An electrolytic cell contains a solution of `Ag_2SO_4` and have platinum electrodes. A current is passed until 1.6 g 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^(+)=e^(-)toAg`
At anode : `2OH^(-)toH_2O+1//2O_2+2e^(-)`
`E_(Ag)=108/1=108,E_(o_2)=((1/2xx32)/2)=8`
`w_(Ag)/E_(Ag)=w_(o_2)/E_(o_2),w_(Ag)=(1.6xx108)/8=21.6`