Copper sulphate solution is electrolysed between two platinum electrodes. $A$ current is passed until $1.6 \ g$ of oxygen is liberated at the anode. The amount of copper deposited at the cathode during the same period is .............. $g$ [At. mass of $Cu = 63.6$]

How long would it take to reduce completely $80 \ mL$ of $0.1 \ M$ $Fe_2(SO_4)_3$ to $FeSO_4$ by passing a current of $2 \ A$?

In the process of electroplating,$m \ g$ of silver is deposited when $4 \ A$ of current flows for $2 \ \text{minutes}$. The amount (in $g$) of silver deposited by $6 \ A$ of current flowing for $40 \ \text{seconds}$ will be:

When two electrolytic cells containing $NiSO_4$ and $CuSO_4$ solutions are connected in series and an electric current is passed,$1.6 \ g$ of $Cu$ is deposited on the electrode. How much $Ni$ metal is produced? If a cell containing $AgNO_3$ solution is connected instead of $NiSO_4$,how many grams of silver will be obtained? [Atomic mass: $Cu = 63.5 \ g/mol$,$Ni = 58.7 \ g/mol$,$Ag = 108 \ g/mol$]

Faraday's laws of electrolysis state that the mass deposited on an electrode is proportional to:

$A$ current of $0.01 \ A$ is passed through a solution of a metal bromide $(MBr_x)$. The only reaction at the cathode is the deposition of metal $(M)$. After $3 \ hours$,$0.072 \ g$ of metal $(M)$ is deposited. The value of '$x$' is approximately (Given: Atomic weight of $M = 192$)