Calculate the current in ampere required to deposit $4.8 \ g$ of $Cu$ from its salt solution in $30 \ minutes$. $\left[ \text{Molar mass of } Cu = 63.5 \ g \ mol^{-1} \right]$ (in $A$)

Electrolytic cells containing $Zn$ and $Al$ salt solutions are connected in series. If $6.5 \ g$ of $Zn$ is deposited in one cell,calculate the mass of $Al$ deposited in the second cell (molar mass: $Zn=65 \ g \ mol^{-1}$,$Al=27 \ g \ mol^{-1}$) by passing a definite quantity of electricity. (in $g$)

$A$ solution of copper sulphate $(CuSO_4)$ is electrolysed for $10 \, \text{minutes}$ with a current of $1.5 \, \text{amperes}$. The mass of copper deposited at the cathode (atomic mass of $Cu = 63 \, u$) is ............ $g$.

If nearly $10^5 \ C$ liberate $1 \ gm$ equivalent of aluminium,then the amount of aluminium (equivalent weight $9$) deposited through electrolysis in $20 \ minutes$ by a current of $50 \ A$ will be .............. $gm$.

One Faraday of electricity when passed through a solution of copper sulphate deposits:

Aluminium oxide may be electrolysed at $1000 \, ^\circ C$ to furnish aluminium metal (At. Mass $= 27 \, amu$; $1 \, F = 96,500 \, C$). The cathode reaction is $Al^{3+} + 3e^- \to Al^0$. To prepare $5.12 \, kg$ of aluminium metal by this method would require: