EMF of the following cell at 298 K in V is $\mathrm{x}\times {10}^{-2}.$

$\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left(0.1\mathrm{M}\right)\right|\left|{\mathrm{Ag}}^{+}\left(0.01\mathrm{M}\right)\right|\mathrm{Ag}$

The value of x is ________. (Round off to the nearest integer)   

[Give: ${\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{\circ }=-0.76\mathrm{V};{\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\circ }=+0.80\mathrm{V};\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059$]

Considering the given cell-representation:

Oxidation at anode:

Zn(s)→Zn²⁺(aq)+2e⁻
Reduction at cathode: 

Ag⁺(aq)+e⁻→Ag(s)

 $\begin{array}{l}{\mathrm{E}}_{\text{cell }}^{\mathrm{o}}={\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\mathrm{o}}-{\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{\mathrm{o}}\\ {\mathrm{E}}_{\text{cell }}^{\mathrm{o}}=\left(0.80-(-0.76)\right) \mathrm{V}=+1.56\mathrm{V}\\ \mathrm{Zn}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to \mathrm{Zn}\left(\mathrm{aq}\right)+2\mathrm{Ag}\left(\mathrm{s}\right)(\mathrm{n}=2)\\ {\mathrm{Q}}_{\mathrm{R}}=\frac{\left[{\mathrm{Zn}}^{2+}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}\\ {\mathrm{E}}_{\text{cell }}={\mathrm{E}}_{\text{cell }}^{\mathrm{o}}-\frac{0.059}{\mathrm{n}}\log {\mathrm{Q}}_{\mathrm{R}}\\ =1.56-\frac{0.059}{2}\log \frac{0.1}{0.01}\\ =1.56-0.08865 \\ =1.47135 \mathrm{V}\\ =147.135\times {10}^{-2} \mathrm{V}\\ \cong 147\times {10}^{-2} \mathrm{V}\end{array}$