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Solve the following.
Answer the following: (1) Calculate the charge required in coulombs to deposit of copper at cathode…
Answer the following:
(a)[1.5]
(1) Calculate the charge required in coulombs to deposit $6\cdot35\text{ g}$ of copper at cathode from $\text{CuSO}_4$ solution.
(2) How much time is required for deposition of copper if an electric current of $3\cdot5\text{ ampere}$ is passed through the electrolyte mentioned above?
(Atomic Mass of $\text{Cu} = 63\cdot5\text{ g mol}^{-1}$)
(b)[1.5]
Can an aluminium vessel be used to store dil. $\text{HCl}$? Give a reason for your answer by referring to the values given below.
$E^\circ_{\text{Al}^{3+}/\text{Al}} = -1\cdot66\text{ V}; E^\circ_{\text{H}^+/\frac{1}{2}\text{H}_2} = 0\cdot0\text{ V}$
Answer
Answer (a)
AIFormulas used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, M = 63.5 g mol^-1, n = 2.0, w = 6.35 g, so Q = 19300 C Faraday's first law: w = M*Q/(n*F) with I = 3.5 A, M = 63.5 g mol^-1, n = 2.0, w = 6.35 g, so t = 5514 s
(1) Moles of Cu $= \dfrac{6.35}{63.5} = 0.1$ mol. $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$, so charge $= 0.1 \times 2 \times 96500 = 19300\ \mathrm{C}$.
(2) $t = \dfrac{Q}{I} = \dfrac{19300}{3.5} = 5514\ \mathrm{s}$ (about 92 minutes).
Final answer: 19300 C
Answer (b)
AIEquation: $2\mathrm{Al} + 6\mathrm{HCl} \longrightarrow 2\mathrm{AlCl_{3}} + 3\mathrm{H_{2}}$
No. Because $E^\circ_{Al^{3+}/Al} = -1.66\ \mathrm{V}$ is much lower than $E^\circ_{H^+/H_2} = 0.0\ \mathrm{V}$, aluminium is a stronger reducing agent than hydrogen and is oxidised by dil. HCl, liberating hydrogen, so the vessel would be corroded.
$E^\circ_{cell} = 0.0 - (-1.66) = +1.66\ \mathrm{V}$ (positive, reaction feasible).
From ISC 2026 Improvement Chemistry Paper 1, question 18(i).
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