FAD1018: Rapid-Fire Drill Pack — Electrochemistry

Objective: Master cell potentials, Nernst equation, Faraday's laws, and electrolysis product prediction through high-volume mechanical repetition.
Target: 2 minutes per problem. If you stall >3 minutes, skip and mark it.
Total problems: 48
Estimated time: 1 hour 40 minutes (96 min)

Cheat Sheet (Memorize First)

Core Equations

Equation What it does
$E°_{cell} = E°_{cathode} - E°_{anode}$ Standard cell potential
$E_{cell} = E°_{cell} - \dfrac{0.0592}{n} \log Q$ Nernst (at 25°C)
$\Delta G° = -nFE°_{cell}$ Gibbs free energy
$E°_{cell} = \dfrac{0.0592}{n} \log K$ Equilibrium constant (25°C)
$m = \dfrac{ItM}{nF}$ Mass from electrolysis
$Q = It$ Charge from current × time
$\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ Arrhenius (for temp-dependent problems)

Constants

$$F = 96{,}485\ \text{C·mol}^{-1} \approx 96{,}500\ \text{C·mol}^{-1}$$
$$R = 8.314\ \text{J·mol}^{-1}\text{·K}^{-1}$$
$0.0592 = \dfrac{RT}{F}\ln 10$ at $T = 298\ \text{K}$

Standard Reduction Potentials (common)

Half-reaction $E°$ (V)
$\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$ +0.80
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$ +0.34
$2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g)$ 0.00
$\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}(s)$ −0.13
$\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}(s)$ −0.25
$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}(s)$ −0.44
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s)$ −0.76
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s)$ −1.66
$\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}(s)$ −2.37
$\text{Na}^+ + e^- \rightarrow \text{Na}(s)$ −2.71
$\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}(s)$ −2.87

Electrochemical Series (Ease of Discharge)

Cations at cathode (hardest → easiest):
K⁺ < Na⁺ < Ca²⁺ < Mg²⁺ < Al³⁺ < Zn²⁺ < Fe²⁺ < Sn²⁺ < Pb²⁺ < H⁺ < Cu²⁺ < Ag⁺

Anions at anode (hardest → easiest):
F⁻ < SO₄²⁻ < NO₃⁻ < Cl⁻ < Br⁻ < I⁻ < OH⁻

Key Mnemonics

  • OIL RIG: Oxidation Is Loss, Reduction Is Gain
  • RED CAT, AN OX: REDuction at CAThode, OXidation at ANode
  • GAL-AN: GALvanic → ANode is Negative
  • QuIM NoF: $m = \frac{Q \cdot M}{n \cdot F} = \frac{I \cdot t \cdot M}{n \cdot F}$

Selective Discharge: Three Factors

  1. E° values (primary — harder to discharge if more negative for cations, more positive for anions)
  2. Concentration (high concentration can override factor 1)
  3. Electrode type (active vs inert)

Overpotential

  • $\text{O}_2$ production at anode: ~0.4–0.6 V overpotential
  • $\text{H}_2$ production at cathode: ~0.1–0.3 V overpotential (on most metals)
  • $\text{Cl}_2$, $\text{Br}_2$, $\text{I}_2$: minimal overpotential on graphite/Pt

Part A: Cell Potential — Identify & Calculate

Target: 90 seconds per problem.

Set A1 — Identify Anode & Cathode (8 problems)

For each pair of half-reactions, identify the cathode, the anode, and calculate $E°_{cell}$.

  1. $\text{Ag}^+/\text{Ag}$ (+0.80 V) and $\text{Zn}^{2+}/\text{Zn}$ (−0.76 V)

  2. $\text{Cu}^{2+}/\text{Cu}$ (+0.34 V) and $\text{Mg}^{2+}/\text{Mg}$ (−2.37 V)

  3. $\text{Al}^{3+}/\text{Al}$ (−1.66 V) and $\text{Fe}^{2+}/\text{Fe}$ (−0.44 V)

  4. $\text{Na}^+/\text{Na}$ (−2.71 V) and $\text{Ni}^{2+}/\text{Ni}$ (−0.25 V)

  5. $\text{Pb}^{2+}/\text{Pb}$ (−0.13 V) and $\text{Ca}^{2+}/\text{Ca}$ (−2.87 V)

  6. $\text{Ag}^+/\text{Ag}$ (+0.80 V) and $\text{Pb}^{2+}/\text{Pb}$ (−0.13 V)

  7. $\text{Cu}^{2+}/\text{Cu}$ (+0.34 V) and $\text{Al}^{3+}/\text{Al}$ (−1.66 V)

  8. $\text{Zn}^{2+}/\text{Zn}$ (−0.76 V) and $\text{Fe}^{2+}/\text{Fe}$ (−0.44 V) — Which is the cathode?

