A current was passed for two hour through a solution of an acid that liberated `11.2` litre of oxygen at NTP at anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same time?

Options

A. 16 g
B. 63 g
C. 31.5 g
D. 8 g

Answer

B

Explanation

Moles of `O_(2)` gas at anode`=11.2//22.4 = 1//2`
Wt. of `O_(2)` gas `=1//2 xx 32 g = 16 g`
`W_(1)/W_(2) = E_(1)/E_(2)`
`16/W_(2) = (16//2)/(63//2)`
`W_(2) = 63 g`