$250 \mathrm{mL}$ 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 : $\left({\text{E}}_{{\text{Ag}}^{+}/\text{Ag}}^0=0.80\text{V},{\text{E}}_{{\text{Au}}^{+}/\text{Au}}^0=1.69\text{V}\right)$

Match List - I with List - II :

An acidic solution of dichromate is electrolysed for $8$ minutes using $2$ A current. As per the following equation

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$
The amount of ${\mathrm{Cr}}^{3+}$obtained was $0.104\mathrm{g}$. The efficiency of the process $(\mathrm{in}\%)$ is (Take : $\mathrm{F}=96000 \mathrm{C}$, At. mass of chromium $=52$)

The photoelectric current from Na (work function, ${\mathrm{w}}_0=2.3\mathrm{eV}$)  is stopped by the output voltage of the cell

$\mathrm{Pt}\left(\mathrm{s}\right)\Vert {\mathrm{H}}_2(\mathrm{g},1\mathrm{bar})\left|\mathrm{HCl}\right(\mathrm{aq}·,\mathrm{pH}=1\left)\right|\mathrm{AgCl}\left(\mathrm{s}\right)\mid \mathrm{Ag}\left(\mathrm{s}\right)$
the $pH$ of aq. $\mathrm{HCl}$ required to stop the photoelectric current from $\mathrm{K}\left({\mathrm{w}}_0=2.25\mathrm{eV}\right)$, all other conditions remaining the same, is $\dots \dots \dots .\times {10}^{-2}$ (to the nearest integer).

Given $2.303\frac{\mathrm{RT}}{\mathrm{F}}=0.06\mathrm{V};{\mathrm{E}}_{\mathrm{AgCl}/\mathrm{Ag}/Cl}^0=0.22\mathrm{V}$