An electrolytic cell contains a solution of `Ag_(2)SO_(4)` and have platinum electrodes. A current is passed until 1.6gm of `O_(2)` has been liberated at anode. The amount of silver deposited at cathode would be

Options

A. 107.88 g
B. 1.6 g
C. 0.8 g
D. 21.60 g

Answer

D

Explanation

At cathode: `Ag^(+) (aq) + e^(-) to Ag(s)`
At anode : `2OH^(-) (aq) to H_(2)O (l) + 1/2 O_(2)(g) + 2e^(-)`
`E_(Ag) = 108/1 = 108, E_(O_(2)) = (1/2 xx 32)/2 = 8`
`W_(Ag)/E_(Ag) = W_(O_(2))/E_(O_(2))`
`W_(Ag) = (1.6 xx 108)/8 = 21.6 g`