Score: ___/8

Set A2 — Determine Spontaneity (6 problems)

For each calculated $E°_{cell}$, state whether the reaction is spontaneous (galvanic) or non-spontaneous (electrolytic).

  1. $\text{Mg}^{2+}/\text{Mg}$ and $\text{Ag}^+/\text{Ag}$ → $E°_{cell} = ?$ → Spontaneous?

  2. $\text{Zn}^{2+}/\text{Zn}$ and $\text{Fe}^{2+}/\text{Fe}$ → $E°_{cell} = ?$ → Spontaneous?

  3. $\text{Ca}^{2+}/\text{Ca}$ and $\text{Na}^+/\text{Na}$ → $E°_{cell} = ?$ → Spontaneous?

  4. $\text{Ni}^{2+}/\text{Ni}$ and $\text{Ag}^+/\text{Ag}$ → $E°_{cell} = ?$ → Spontaneous?

  5. $\text{Al}^{3+}/\text{Al}$ and $\text{Pb}^{2+}/\text{Pb}$ → $E°_{cell} = ?$ → Spontaneous?

  6. Reverse: If $E°_{cell} = -0.51$ V for $\text{Fe}^{2+}/\text{Fe}$ and $\text{Zn}^{2+}/\text{Zn}$, which way would electrons have to be forced by an external source?

Score: ___/6

Part B: Cell Notation & Half-Reactions

Target: 2 minutes per problem.

Set B1 — Write Cell Notation (5 problems)

Write the standard cell notation (cell diagram) for each galvanic cell.

  1. Zn(s) electrode in Zn²⁺(aq) connected via salt bridge to Cu(s) in Cu²⁺(aq)

  2. Mg(s) in Mg²⁺(aq) connected to Ag(s) in Ag⁺(aq) — inert cathode needed?

  3. A cell using $\text{Fe}^{2+}/\text{Fe}$ (−0.44 V) as anode and $\text{Cu}^{2+}/\text{Cu}$ (+0.34 V) as cathode

  4. A cell where $\text{Al}^{3+}$ is reduced at the cathode and $\text{Mg}$ is oxidised at the anode

  5. The hydrogen electrode ($P_{\text{H}_2} = 1\ \text{atm}$, $[\text{H}^+] = 1\ \text{M}$) connected to a $\text{Zn}^{2+}/\text{Zn}$ half-cell (measuring Zn electrode potential)

Score: ___/5

Set B2 — Write Half-Reactions (5 problems)

Write balanced half-reactions for each electrode process.

  1. Electrolysis of molten $\text{NaCl}$: what happens at cathode? at anode?

  2. Electrolysis of aqueous $\text{CuSO}_4$ with inert Pt electrodes: cathode half-reaction? anode half-reaction?

  3. Electrolysis of aqueous $\text{CuSO}_4$ with copper electrodes: cathode half-reaction? anode half-reaction?

  4. Discharge of a lead-acid battery: $\text{Pb}(s) \rightarrow$ ? at anode; $\text{PbO}_2(s) \rightarrow$ ? at cathode (acidic medium)

  5. Trap: In a galvanic cell with $\text{Zn}/\text{Zn}^{2+}$ and $\text{Cu}/\text{Cu}^{2+}$, what actually happens at each electrode (write the half-equations that occur in practice)?

Score: ___/5

Part C: Nernst Equation

Target: 2.5 minutes per problem.

Set C1 — Direct Nernst Application (6 problems)

Use the Nernst equation at 25°C: $E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q$

  1. For $\text{Zn}(s) \vert \text{Zn}^{2+}(0.10\ \text{M}) \parallel \text{Cu}^{2+}(0.010\ \text{M}) \vert \text{Cu}(s)$, calculate $E_{cell}$.

Hint: The cell reaction is $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, $n=2$, $E°_{cell} = 1.10\ \text{V}$, $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$

  1. For $\text{Zn}(s) \vert \text{Zn}^{2+}(0.010\ \text{M}) \parallel \text{Cu}^{2+}(0.10\ \text{M}) \vert \text{Cu}(s)$, calculate $E_{cell}$.

  2. A galvanic cell: $\text{Pb}(s) \vert \text{Pb}^{2+}(?\ \text{M}) \parallel \text{Ag}^+(0.50\ \text{M}) \vert \text{Ag}(s)$ measures $E_{cell} = 1.00\ \text{V}$. Given $E°_{\text{Pb}^{2+}/\text{Pb}} = -0.13\ \text{V}$, $E°_{\text{Ag}^+/\text{Ag}} = +0.80\ \text{V}$, find $[\text{Pb}^{2+}]$.

