Electrochemistry 1 Question 10
10. Two Faraday of electricity is passed through a solution of $\mathrm{CuSO}_{4}$. The mass of copper deposited at the cathode is (at. mass of $\mathrm{Cu}=63.5 \mathrm{u}$ )
(2015 Main)
(a) $0 \mathrm{~g}$
(b) $63.5 \mathrm{~g}$
(c) $2 \mathrm{~g}$
(d) $127 \mathrm{~g}$
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Answer:
Correct Answer: 10. (b)
Solution:
- Given, $Q=2 F$
Atomic mass of $\mathrm{Cu}=63.5 \mathrm{u}$
Valency of the metal $Z=2$
We have, $\quad \mathrm{CuSO}{4} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{SO}{4}^{2-}$
$$ \underset{1 \mathrm{~mol}}{\mathrm{Cu}^{2+}}+\underset{\substack{2 \mathrm{~mol} \ 2 F}}{2 e^{-}} \longrightarrow \underset{1 \mathrm{~mol}=63.5 \mathrm{~g}}{\mathrm{Cu}} $$
Alternatively.
$$ W=Z Q=\frac{E}{F} \cdot 2 F=2 E=\frac{2 \times 63.5}{2}=63.5 $$
