Electrochemistry - Result Question 10
10. Two Faraday of electricity is passed through a solution of $CuSO _4$. The mass of copper deposited at the cathode is (at. mass of $Cu=63.5 u$ )
(2015 Main)
(a) $0 $ $g$
(b) $63.5 $ $g$
(c) $2 $ $g$
(d) $127 $ $g$
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Answer:
Correct Answer: 10. (b)
Solution:
- Given, $Q=2 F$
Atomic mass of $Cu=63.5 u$
Valency of the metal $Z=2$
We have, $\quad CuSO _4 \longrightarrow Cu^{2+}+SO _4^{2-}$
$ \underset{\substack{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 $
BETA