  3. For the cell $\text{Fe}(s) \vert \text{Fe}^{2+}(0.050\ \text{M}) \parallel \text{Ni}^{2+}(0.50\ \text{M}) \vert \text{Ni}(s)$, calculate $E_{cell}$.

  4. Concentration cell: $\text{Cu}(s) \vert \text{Cu}^{2+}(0.0010\ \text{M}) \parallel \text{Cu}^{2+}(1.0\ \text{M}) \vert \text{Cu}(s)$. Calculate $E_{cell}$. Which side is the anode?

  5. A concentration cell of $\text{Ag}^+/\text{Ag}$ has $E_{cell} = 0.0592\ \text{V}$ at 25°C. If one half-cell has $[\text{Ag}^+] = 1.0\ \text{M}$, what is $[\text{Ag}^+]$ in the other half-cell?

Score: ___/6

Set C2 — Nernst with pH & Gas Electrodes (4 problems)

  1. For the cell $\text{Pt}(s) \vert \text{H}2(1\ \text{atm}) \vert \text{H}^+(?\ \text{M}) \parallel \text{Cu}^{2+}(1.0\ \text{M}) \vert \text{Cu}(s)$ with $E{cell} = 0.46\ \text{V}$, find the pH of the unknown solution.

Hint: SHE half-cell $E = 0 - 0.0592\log\frac{1}{[\text{H}^+]}$ under non-standard conditions

  1. A hydrogen electrode ($P_{\text{H}2} = 2.0\ \text{atm}$) in a solution of unknown $[\text{H}^+]$ is paired with a standard $\text{Ag}^+/\text{Ag}$ half-cell. $E{cell} = 0.92\ \text{V}$. Find the pH.

  2. Trap: For the cell $\text{Pt} \vert \text{H}_2(1\ \text{atm}) \vert \text{H}^+(0.10\ \text{M}) \parallel \text{H}^+(0.0010\ \text{M}) \vert \text{H}2(1\ \text{atm}) \vert \text{Pt}$, calculate $E{cell}$. Which side does the spontaneous reaction produce $\text{H}_2$?

  3. For Problem 33, what would $E_{cell}$ be if the pressures of $\text{H}_2$ on the two sides were 0.50 atm and 2.0 atm respectively (same $[\text{H}^+]$)?

Score: ___/4

Part D: Faraday's Laws — Quantitative Electrolysis

Target: 2 minutes per problem.

Set D1 — Mass from Current & Time (7 problems)

Use $m = \dfrac{ItM}{nF}$ with $F = 96{,}500\ \text{C·mol}^{-1}$.

  1. Calculate the mass of copper deposited when 2.0 A passes through $\text{CuSO}4(aq)$ for 30 minutes. ($M{\text{Cu}} = 63.5\ \text{g·mol}^{-1}$)

  2. How many grams of silver are deposited when 1.5 A passes through $\text{AgNO}3(aq)$ for 1 hour? ($M{\text{Ag}} = 107.9\ \text{g·mol}^{-1}$)

  3. Trap: Calculate the mass of aluminium produced when 10.0 A passes through molten $\text{Al}_2\text{O}3$ for 2 hours. ($M{\text{Al}} = 27.0\ \text{g·mol}^{-1}$, careful with $n$!)

  4. A current of 3.0 A deposits 1.18 g of nickel from $\text{NiSO}4(aq)$ solution. How long was the current applied? ($M{\text{Ni}} = 58.7\ \text{g·mol}^{-1}$)

  5. Reverse: What current is needed to deposit 5.00 g of zinc in 1 hour from $\text{ZnSO}4(aq)$? ($M{\text{Zn}} = 65.4\ \text{g·mol}^{-1}$)

  6. During the electrolysis of water, $\text{H}_2$ and $\text{O}2$ are produced. If the same current passes for the same time, what is the mass ratio $m{\text{O}2}/m{\text{H}2}$ produced? ($M{\text{O}2} = 32.0$, $M{\text{H}_2} = 2.0$; hint: compare $M/n$ for each)

  7. In the Downs cell, what mass of sodium is produced per hour at 25,000 A? ($M_{\text{Na}} = 23.0\ \text{g·mol}^{-1}$)

Score: ___/7

Set D2 — Gas Volume from Electrolysis (4 problems)

At STP (0°C, 1 atm), 1 mole of gas = 22.4 L.

