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 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 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⁻
• B2H₂O + 2e⁻ → H₂ + 2OH⁻
• C2H₂ + 4OH⁻ → 4H₂O + 4e⁻✓
• DCH₃OH + 6OH⁻ → CO₂ + 5H₂O + 6e⁻

Answer: At the negative electrode hydrogen is oxidised (loses electrons). In alkaline solution this is 2H₂ + 4OH⁻ → 4H₂O + 4e⁻. The reduction of oxygen, O₂ + 2H₂O + 4e⁻ → 4OH⁻, is the cathode reaction.

What happens to a species that is oxidised?

• AIt gains electrons and its oxidation state rises
• BIt loses electrons and its oxidation state falls
• CIt loses electrons and its oxidation state rises✓
• DIt gains electrons and its oxidation state falls

Answer: Oxidation is loss of electrons, which makes the oxidation state more positive. 'It gains electrons and its oxidation state falls' describes reduction — OIL RIG remains the quickest way to keep them straight.

What is the oxidation state of manganese in MnO₄⁻?

• A+2
• B+4
• C+7✓
• D−1

Answer: Each oxygen is −2, giving −8 in total, and the whole ion carries −1. So manganese must be +7. That high oxidation state is why manganate(VII) is such a powerful oxidising agent.

In an electrochemical cell, what happens at the negative electrode?

• AReduction, with the more negative electrode potential
• BOxidation, with the more positive electrode potential
• CReduction, with the more positive electrode potential
• DOxidation, with the more negative electrode potential✓

Answer: The half-cell with the more negative potential releases electrons, so oxidation happens there and it becomes the negative terminal. Those electrons flow through the circuit to the positive electrode, where reduction occurs.

What conditions define a standard electrode potential?

• A298 K, 100 kPa and 1.00 mol dm⁻³ solutions✓
• B273 K, 100 kPa and 1.00 mol dm⁻³ solutions
• C298 K, 101 kPa and 0.10 mol dm⁻³ solutions
• D373 K, 100 kPa and 1.00 mol dm⁻³ solutions

Answer: Values are quoted against the standard hydrogen electrode under these conditions. Changing concentration or temperature shifts the potential, which is why the conditions must be stated.

How is the emf of a cell calculated from standard electrode potentials?

• AE of the negative electrode minus E of the positive electrode
• BThe sum of the two standard electrode potentials
• CThe average of the two standard electrode potentials
• DE of the positive electrode minus E of the negative electrode✓

Answer: Subtracting the less positive from the more positive gives a positive emf, which indicates a feasible reaction. A negative value means the reaction as written will not proceed spontaneously.

What is the purpose of the salt bridge in an electrochemical cell?

• AAllowing electrons to pass directly between the two half-cells
• BCompleting the circuit while keeping the solutions separate✓
• CPreventing any ions at all from moving between the two half-cells
• DSupplying additional energy to drive the cell reaction

Answer: Ions migrate through the bridge to balance the charge building in each half-cell, so current keeps flowing. Electrons travel through the external wire, never through the bridge.

Why is a platinum electrode used in a half-cell containing only ions?

• AIt reacts with the ions to generate a potential
• BIt dissolves slowly to release electrons steadily
• CIt is inert and conducts electrons to the solution✓
• DIt catalyses the reaction occurring in the half-cell

Answer: Where both species of a redox couple are in solution, such as Fe²⁺/Fe³⁺, a conducting but unreactive surface is needed for electron transfer. Platinum provides it without taking part.

What limits the usefulness of standard electrode potentials for predicting reactions?

• AThey indicate rate but say nothing about feasibility
• BThey indicate feasibility but say nothing about rate✓
• CThey apply only to reactions involving metals
• DThey apply only at temperatures above 373 K

Answer: A large positive emf shows the reaction is thermodynamically feasible, but a high activation energy can still make it immeasurably slow. Non-standard conditions also shift the actual potentials.

What is a disproportionation reaction?

• ATwo different species are both oxidised
• BTwo different species are both reduced
• COne species is both oxidised and reduced✓
• DOne species is oxidised by a strong oxidising agent

Answer: In disproportionation a single element in one oxidation state ends up in both a higher and a lower one — as when chlorine reacts with cold dilute sodium hydroxide to give chloride and chlorate(I).

