R3.2 Exam Practice

A student places strips of four metals (A–D) into solutions of the other metals' ions. A tick (✓) indicates a reaction occurred:

(a) Deduce the order of reactivity from most to least reactive. [1]

(b) Write the ionic equation for the reaction between metal A and B²⁺(aq). [1]

(c) Identify which species is oxidised and which is reduced in this reaction. [1]

(d) Explain why metal D does not react with any of the other metal ion solutions. [2]

Show Mark Scheme

(a) A > B > C > D (most to least reactive) [1]

(b) A(s) + B²⁺(aq) → A²⁺(aq) + B(s) [1]

(c) A is oxidised (loses electrons); B²⁺ is reduced (gains electrons) [1]

(d) D is the least reactive metal [1]; it cannot displace any of the other metals from their ions because it is a weaker reducing agent / has a less negative electrode potential [1]

Examiner tip: A more reactive metal displaces a less reactive metal from solution. The more reactive metal is always the one being oxidised (acting as the reducing agent).

A voltaic cell is constructed using a zinc electrode in Zn²⁺(aq) and a copper electrode in Cu²⁺(aq).
E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.

(a) Calculate the standard cell potential, E°_cell. [1]

(b) Identify the anode and cathode. [1]

(c) State the direction of electron flow in the external circuit. [1]

(d) State the function of the salt bridge. [1]

Show Mark Scheme

(a) E°_cell = E°_cathode − E°_anode = (+0.34) − (−0.76) = +1.10 V [1]

(b) Anode: zinc electrode; Cathode: copper electrode [1]

(c) Electrons flow from zinc (anode) to copper (cathode) through the external circuit [1]

(d) The salt bridge completes the circuit by allowing ions to flow between the two half-cells, maintaining electrical neutrality [1]

Examiner tip: E°_cell = E°_cathode − E°_anode. The more negative electrode is always the anode in a voltaic cell. Oxidation occurs at the anode (AN OX), reduction at the cathode (RED CAT).

(a) State the oxidation state of the indicated element in each species: (i) Cr in Cr₂O₇²⁻, (ii) N in NO₃⁻, (iii) S in H₂SO₃. [3]

(b) In the reaction 2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺, identify the oxidising agent and the reducing agent. [2]

Show Mark Scheme

(a)(i) Cr in Cr₂O₇²⁻: +6 [1]

(a)(ii) N in NO₃⁻: +5 [1]

(a)(iii) S in H₂SO₃: +4 [1]

(b) Oxidising agent: Fe³⁺ (it is reduced, gaining an electron) [1]; Reducing agent: Sn²⁺ (it is oxidised, losing electrons) [1]

Examiner tip: The oxidising agent IS the species that gets REDUCED (and vice versa). This is the most commonly confused concept in redox chemistry.

Use the data booklet values: E°(Ag⁺/Ag) = +0.80 V, E°(Ni²⁺/Ni) = −0.26 V, E°(Fe³⁺/Fe²⁺) = +0.77 V.

(a) Predict whether Ni will reduce Ag⁺ ions. Write the overall equation if the reaction is spontaneous. [2]

(b) Calculate the E°_cell for a cell using Ni and Ag electrodes. [1]

(c) Predict whether Fe²⁺ can reduce Ag⁺ to Ag. Justify your answer. [2]

Show Mark Scheme

(a) Yes, Ni is more reactive (more negative E°), so it will reduce Ag⁺ [1]; Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s) [1]

(b) E°_cell = +0.80 − (−0.26) = +1.06 V [1]

(c) Yes. E°(Ag⁺/Ag) = +0.80 V > E°(Fe³⁺/Fe²⁺) = +0.77 V [1]; so Ag⁺ is a stronger oxidising agent than Fe³⁺ and can oxidise Fe²⁺ to Fe³⁺ (E°_cell = +0.80 − 0.77 = +0.03 V > 0) [1]

Examiner tip: A reaction is spontaneous when E°_cell > 0. The species with the more positive E° will act as the oxidising agent (be reduced).

An electrolysis cell is used to plate copper onto a cathode from CuSO₄(aq) using a current of 2.50 A for 30.0 minutes.

(a) Write the half-equation for the reaction at the cathode. [1]

(b) Calculate the total charge passed. [1]

(c) Calculate the mass of copper deposited. (F = 96 500 C mol⁻¹, A_r(Cu) = 63.55) [2]

Show Mark Scheme

(a) Cu²⁺(aq) + 2e⁻ → Cu(s) [1]

(b) Q = It = 2.50 × (30.0 × 60) = 4500 C [1]

(c) Moles of electrons = 4500 / 96 500 = 0.04663 mol; moles of Cu = 0.04663 / 2 = 0.02332 mol [1]; mass = 0.02332 × 63.55 = 1.48 g [1]

Examiner tip: Always convert time to seconds before using Q = It. Remember to divide moles of electrons by the number of electrons in the half-equation (here 2 for Cu²⁺).

Links to: R2.1 Amount of Substance (mole calculations)