The mass of copper that will be deposited at cathode in electrolysis of 0.2 M solution of copper sulphate when a quantity of electricity equal to that required to liberate 2.24 L of hydrogen from 0.1 M aqueous `H_2SO_4` is passed will be (At. mass of Cu = 63.5)

Options

A. 1.59 g
B. 3.18 g
C. 6.35 g
D. 12.70 g

Answer

C

Explanation

Applying, Faraday.s second law
`w_(Cu)/w_(H_2)=E_(Cu)/E_(H_2)`
22.4 L of `H_2` gas weight =2 g
`therefore` 2.24 L of `H_2` will weigh = `2/22.4xx2.24=0.2` g
`w_(Cu)=31.75/1xx0.2=6.35` g