$250 \mathrm{~mL}$ of a waste solution obtained from the workshop of a goldsmith contains $0.1 \mathrm{M} \mathrm{AgNO}_{3}$ and $0.1 \mathrm{M} \mathrm{AuCl}$. The solution was electrolyzed at $2 \mathrm{~V}$ by passing a current of $1 \mathrm{~A}$ for 15 minutes. The metal/metals electrodeposited will be :-
$\left(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=0.80 \mathrm{~V}, \mathrm{E}_{\mathrm{Au}^{+} / \mathrm{Au}}^{0}=1.69 \mathrm{~V}\right)$
Correct Option: , 4
As voltage is '2V' so both $\mathrm{Ag}^{+} \& \mathrm{Au}^{+}$will reduce and their equal gm equivalent will reduce so
gmeq $\mathrm{Ag}=$ gmeq of $\mathrm{Au}$
$\frac{\mathrm{Wt}_{\mathrm{Ag}}}{\mathrm{E}_{\mathrm{qwt}_{\mathrm{Ag}}}}=\frac{\mathrm{Wt}_{\mathrm{Au}}}{\mathrm{E}_{\mathrm{qwt}_{\mathrm{Au}}}}$
So $\frac{\mathrm{wt}_{\mathrm{Ag}}}{\mathrm{wt}_{\mathrm{Au}}}=\frac{\mathrm{E}_{\mathrm{qwt}_{\mathrm{Aq}}}}{\mathrm{E}_{\mathrm{qwt}_{\mathrm{Au}}}}=\frac{\text { At } \mathrm{wt}_{\mathrm{Ag}}}{\text { Atwt }_{\mathrm{Au}}}$
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