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Solve the following.
Three electrolytic cells (X), (Y) and (Z) containing solutions of , and respectively are connected…
Three electrolytic cells (X), (Y) and (Z) containing solutions of $\text{AgNO}_3$, $\text{CuSO}_4$ and $\text{ZnSO}_4$ respectively are connected in series. A steady current of $1\cdot5\text{ ampere}$ is passed through these electrolytic cells until $1\cdot45\text{ g}$ of silver is deposited at the cathode of cell (X).
(Atomic weight of $\text{Ag} = 108$, $\text{Cu} = 63\cdot5$ and $\text{Zn} = 65\cdot3$)
(a)[1.5]
How much of charge is given to the electrolyte solution?
(b)[1.5]
What mass of copper and zinc is deposited at the respective cathode?
Answer
Answer (a)
Official answer keyFormula used: Faraday's first law: w = M*Q/(n*F) with M = 108 g mol^-1, n = 1, w = 1.45 g, so Q = 1295.6 C
$\mathrm{Ag^+ + e^- \rightarrow Ag}$: 108 g of Ag is deposited by 96500 C.
$Q = \dfrac{96500\times 1.45}{108} = 1295.6\text{ C}$
Final answer: 1295.6 C
Answer (b)
Official answer keyCell Y: $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$; $2\times 96500$ C deposit 63.5 g of Cu.
Mass of Cu $= \dfrac{63.5\times 1295.6}{2\times 96500} = 0.426\text{ g}$
Cell Z: $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$; $2\times 96500$ C deposit 65.3 g of Zn.
Mass of Zn $= \dfrac{65.3\times 1295.6}{2\times 96500} = 0.438\text{ g}$
Final answer: 0.426 g
From ISC 2027 Specimen Chemistry Paper 1, question 17(i).