First slide

Faraday's Laws

Question

On passing current through molten KCl, 19.5 g of K is deposited. The amount of Al deposited by the same quantity of electricity when passed through molten AlCl₃ is

Moderate

A

4.5 g
B

9 g
C

13.5 g
D

2.7 g

Solution

According to Faraday's II law:
$\frac{{{m_{Al}}}}{{{m_K}}} = \frac{{{E_{Al}}}}{{{E_K}}} \Rightarrow {m_{Al}} = \frac{{19.5 \times 9}}{{39}} = 4.5gm$

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