Calculate the time required in seconds to deposit $6.35 \ g$ of copper from its salt solution by passing a $5 \ A$ current. [Molar mass of $Cu = 63.5 \ g \ mol^{-1}$]

$A$ current is passed for $2 \, \text{hours}$ through an acidic solution,liberating $11.2 \, L$ of oxygen at $NTP$ at the anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same duration? .......... $g$

What is the charge required for the reduction of two moles of $Cu^{2+}$ to $Cu$?

If the total electricity required to deposit $1$ mole of a metal $M$ is equal to that of $10.7$ amperes of current for $10$ hours,the equivalent weight of the metal is (atomic weight $= M \ u$):

In the electrolysis of a $CuSO_4$ solution,how many grams of $Cu$ are plated out on the cathode,in the time that is required to liberate $5.6 \ L$ of $O_{2(g)}$,measured at $1 \ atm$ and $273 \ K$,at the anode (in $g$)?

Potassium chlorate is prepared by the electrolysis of $KCl$ in basic solution:
$6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$
If only $60\%$ of the current is utilized in the reaction,the time (rounded to the nearest hour) required to produce $10 \ g$ of $KClO_{3}$ using a current of $2 \ A$ is:
(Given: $F = 96,500 \ C \ mol^{-1}$,molar mass of $KClO_{3} = 122 \ g \ mol^{-1}$)