Question

The mass of Cu deposited when the same quantity of electricity is passed, which deposits 1.08gm of Ag through AgNO3

Answer 1

Given

Mass of Ag deposited = 1.08 gm.

To find

Mass of Cu that can be deposited using same quantity of electricity.

Law to be used :- Faraday's second law

The ratio of amount of different substance deposited is equal to ratio of their chemical equivalent weight.

Formula

Equivalent weight = $\frac{molar \: mass}{n \: factor}$

For example :- In the reaction ${Ag}^{ + } + {e}^{ - } = Ag$

The equivalent weight of Ag is 108/1 = 108 ( N factor = 1)

SOLUTIONS

Equivalent weight of Cu = 63.5/2 = 31.75 .

Thus, using Faraday's second law,

$\frac{amount\: of \: Cu}{amount \: of \: Ag} = \frac{31.75}{108} \\ \\ = > \frac{m}{1.08} = \frac{31.75}{108 } \\ \\ = > m \: = \frac{31.72}{100} \\ \\ = > m = 0.3175 \: gram$

Hence, mass of Cu deposited is 0.3175 gram.