A-LevelChemistryElectrochemistry9701/43 · Oct/Nov 2019 · Q112 Marks

An electrochemical cell is constructed using two half-cells. • an Sn4+/Sn2+ half-cell • an Al3+/Al half-cell

(a)State the material used for the electrode in each half-cell. • Sn4+/Sn2+ half-cell • Al3+/Al half-cell[1]
(b)(i)The cell is operated at 298 K. The Al3+/Al half-cell has standard concentrations. The Sn⁴+/Sn2+ half-cell has [Sn⁴+] = 0.300 mol dm⁻³ and [Sn²+] = 0.150 mol dm⁻³. Use the Nernst equation to calculate the electrode potential, E, of the Sn⁴+/Sn²+ half-cell under these conditions.[2]
(b)(ii)Calculate the Ecell under these conditions.[1]
(b)(iii)Write an equation for the overall cell reaction that occurs.[2]
(c)Aluminium is produced industrially by electrolysis of a melt containing large amounts of Al3+ ions. Calculate the mass of aluminium that is obtained when a current of 300000A is passed for 24 hours. Give your answer to three significant figures.[4]
(d)Explain why chromium metal cannot be obtained by the electrolysis of dilute aqueous chromium(II) sulfate. Your answer should include data from the Data Booklet.[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[1] Platinum / Pt Aluminium / Al BOTH
[2] M1: use of or quoting a valid Nernst equation E = E° + 0.0590 / z log [ox] / [red] OR E = 0.15 + (0.0590 / 2)log2 M2: E = (+)0.16 (0.159) V minimum 2 sig. fig. correct answer scores 2 marks
[1] Ecell = 0.16-(-1.66) = +1.82 V minimum 3 sig. fig.
[2] 2Al + 3Sn4+ → 2Al3+ + 3Sn2+ M1: species M2: balancing
[4] M1: number of C (= 300 000 × 60 × 60 × 24) = 2.59 × 1010 (C) M2: number of F(= 2.592 × 1010/ 9.65 × 104) = 2.69 ×105 (moles of electrons) M3: moles of Al(= 2.69 × 105/3) = 8.95 × 104 M4: mass of Al(= 8.95 × 104 × 27) = 2420 kg correct answer scores 4 marks
[2] M1: (Cr2+ + 2e¯⇌ Cr) Eº = −0.91 and (2H+ + 2e¯ = H2) Eº = 0.00 seen M2: hydrogen formed instead / hydrogen (ions) easier to reduce / hydrogen has more positive E

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