  1. Electrolysis of dilute $\text{H}_2\text{SO}_4$ with Pt electrodes produces $\text{H}_2$ at the cathode and $\text{O}_2$ at the anode. If 2.0 A flows for 30 minutes, what volume of $\text{H}_2$ (at STP) is produced?

Hint: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$, so $n=2$

  1. For the same setup as Problem 42, what volume of $\text{O}_2$ is produced? Why is it half the volume of $\text{H}_2$?

  2. Electrolysis of aqueous $\text{NaCl}$ (brine) produces $\text{Cl}_2$ at the anode. What volume of $\text{Cl}_2$ (at STP) is produced by 5.0 A in 1 hour?

  3. Combined: During electrolysis of dilute $\text{H}_2\text{SO}_4$, the total gas volume ($\text{H}_2 + \text{O}_2$) collected is 336 mL at STP. What was the total charge passed?

Score: ___/4

Part E: Selective Discharge & Product Prediction

Target: 2 minutes per problem.

Set E1 — Predict Products (8 problems)

For each, predict the products at the cathode and anode. Assume inert electrodes (Pt or graphite) unless stated otherwise.

  1. Electrolysis of molten $\text{PbBr}_2$

  2. Electrolysis of aqueous $\text{NaCl}$ (concentrated brine)

  3. Electrolysis of aqueous $\text{NaCl}$ (very dilute)

  4. Electrolysis of aqueous $\text{CuSO}_4$ with inert Pt electrodes

  5. Electrolysis of aqueous $\text{CuSO}_4$ with copper electrodes

  6. Electrolysis of aqueous $\text{Na}_2\text{SO}_4$ with Pt electrodes

  7. Electrolysis of aqueous $\text{KI}$ with graphite electrodes — consider overpotential

  8. Electrolysis of aqueous $\text{AgNO}_3$ with Pt electrodes

Score: ___/8

Set E2 — Selective Discharge Reasoning (5 problems)

For each, explain why the predicted product forms, referencing the three factors (E°, concentration, electrode type).

  1. In the electrolysis of concentrated aqueous $\text{NaCl}$, why is $\text{Cl}_2$ produced at the anode instead of $\text{O}_2$?

  2. In the electrolysis of dilute aqueous $\text{NaCl}$, the anode product changes. What does it change to and why?

  3. Why is $\text{H}_2$ (not Na metal) produced when aqueous $\text{NaCl}$ is electrolysed?

  4. In the electrolysis of aqueous $\text{CuSO}_4$ with Cu electrodes, the anode loses mass while with Pt electrodes, $\text{O}_2$ is evolved. Explain the difference.

  5. Predict: A solution contains 0.010 M $\text{Cu}^{2+}$ and 1.0 M $\text{Zn}^{2+}$. When electrolysed with Pt electrodes at increasing voltage, which metal deposits first? (Hint: concentration factor)

Score: ___/5

Part F: Thermodynamics & Equilibrium

Target: 2 minutes per problem.

Set F1 — ΔG° & K Calculations (6 problems)

  1. Calculate $\Delta G°$ for the Zn-Cu Daniell cell ($E°_{cell} = 1.10\ \text{V}$, $n=2$).

  2. Calculate the equilibrium constant $K$ at 25°C for the reaction $\text{Zn} + \text{Cu}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cu}$.

  3. For a cell with $E°_{cell} = 0.46\ \text{V}$ and $n=2$, calculate $K$.

  4. Reverse: A reaction has $K = 1.0 \times 10^{15}$ and $n=2$. What is $E°_{cell}$?

  5. A reaction has $\Delta G° = -150\ \text{kJ·mol}^{-1}$ and $n=2$. What is $E°_{cell}$?

  6. Trap: For the reaction $2\text{Ag}^+ + \text{Cu} \rightarrow 2\text{Ag} + \text{Cu}^{2+}$, given $E°_{\text{Ag}^+/\text{Ag}} = +0.80\ \text{V}$ and $E°_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$, calculate $K$. Note: $n=2$ for this reaction.

Score: ___/6

Set F2 — Nernst at Equilibrium (3 problems)

  1. Prove that when $E_{cell} = 0$, the Nernst equation reduces to $E°_{cell} = \frac{0.0592}{n} \log K$.

  2. For a galvanic cell with $E°_{cell} = 0.78\ \text{V}$ ($n=2$), at what value of $Q$ does the cell stop producing current ($E_{cell} = 0$)?

