A solution of `Ni(NO_3)_2` is electrolysed between platinam electrodes using a current of `5 am^-` peres for 20 minutes. What mass of `N i` is `d e` posited at the cathode?

Options

A. `75.5:1`
B. `65.2:1`
C. `55.2:1`
D. `45.2:1`

Answer

A

Explanation

The reactions occuring at the electrodes are: Cathode `Sn^(2+) + 2e^(-) rarr Sn`, Anode `Sn^(2) rarr Sn^(4+) + 2e^(-)`. If 0.119g of Sn is deposited at the cathode, then 0.119 g of `Sn^(4+)` will be formed at the anode. Thus, a total of `2 xx 0.119g` of `Sn^(2+)` is lost from the solution. Hence mass of the remaining `SnCl_(2)`
`=20g - (M_(SnCl_(2)))/(M_(sn)) xx m_(Sn)= 20g - (189.6)/(119) xx 2 xx 0.119g = (20- 0.38)g= 19.62g`
Mass of `SnCl_(4)` formed `=(M_(SnCl_(4))/(M_(sn)) xx m_(Sn) = (260.6)/(119) xx 0.119g = 0.26g (m(SnCl_(2)))/(m(SnCl_(4)))= (19.62)/(0.26)= 75.5`