Question

A solution of $Ni(N{O}_3{)}_2$ is electrolysed between platinum electrodes using a current of 5 amperes for 20 min. What mass of Ni is deposited at the cathode ? (Atomic mass of Ni = 58.7)

Answer 1

Quantity of elctricity used = Current in amperes $\times$ Time in second $=5A\times (20\times 60)s=6000C$
Chemical equation for the reaction during electrolysis :
$Ni\left(N{O}_3{)}_2\right(aq)+2H^{+}+\underset{2F}{2e^{-}}\to \underset{58.7g}{Ni}+2HN{O}_3$
Charge Q on n moles of electrons is given by : Q =nF
Charge required to deposit 1 mole nickel
$=2F=2\times 96500C=1.93\times {10}^{5}C$
Molar mass of nickel $=58.7gmo{l}^{-1}$
$1.93\times {10}^{5}Cofchargeproducesnickel=58.7g$
6000 C of charge produces nickel $=\frac{\left(58.7g\right)\times \left(6000C\right)}{(1.93\times {10}^{5}C)}=1.825g$