Electrochemistry - Result Question 61

####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:

  1. Given, $Q=2 F$

Atomic mass of $Cu=63.5 u$

Valency of the metal $Z=2$

We have, $\quad CuSO _4 \longrightarrow Cu^{2+}+SO _4^{2-}$

$$ \underset{1 mol}{Cu^{2+}}+\underset{\substack{2 mol \ 2 F}}{2 e^{-}} \longrightarrow \underset{1 mol=63.5 g}{Cu} $$

Alternatively.

$$ W=Z Q=\frac{E}{F} \cdot 2 F=2 E=\frac{2 \times 63.5}{2}=63.5 $$



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