysed between inert electrodes. A current of $1.50\ \text{A}$ is passed for $45.0\ \text{min}$ .

(c) State Faraday's laws and compute the mass of Cu deposited at the cathode. ( $M_{\text{Cu}} = 63.55\ \text{g mol}^{-1}$ .) (5)

(d) Compute the volume of $\text{O}_2$ gas (at STP, $22.4\ \text{L mol}^{-1}$ ) liberated at the anode over the same time, assuming water oxidation $2\text{H}_2\text{O} \to \text{O}_2 + 4\text{H}^+ + 4e^-$ . (4)

(e) Cross-domain. If instead this charge is used to charge a lead-acid cell segment, estimate the electrical energy (in J) delivered assuming a constant terminal voltage of $2.05\ \text{V}$ . (4)

Question 3 — Fuel cell, corrosion & argument (18 marks)

(a) For the H₂/O₂ fuel cell in alkaline medium, write both electrode reactions and the overall reaction. Given $E^\circ_{\text{cell}} = 1.23\ \text{V}$ and $n = 4$ per O₂, compute $\Delta G^\circ$ per mole of O₂ (kJ). (5)

(b) Explain the spacecraft relevance of the H₂/O₂ fuel cell in two distinct points beyond "it produces electricity." (3)

(c) Prove/reason. Iron ( $E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\ \text{V}$ ) corrodes. Using electrode potentials, prove that attaching zinc ( $E^\circ = -0.76\ \text{V}$ ) provides cathodic protection whereas attaching tin ( $E^\circ(\text{Sn}^{2+}/\text{Sn}) = -0.14\ \text{V}$ ) would accelerate corrosion of iron once the coating is scratched. (6)

(d) Distinguish galvanization from cathodic protection by impressed current in one functional sentence each. (4)

Answer keyMark scheme & solutions

Question 1

(a) (4)

Anode (oxidation): $\text{Zn} \to \text{Zn}^{2+} + 2e^-$ (1)
Cathode (reduction): $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ (1)
Overall: $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$ (1)
$n = 2$ (1) Why: Zn is more easily oxidised (more negative $E^\circ$ ) so it is the negative anode.

(b) (4)

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\ \text{V}$ (2)
$\Delta G^\circ = -nFE^\circ = -(2)(96485)(1.10) = -212{,}267\ \text{J} \approx -212.3\ \text{kJ mol}^{-1}$ (1)
$\Delta G^\circ < 0 \Rightarrow$ spontaneous. (1)

(c) (5)

Set $\Delta G^\circ = -nFE^\circ$ equal to $\Delta G^\circ = -RT\ln K$ : $-nFE^\circ = -RT\ln K$ (2)
$\Rightarrow \ln K = \dfrac{nFE^\circ}{RT}$ (1)
Numerics: $\ln K = \dfrac{2 \times 96485 \times 1.10}{8.314 \times 298.15} = \dfrac{212267}{2478.8} = 85.63$ (1)
$K = e^{85.63} \approx 1.5\times10^{37}$ (1) Why: Large $K$ confirms reaction goes essentially to completion.

(d) (4)

$Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.100}{0.00100} = 100$ (1)
Nernst: $E = E^\circ - \dfrac{0.05916}{n}\log Q = 1.10 - \dfrac{0.05916}{2}\log(100)$ (2)
$= 1.10 - 0.02958\times 2 = 1.10 - 0.05916 = 1.041\ \text{V}$ (1)

(e) (5)

import math
def cell_emf(Zn, Cu):
    E0 = 1.10
    n = 2
    Q = Zn / Cu
    return E0 - (0.05916/n)*math.log10(Q)

(3 for correct function: E°, Q, Nernst term)

At equilibrium $E=0$ : $E^\circ = \dfrac{0.05916}{2}\log Q \Rightarrow \log Q = \dfrac{2\times1.10}{0.05916}=37.19$ (1)
$[\text{Zn}^{2+}]/[\text{Cu}^{2+}] = 10^{37.19} \approx 1.5\times10^{37} = K$ (1) (equals $K$ , as expected).

