By passing electric current,$NaClO_{3}$ is converted into $NaClO_{4}$ according to the following equation:
$NaClO_{3} + H_{2}O \rightarrow NaClO_{4} + H_{2}$
How many moles of $NaClO_{4}$ will be formed when $3 \ F$ of charge is passed through $NaClO_{3}$?

$10 \, A$ current is passed through $CuSO_4$ solution for $193 \, \text{minutes}$ at $1 \, \text{bar}$ pressure and $300 \, K$ temperature. If the cell has $80 \%$ efficiency,find the weight of $Cu$ deposited and the volume of $O_2$ evolved. $[Cu = 63.5 \, u, R = 0.08314 \, \text{L bar K}^{-1} \text{mol}^{-1}]$

The value of one Faraday is .............. $C \ mol^{-1}$.

$CuSO_4$ solution is electrolysed for $15 \ minutes$ to deposit $0.4725 \ g$ of copper at the cathode. The current in amperes required is (Faraday $= 96,500 \ C \ mol^{-1}$,atomic weight of copper $= 63$)

Match the following and select the correct option for the quantity of electricity,in $C \ mol^{-1}$,required to deposit various metals at the cathode:

List-$I$	List-$II$
$a. \ Ag^{+}$	$i. \ 386000 \ C \ mol^{-1}$
$b. \ Mg^{2+}$	$ii. \ 289500 \ C \ mol^{-1}$
$c. \ Al^{3+}$	$iii. \ 96500 \ C \ mol^{-1}$
$d. \ Ti^{4+}$	$iv. \ 193000 \ C \ mol^{-1}$

$Assertion (A)$: $A$ current of $96.5 \ A$ is passed into aqueous $AgNO_3$ solution for $100 \ s$. The weight of silver deposited is $10.8 \ g$ (At. wt. of $Ag = 108$).
$Reason (R)$: The mass of a substance deposited during the electrolysis of an electrolyte is inversely proportional to the quantity of electricity passing through the electrolyte.
The correct answer is :