A solution of `Ni(NO_(3))_(2)` is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode?

Options

A. `0.20`
B. `0.05`
C. `0.10`
D. `0.15`

Answer

B

Explanation

`Ni^(2+) + 2e^(-) to Ni`
2 F of electricity deposit nickel = 1 mol
0.1 F of electricity will deposit nickel = `(1)/(2) xx 0.1`
=`0.05` mol