  3. A concentration cell has $E_{cell} = 0$ when both half-cells have equal concentration. If $[\text{Cu}^{2+}]_1 = 1.0\ \text{M}$, what must $[\text{Cu}^{2+}]2$ be for $E{cell} = 0.0296\ \text{V}$?

Score: ___/3

Final Scorecard

Part Sets Problems Raw Score
A — Cell Potential A1, A2 14 ___/14
B — Cell Notation B1, B2 10 ___/10
C — Nernst Equation C1, C2 10 ___/10
D — Faraday's Laws D1, D2 11 ___/11
E — Selective Discharge E1, E2 13 ___/13
F — Thermodynamics F1, F2 9 ___/9
TOTAL 67 ___/67

Proficiency Benchmarks

  • ≤ 47/67 (< 70%) — Needs work. Re-read the Reading Note and re-drill your weakest parts.
  • 48–56/67 (70–84%) — Solid. You understand the concepts. Focus on fixing specific error patterns.
  • 57–63/67 (85–94%) — Strong. Fast and accurate. Review only missed problems.
  • 64–67/67 (≥ 95%) — Exam-ready. Any mistake now is a careless slip.

Speed Benchmarks

  • < 75 min — Excellent mechanical fluency. You're faster than exam pace.
  • 75–100 min — Good. Review the sets that slowed you down.
  • > 100 min — Too slow for exam conditions. Time yourself on specific weak sets tomorrow.

Error Log Template

After grading, list every wrong problem number with a one-word reason:

Problem Reason
e.g. 27 forgot log
e.g. 37 n=3 not 2

Top 3 error patterns (fill after grading): 1. 2. 3.

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A

  1. Cathode: $\text{Ag}^+/\text{Ag}$, Anode: $\text{Zn}^{2+}/\text{Zn}$, $E°_{cell} = 0.80 - (-0.76) = 1.56\ \text{V}$
  2. Cathode: $\text{Cu}^{2+}/\text{Cu}$, Anode: $\text{Mg}^{2+}/\text{Mg}$, $E°_{cell} = 0.34 - (-2.37) = 2.71\ \text{V}$
  3. Cathode: $\text{Fe}^{2+}/\text{Fe}$, Anode: $\text{Al}^{3+}/\text{Al}$, $E°_{cell} = (-0.44) - (-1.66) = 1.22\ \text{V}$
  4. Cathode: $\text{Ni}^{2+}/\text{Ni}$, Anode: $\text{Na}^+/\text{Na}$, $E°_{cell} = (-0.25) - (-2.71) = 2.46\ \text{V}$
  5. Cathode: $\text{Pb}^{2+}/\text{Pb}$, Anode: $\text{Ca}^{2+}/\text{Ca}$, $E°_{cell} = (-0.13) - (-2.87) = 2.74\ \text{V}$
  6. Cathode: $\text{Ag}^+/\text{Ag}$, Anode: $\text{Pb}^{2+}/\text{Pb}$, $E°_{cell} = 0.80 - (-0.13) = 0.93\ \text{V}$
  7. Cathode: $\text{Cu}^{2+}/\text{Cu}$, Anode: $\text{Al}^{3+}/\text{Al}$, $E°_{cell} = 0.34 - (-1.66) = 2.00\ \text{V}$
  8. Cathode: $\text{Fe}^{2+}/\text{Fe}$ (−0.44 > −0.76), Anode: $\text{Zn}^{2+}/\text{Zn}$, $E°_{cell} = (-0.44) - (-0.76) = 0.32\ \text{V}$
  9. $E°_{cell} = 0.80 - (-2.37) = 3.17\ \text{V}$, Spontaneous (galvanic)
  10. $E°_{cell} = (-0.44) - (-0.76) = 0.32\ \text{V}$, Spontaneous (galvanic)
  11. $E°_{cell} = (-2.71) - (-2.87) = 0.16\ \text{V}$, Spontaneous (galvanic) but very small
  12. $E°_{cell} = 0.80 - (-0.25) = 1.05\ \text{V}$, Spontaneous (galvanic)
  13. $E°_{cell} = (-0.13) - (-1.66) = 1.53\ \text{V}$, Spontaneous (galvanic)
  14. Electrons forced from Fe (more negative E°) to Zn → Fe oxidises, Zn²⁺ reduces. External power source needed.

