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When same quantity of electricity is passed through fused Cupric chloride(A) and Cuprous chloride(B), the weight of Copper deposited from ‘A’ is ‘X’ and ‘B' is ’Y'. Ratio of X to Y is

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3:2

1:2

2:3

2:1

answer is A.

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Detailed Solution

$According to Faraday' s II law W_{I} : W_{2} = E_{1} : E_{2}; = \frac{63.6}{2} : \frac{63.6}{1} = 1 : 2$

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