E⦵ values: VO₂⁺/VO²⁺ +1.00 V; VO²⁺/V³⁺ +0.34 V; V³⁺/V²⁺ −0.26 V; Sn⁴⁺/Sn²⁺ +0.15 V.

Excess acidified Sn²⁺(aq) is added to a solution of VO₂⁺ ions. What is the final vanadium species?

• AV²⁺
• BVO²⁺
• CVO₂⁺
• DV³⁺✓

Answer: Sn²⁺ can reduce any couple with a more positive E⦵ than +0.15 V. VO₂⁺ → VO²⁺ (+1.00 − 0.15 = +0.85 V) and VO²⁺ → V³⁺ (+0.34 − 0.15 = +0.19 V) are both feasible. V³⁺ → V²⁺ would give −0.26 − 0.15 = −0.41 V, which is not feasible, so the reduction stops at V³⁺. Zinc (E⦵ −0.76 V) would be needed to reach V²⁺.

To standardise a potassium manganate(VII) solution, 0.150 g of sodium ethanedioate, Na₂C₂O₄ (Mr 134.0), was dissolved in acid and titrated hot. The titre was 22.40 cm³. 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O

What is the concentration of the manganate(VII) solution?

• A0.0200 mol dm⁻³✓
• B0.0500 mol dm⁻³
• C0.125 mol dm⁻³
• D2.00 × 10⁻⁵ mol dm⁻³

Answer: C₂O₄²⁻ = 0.150 ÷ 134.0 = 1.119 × 10⁻³ mol. MnO₄⁻ = 1.119 × 10⁻³ × 2/5 = 4.48 × 10⁻⁴ mol. Concentration = 4.48 × 10⁻⁴ ÷ 0.02240 dm³ = 0.0200 mol dm⁻³. Using the ratio upside down (5/2) gives 0.125; ignoring the ratio gives 0.0500.

In acidic solution, iodate(V) ions, IO₃⁻, are reduced to iodine, I₂.

Which is the balanced half-equation for this reduction?

• A2IO₃⁻ + 12H⁺ + 10e⁻ → I₂ + 6H₂O✓
• B2IO₃⁻ + 12H⁺ + 5e⁻ → I₂ + 6H₂O
• CIO₃⁻ + 6H⁺ + 5e⁻ → I₂ + 3H₂O
• D2IO₃⁻ + 6H⁺ + 10e⁻ → I₂ + 3H₂O

Answer: Iodine goes from +5 in IO₃⁻ to 0 in I₂, a gain of 5 electrons per I atom; two I atoms make one I₂, so 10e⁻. The 6 O atoms become 6H₂O, which needs 12H⁺. Charge check: left = −2 + 12 − 10 = 0, right = 0. Using 5e⁻ balances the iodine atoms but leaves the charges unbalanced.

10.0 cm³ of a household bleach was diluted to 250 cm³. A 25.0 cm³ portion was acidified and added to excess potassium iodide, then the iodine formed was titrated with 0.100 mol dm⁻³ sodium thiosulfate; the titre was 18.60 cm³. ClO⁻ + 2I⁻ + 2H⁺ → Cl⁻ + I₂ + H₂O I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

What is the concentration of ClO⁻ in the undiluted bleach?

• A1.86 mol dm⁻³
• B0.0930 mol dm⁻³
• C0.930 mol dm⁻³✓
• D0.0372 mol dm⁻³

Answer: S₂O₃²⁻ = 0.01860 × 0.100 = 1.86 × 10⁻³ mol, so I₂ = 9.30 × 10⁻⁴ mol = ClO⁻ in 25.0 cm³. In 250 cm³: 9.30 × 10⁻³ mol, all of which came from 10.0 cm³ of bleach, so [ClO⁻] = 9.30 × 10⁻³ ÷ 0.0100 = 0.930 mol dm⁻³. Skipping the 2 : 1 thiosulfate ratio gives 1.86; giving the diluted concentration instead gives 0.0372.