The electrochemical equivalent of a metal is $3.3 \times 10^{-7} \ kg/C$. The mass of the metal liberated at the cathode when a $3 \ A$ current is passed for $2 \ s$ will be:

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 $KClO_{3} = 122 \ g \ mol^{-1}$)

$250 \ mL$ of a waste solution obtained from the workshop of a goldsmith contains $0.1 \ M \ AgNO_3$ and $0.1 \ M \ AuCl$. The solution was electrolyzed at $2 \ V$ by passing a current of $1 \ A$ for $15 \ minutes$. The metal/metals deposited will be
$(E^0_{Ag^+/Ag} = 0.80 \ V, E^0_{Au^+/Au} = 1.69 \ V)$

An acidic solution of dichromate is electrolyzed for $8 \ min$ using $2 \ A$ current. As per the following equation: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$. The amount of $Cr^{3+}$ obtained was $0.104 \ g$. The efficiency of the process (in $\%$) is (Take: $F = 96000 \ C$,At. mass of chromium $= 52$)

What current strength in $amp$ will be required to liberate $10 \ g$ of chlorine from $NaCl$ solution in one hour?

$38.6 \ A$ of current is passed for $100 \ s$ through an aqueous $CuSO_4$ solution using platinum electrodes. The mass of copper deposited from the solution and the volume of gas liberated at $STP$ are respectively (molar mass of $Cu = 63.54 \ g \ mol^{-1}$)