Redox & Electrochemistry

What is the oxidation state of chromium in the dichromate(VI) ion, Cr₂O₇²⁻?

• A+3
• B+6✓
• C+7
• D+12

Answer: Oxygen is −2, so 7 × (−2) = −14. The overall charge is −2, so the two Cr atoms must total +12, giving each chromium an oxidation state of +6.

In the reaction of acidified manganate(VII) ions with iron(II) ions, MnO₄⁻ is reduced to Mn²⁺. What is the half-equation for this reduction?

• AMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O✓
• BMnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O
• CMnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
• DMnO₄⁻ + 8H⁺ + 3e⁻ → Mn²⁺ + 4H₂O

Answer: Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺, a gain of 5 electrons. Balancing the 4 oxygens with 4H₂O requires 8H⁺ on the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

When acidified potassium dichromate(VI) is used to oxidise ethanol, the orange Cr₂O₇²⁻ is reduced. To which species and oxidation state is the chromium reduced?

• ACrO₄²⁻, +6
• BCr²⁺, +2
• CCr metal, 0
• DCr³⁺, +3✓

Answer: Acidified dichromate(VI) is reduced from orange Cr₂O₇²⁻ (Cr is +6) to the green Cr³⁺ ion (Cr is +3) as it oxidises the alcohol.

Standard electrode potentials: Zn²⁺/Zn = −0.76 V; Cu²⁺/Cu = +0.34 V. For the cell made from these two half-cells, what is the standard cell potential (e.m.f.)?

• A+0.42 V
• B−1.10 V
• C+1.10 V✓
• D+0.34 V

Answer: E°cell = E°(positive/cathode) − E°(negative/anode) = (+0.34) − (−0.76) = +1.10 V. The more positive Cu²⁺/Cu electrode is the cathode.

Using standard electrode potentials Fe³⁺/Fe²⁺ = +0.77 V, I₂/I⁻ = +0.54 V and Br₂/Br⁻ = +1.09 V, which statement about the feasibility of reactions with Fe³⁺ is correct under standard conditions?

• AFe³⁺ oxidises Br⁻ to Br₂ but not I⁻ to I₂
• BFe³⁺ oxidises both I⁻ and Br⁻
• CFe³⁺ oxidises neither I⁻ nor Br⁻
• DFe³⁺ oxidises I⁻ to I₂ but not Br⁻ to Br₂✓

Answer: A reaction is feasible when E°cell is positive. For Fe³⁺ oxidising I⁻: E°cell = 0.77 − 0.54 = +0.23 V (feasible). For Fe³⁺ oxidising Br⁻: 0.77 − 1.09 = −0.32 V (not feasible). So Fe³⁺ oxidises I⁻ but not Br⁻.

Which species is acting as the reducing agent in the reaction: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ ?

• ACl₂
• BBr⁻✓
• CCl⁻
• DBr₂

Answer: Br⁻ is oxidised (Br goes from −1 to 0 in Br₂), so it loses electrons and is therefore the reducing agent. Cl₂ is reduced and acts as the oxidising agent.

In a standard hydrogen electrode used as a reference, which set of conditions is correct?

• AH₂ at 100 kPa, 1.0 mol dm⁻³ OH⁻(aq), 273 K, Cu electrode
• BH₂ at 100 kPa, 1.0 mol dm⁻³ H⁺(aq), 298 K, Pt electrode✓
• CH₂ at 10 kPa, 0.1 mol dm⁻³ H⁺(aq), 298 K, Pt electrode
• DH₂ at 100 kPa, 2.0 mol dm⁻³ H⁺(aq), 373 K, Pt electrode

Answer: The standard hydrogen electrode operates with H₂ gas at 100 kPa, [H⁺] = 1.0 mol dm⁻³, a temperature of 298 K (25 °C), and an inert platinum electrode. It is assigned E° = 0.00 V.

A hydrogen–oxygen fuel cell operating in alkaline conditions has the overall reaction 2H₂ + O₂ → 2H₂O. Which is the correct half-equation occurring at the negative electrode (anode) in alkaline solution?

• AO₂ + 2H₂O + 4e⁻ → 4OH⁻
• B2H⁺ + 2e⁻ → H₂
• C2H₂ + 4OH⁻ → 4H₂O + 4e⁻✓
• D4OH⁻ → O₂ + 2H₂O + 4e⁻

Answer: At the negative electrode hydrogen is oxidised (loses electrons). In alkaline solution this is 2H₂ + 4OH⁻ → 4H₂O + 4e⁻. Option A is the cathode (reduction) reaction.