First slide

Faraday's Laws

Question

Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode is (atomic mass of Cu = 63.5 u)

Easy

A

0 g
B

63.5 g
C

2 g
D

127 g

Solution

Given, Q - 2F, Atomic mass of Cu - 63.5 u
Valency of the metal Z = 2
We have, ${CuSO}_{4} ⟶ {Cu}^{2 +} + {SO}_{4}^{2}$
$\underset{1 mol}{{Cu}^{2 +}} + \underset{2 mol}{2 e^{-}} ⟶ \underset{1 mol = 63 .5 g}{Cu}$
Alternatively, $w = zQ = \frac{E}{F} \cdot 2 F = 2 E = \frac{2 \times 63 .5}{2} = 63 .5 g$

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