Match List - I with List - II :

	List - I (Parameter)		List - II (Unit)
(a)	Cell constant	(i)	$\mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$
(b)	Molar conductivity	(ii)	Dimensionless
(c)	Conductivity	(iii)	${\mathrm{m}}^{-1}$
(d)	Degree of dissociation of electrolyte	(iv)	${\mathrm{\Omega }}^{-1}{ \mathrm{m}}^{-1}$

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

For the cell $\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)(0.1\mathrm{M})\Vert {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)(0.01\mathrm{M})\right|\mathrm{Ag}\left(\mathrm{s}\right)$ the cell potential ${\mathrm{E}}_1=0.3095 \mathrm{V}$. For the cell $\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)(0.01\mathrm{M})\Vert {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)(0.001\mathrm{M})\right|\mathrm{Ag}\left(\mathrm{s}\right)$ the cell potential $=\mathrm{x}\times {10}^{-2} \mathrm{V}.$ Find value of x (Round off the Nearest Integer).

[ Use $:\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{J}$ ]