Question 2

(a) (3)

Both electrodes identical; EMF arises from difference in $\text{Cu}^{2+}$ activity (reduction more favourable where ions concentrated). (2)
Cathode = concentrated ( $c_2$ ) side; reduction proceeds there to lower concentration. (1)

(b) (4)

$E_{\text{cell}} = \dfrac{0.05916}{n}\log\dfrac{c_2}{c_1}$ , $n=2$ (2)
$= \dfrac{0.05916}{2}\log\dfrac{1.00}{0.0100} = 0.02958\times2 = 0.05916\ \text{V} \approx 0.0592\ \text{V}$ (2)

(c) (5)

Faraday: (1st) mass $\propto$ charge; (2nd) equal charge deposits masses $\propto$ equivalent mass. (1)
$Q = It = 1.50 \times (45.0\times60) = 1.50\times2700 = 4050\ \text{C}$ (1)
moles $e^- = 4050/96485 = 0.04197$ mol (1)
$\text{Cu}^{2+}+2e^-\to\text{Cu}$ : moles Cu $= 0.04197/2 = 0.02099$ mol (1)
mass $= 0.02099\times63.55 = 1.334\ \text{g}$ (1)

(d) (4)

$4e^-$ per O₂: moles O₂ $= 0.04197/4 = 0.010492$ mol (2)
Volume $= 0.010492\times22.4 = 0.2350\ \text{L} \approx 235\ \text{mL}$ (2)

(e) (4)

Energy $= Q \times V = 4050 \times 2.05 = 8302.5\ \text{J} \approx 8.30\ \text{kJ}$ (3)
(Charge–energy link $W=QV$ ; ideal, ignoring losses.) (1)

Question 3

(a) (5)

Anode: $\text{H}_2 + 2\text{OH}^- \to 2\text{H}_2\text{O} + 2e^-$ (1)
Cathode: $\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \to 4\text{OH}^-$ (1)
Overall: $2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$ (1)
$\Delta G^\circ = -nFE^\circ = -(4)(96485)(1.23) = -474{,}706\ \text{J} \approx -474.7\ \text{kJ}$ per mol O₂ (2)

(b) (3) — any two:

Product water is potable/usable by crew. (1.5)
High energy density, no moving parts, quiet, high efficiency vs combustion. (1.5)
Continuous operation from stored H₂/O₂ without recharging. (1.5) (max 3)

(c) (6)

Protection requires the sacrificial metal to be more easily oxidised (more negative $E^\circ$ ) so it becomes the anode and Fe becomes cathode (protected). (2)
Zn: $-0.76 < -0.44$ ⇒ Zn oxidises preferentially, Fe protected (cathodic protection). (2)
Sn: $-0.14 > -0.44$ ⇒ Fe is more easily oxidised than Sn; when coating scratched Fe becomes anode and corrodes faster (galvanic acceleration). (2)

(d) (4)

Galvanization: coating iron with zinc, which both physically blocks and sacrificially corrodes to protect the iron. (2)
Impressed-current cathodic protection: an external DC source forces the metal to be the cathode, driving electrons in to suppress its oxidation. (2)

[
  {"claim":"E°cell Daniell = 1.10 V", "code":"Ec=0.34-(-0.76); result = abs(Ec-1.10)<1e-9"},
  {"claim":"ΔG° Daniell ≈ -212.27 kJ/mol", "code":"dG=-2*96485*1.10/1000; result = abs(dG-(-212.267))<0.01"},
  {"claim":"ln K Daniell ≈ 85.63", "code":"import sympy as sp; lnK=2*96485*1.10/(8.314*298.15); result = abs(lnK-85.63)<0.05"},
  {"claim":"E_cell at Q=100 is 1.041 V", "code":"import sympy as sp; E=1.10-(0.05916/2)*sp.log(100,10); result = abs(float(E)-1.041)<0.001"},
  {"claim":"Mass Cu deposited = 1.334 g", "code":"Q=1.5*2700; m=(Q/96485)/2*63.55; result = abs(m-1.334)<0.005"},
  {"claim":"O2 volume = 0.235 L", "code":"Q=1.5*2700; V=(Q/96485)/4*22.4; result = abs(V-0.235)<0.002"},
  {"claim":"Energy = 8302.5 J", "code":"W=1.5*2700*2.05; result = abs(W-8302.5)<0.1"}
]