Part B

  1. $\text{Zn}(s) \vert \text{Zn}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \vert \text{Cu}(s)$
  2. $\text{Mg}(s) \vert \text{Mg}^{2+}(aq) \parallel \text{Ag}^+(aq) \vert \text{Ag}(s)$ — Ag is the electrode, no inert needed
  3. $\text{Fe}(s) \vert \text{Fe}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \vert \text{Cu}(s)$
  4. $\text{Mg}(s) \vert \text{Mg}^{2+}(aq) \parallel \text{Al}^{3+}(aq) \vert \text{Al}(s)$
  5. $\text{Zn}(s) \vert \text{Zn}^{2+}(1\ \text{M}) \parallel \text{H}^+(1\ \text{M}) \vert \text{H}_2(1\ \text{atm}) \vert \text{Pt}(s)$
  6. Cathode: $\text{Na}^+ + e^- \rightarrow \text{Na}(l)$, Anode: $2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$
  7. Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$, Anode: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ (or $4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-$)
  8. Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$, Anode: $\text{Cu}(s) \rightarrow \text{Cu}^{2+} + 2e^-$ (active electrode dissolves)
  9. Anode: $\text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2e^-$, Cathode: $\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O}$
  10. Anode: $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$, Cathode: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

Part C

  1. $E_{cell} = 1.10 - \dfrac{0.0592}{2}\log\dfrac{0.10}{0.010} = 1.10 - 0.0296\log 10 = 1.10 - 0.0296 = 1.07\ \text{V}$
  2. $E_{cell} = 1.10 - \dfrac{0.0592}{2}\log\dfrac{0.010}{0.10} = 1.10 - 0.0296\log 0.10 = 1.10 - 0.0296(-1) = 1.13\ \text{V}$
  3. $E°_{cell} = 0.80 - (-0.13) = 0.93\ \text{V}$. Reaction: $\text{Pb} + 2\text{Ag}^+ \rightarrow \text{Pb}^{2+} + 2\text{Ag}$, $n=2$, $Q = \dfrac{[\text{Pb}^{2+}]}{[\text{Ag}^+]^2} = \dfrac{[\text{Pb}^{2+}]}{(0.50)^2}$. $1.00 = 0.93 - \dfrac{0.0592}{2}\log\dfrac{[\text{Pb}^{2+}]}{0.25}$. $\log\dfrac{[\text{Pb}^{2+}]}{0.25} = \dfrac{2(0.93-1.00)}{0.0592} = -2.365$. $\dfrac{[\text{Pb}^{2+}]}{0.25} = 10^{-2.365} = 4.31 \times 10^{-3}$. $[\text{Pb}^{2+}] = 1.08 \times 10^{-3}\ \text{M}$
  4. $E°_{cell} = (-0.25) - (-0.44) = 0.19\ \text{V}$. Reaction: $\text{Fe} + \text{Ni}^{2+} \rightarrow \text{Fe}^{2+} + \text{Ni}$, $Q = \frac{0.050}{0.50} = 0.10$. $E_{cell} = 0.19 - \dfrac{0.0592}{2}\log 0.10 = 0.19 - 0.0296(-1) = 0.22\ \text{V}$
  5. $E°_{cell} = 0$, $E_{cell} = 0 - \frac{0.0592}{2}\log\frac{0.0010}{1.0} = 0 - 0.0296(-3) = 0.0888\ \text{V}$. Anode: the more dilute side (0.0010 M) because Cu oxidises there to increase concentration.
  6. $0.0592 = 0 - 0.0592\log\frac{[\text{Ag}^+]_{\text{dilute}}}{1.0}$ → $\log[\text{Ag}^+] = -1$ → $[\text{Ag}^+] = 0.10\ \text{M}$
  7. SHE: $E_{\text{SHE}} = 0 - 0.0592\log\frac{1}{[\text{H}^+]} = 0 - 0.0592\ \text{pH}$. $E_{cell} = E_{\text{Cu}} - E_{\text{SHE}} = 0.34 - (0 - 0.0592\ \text{pH}) = 0.34 + 0.0592\ \text{pH}$. $0.46 = 0.34 + 0.0592\ \text{pH}$ → $\text{pH} = \frac{0.12}{0.0592} = 2.03$
  8. $E_{cell} = 0.80 - E_{\text{H}2}$. $E{\text{H}2} = 0 - \dfrac{0.0592}{2}\log\dfrac{P{\text{H}_2}}{[\text{H}^+]^2} = 0 - 0.0296\log\dfrac{2.0}{[\text{H}^+]^2}$. $0.92 = 0.80 - (-0.0296\log\frac{2.0}{[\text{H}^+]^2})$. $0.12 = 0.0296(\log 2.0 - 2\log[\text{H}^+])$. $\log 2.0 - 2\log[\text{H}^+] = 4.054$. $0.301 - 2\log[\text{H}^+] = 4.054$. $-2\log[\text{H}^+] = 3.753$. $\log[\text{H}^+] = -1.877$. $\text{pH} = 1.88$
  9. $E_{cell} = 0 - \dfrac{0.0592}{2}\log\dfrac{[\text{H}^+]^2_{\text{cathode}}/P_{\text{H}2,\text{cathode}}}{[\text{H}^+]^2{\text{anode}}/P_{\text{H}2,\text{anode}}}$. With equal $P{\text{H}2}$: $E{cell} = -\dfrac{0.0592}{2}\log\dfrac{(0.0010)^2}{(0.10)^2} = -0.0296\log 10^{-4} = -0.0296(-4) = 0.118\ \text{V}$. $\text{H}_2$ produced at cathode (more concentrated H⁺ side).
  10. $E_{cell} = -\dfrac{0.0592}{2}\log\dfrac{(0.0010)^2/0.50}{(0.10)^2/2.0} = -0.0296\log\dfrac{2\times10^{-6}/0.50}{0.01/2.0} = -0.0296\log\dfrac{4\times10^{-6}}{5\times10^{-3}} = -0.0296\log(8\times10^{-4}) = -0.0296(-3.097) = 0.092\ \text{V}$

