A solution of silver nitrate was electrolysed using silver electrodes each weighing 12 gms. The electrolysis was carried for 5.36 hours using a current of strength 0.5 amperes. Incorrect statements are

Coulombs passed through cell $=5.36\times 3600\times 0.5=9650$ (rounded off to 9650 from 9648)

$\text{ Cathode reaction, }{\mathrm{Ag}}^{+}+e\to \mathrm{Ag}$

$\text{ Weight of silver deposited at cathode }=\frac{108\times 9650}{96,500}=10.8\mathrm{gms}$

Weight of cathode after reaction (electrolysis) - 10.8 + 12.0 = 22.8 gms

Anode reaction : $\mathrm{Ag}+{\mathrm{NO}}_3^{-}=\mathrm{Ag}{\mathrm{NO}}_3+{\mathrm{e}}^{-}$

At anode, 10.8 grams of anode dissolves.

$\therefore$Weight of anode remaining = 12.0 - 10.8 = 1.2 gms

$\text{ Moles of }{e}^{-}\text{passed through cell }=\frac{9650}{96,500}=0.1$