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?

Answer

0.05

Explanation

According to the Faraday’s law of electrolysis, nF of current is required for the deposition of 1 mole. According to the reaction,
`Ni(NO_(3))_(2) to Ni^(2+) + 2NO_(3)^(-)`
2F of current deposits = 1 mol
`therefore 0.1` F of current deposits = 0.05 mol