Part D

  1. $Q = 2.0 \times 30 \times 60 = 3600\ \text{C}$. $m = \frac{3600 \times 63.5}{2 \times 96500} = 1.18\ \text{g}$
  2. $Q = 1.5 \times 3600 = 5400\ \text{C}$. $n=1$ for $\text{Ag}^+$. $m = \frac{5400 \times 107.9}{1 \times 96500} = 6.04\ \text{g}$
  3. Trap: $n=3$ for $\text{Al}^{3+}$. $Q = 10.0 \times 2 \times 3600 = 72000\ \text{C}$. $m = \frac{72000 \times 27.0}{3 \times 96500} = 6.72\ \text{g}$
  4. $Q = \frac{m \times n \times F}{M} = \frac{1.18 \times 2 \times 96500}{58.7} = 3880\ \text{C}$. $t = \frac{Q}{I} = \frac{3880}{3.0} = 1293\ \text{s} \approx 21.6\ \text{min}$
  5. $Q = \frac{5.00 \times 2 \times 96500}{65.4} = 14750\ \text{C}$. $I = \frac{Q}{t} = \frac{14750}{3600} = 4.10\ \text{A}$
  6. $\text{H}_2$: $n=2$, $M/n = 2.0/2 = 1.0$. $\text{O}_2$: $n=4$, $M/n = 32.0/4 = 8.0$. Ratio = $8.0/1.0 = 8:1$
  7. $Q = 25000 \times 3600 = 9.0 \times 10^7\ \text{C}$. $m = \frac{9.0\times10^7 \times 23.0}{1 \times 96500} = 2.14 \times 10^4\ \text{g} = 21.4\ \text{kg}$
  8. $Q = 2.0 \times 30 \times 60 = 3600\ \text{C}$. Moles e⁻ = $\frac{3600}{96500} = 0.0373\ \text{mol}$. Moles $\text{H}_2 = \frac{0.0373}{2} = 0.01865\ \text{mol}$. Volume = $0.01865 \times 22.4 = 0.418\ \text{L} = 418\ \text{mL}$
  9. $n=4$ for $\text{O}_2$. Moles $\text{O}_2 = \frac{0.0373}{4} = 0.00933\ \text{mol}$. Volume = $0.00933 \times 22.4 = 0.209\ \text{L} = 209\ \text{mL}$. Half the $\text{H}_2$ volume because $\text{H}_2: 2e^-$/molecule, $\text{O}_2: 4e^-$/molecule.
  10. $Q = 5.0 \times 3600 = 18000\ \text{C}$. $n=2$ for $\text{Cl}_2$. Moles $\text{Cl}_2 = \frac{18000}{2 \times 96500} = 0.0933\ \text{mol}$. Volume = $0.0933 \times 22.4 = 2.09\ \text{L}$
  11. $\text{H}_2:\text{O}_2 = 2:1$ by moles. Let moles $\text{H}_2 = 2x$, moles $\text{O}_2 = x$. Total moles = $3x = \frac{0.336}{22.4} = 0.015$. $x = 0.005$. Moles e⁻ = $2 \times \text{moles H}_2 + 4 \times \text{moles O}_2 = 2(0.010) + 4(0.005) = 0.040\ \text{mol}$. $Q = 0.040 \times 96500 = 3860\ \text{C}$

Part E

  1. Cathode: $\text{Pb}(l)$ ($\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$), Anode: $\text{Br}_2(g)$ ($2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$)
  2. Cathode: $\text{H}_2(g)$ ($2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$), Anode: $\text{Cl}_2(g)$ ($2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$)
  3. Cathode: $\text{H}_2(g)$, Anode: $\text{O}_2(g)$ — in very dilute NaCl, $\text{OH}^-$ concentration from water exceeds $\text{Cl}^-$; overpotential can't compensate for the concentration difference.
  4. Cathode: $\text{Cu}(s)$, Anode: $\text{O}_2(g)$ ($2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$)
  5. Cathode: $\text{Cu}(s)$ deposited, Anode: $\text{Cu}(s)$ dissolves ($\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$)
  6. Cathode: $\text{H}_2(g)$, Anode: $\text{O}_2(g)$. $\text{SO}_4^{2-}$ does not discharge; $\text{H}^+$ and $\text{OH}^-$ from water are discharged instead.
  7. Cathode: $\text{H}_2(g)$, Anode: $\text{I}_2(s)$ ($2\text{I}^- \rightarrow \text{I}_2 + 2e^-$). I⁻ is easiest anion to discharge and I₂ overpotential is minimal.
  8. Cathode: $\text{Ag}(s)$ ($\text{Ag}^+ + e^- \rightarrow \text{Ag}$), Anode: $\text{O}_2(g)$ ($2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$). $\text{Ag}^+$ is very easy to reduce (E° = +0.80 V), $\text{NO}_3^-$ does not discharge.
  9. Concentrated NaCl → high [Cl⁻] overrides the electrochemical series, plus $\text{O}_2$ overpotential makes Cl⁻ oxidation more favourable than water oxidation.
  10. In dilute NaCl, [Cl⁻] is much lower. Now [OH⁻] from water autoprotolysis and the concentration factor mean OH⁻ oxidation to O₂ becomes competitive. Overpotential alone cant compensate.
  11. $\text{Na}^+$ ($E° = -2.71\ \text{V}$) is much harder to reduce than $\text{H}^+$ from water ($E° = 0.00\ \text{V}$, or effectively ~−0.41 V at neutral pH). H⁺ is preferentially reduced.
  12. Cu electrodes are active: the Cu anode oxidises ($\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$) because Cu metal is easier to oxidise than water. Pt is inert: water must be oxidised, producing O₂.
  13. $\text{Cu}^{2+}$ deposits first. $E°(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}$ vs $E°(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$. Cu²⁺ is much easier to reduce. However, as Cu²⁺ depletes, its deposition potential shifts per Nernst: $E = 0.34 - \frac{0.0592}{2}\log\frac{1}{0.010} \approx 0.28\ \text{V}$, still far above Zn's. Concentration can only override E° when the potential difference is small.

Part F

  1. $\Delta G° = -nFE°_{cell} = -2 \times 96500 \times 1.10 = -212{,}300\ \text{J·mol}^{-1} = -212\ \text{kJ·mol}^{-1}$
  2. $E°_{cell} = \frac{0.0592}{n} \log K$ → $1.10 = \frac{0.0592}{2} \log K$ → $\log K = 37.16$ → $K = 1.45 \times 10^{37}$
  3. $\log K = \frac{2 \times 0.46}{0.0592} = 15.54$ → $K = 3.47 \times 10^{15}$
  4. $\log K = 15$, $E°_{cell} = \frac{0.0592}{2} \times 15 = 0.444\ \text{V}$
  5. $E°_{cell} = \frac{-\Delta G°}{nF} = \frac{150{,}000}{2 \times 96{,}500} = 0.777\ \text{V}$
  6. $E°_{cell} = 0.80 - 0.34 = 0.46\ \text{V}$. $\log K = \frac{2 \times 0.46}{0.0592} = 15.54$. $K = 3.47 \times 10^{15}$
  7. At $E_{cell} = 0$: $0 = E°_{cell} - \frac{0.0592}{n}\log K$. Rearranging: $E°_{cell} = \frac{0.0592}{n}\log K$. ✓
  8. At $E_{cell} = 0$: $0 = 0.78 - \frac{0.0592}{2}\log Q$ → $\log Q = \frac{2 \times 0.78}{0.0592} = 26.35$ → $Q = 2.24 \times 10^{26}$ (which equals $K$)
  9. $E_{cell} = 0 - \frac{0.0592}{2}\log\frac{[\text{Cu}^{2+}]_2}{1.0}$ → $0.0296 = -0.0296\log[\text{Cu}^{2+}]_2$ → $\log[\text{Cu}^{2+}]_2 = -1$ → $[\text{Cu}^{2+}]_2 = 0.10\ \text{M}$

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