Find the emf of the cell in which the following reaction takes place at 298 K
$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}(0.001\mathrm{M})\to {\mathrm{Ni}}^{2+}(0.001\mathrm{M})+2\mathrm{Ag}\left(\mathrm{s}\right)$
$(\mathrm{Given} \mathrm{that} {\mathrm{E}}_{\mathrm{cell} }^0=10.5\mathrm{V},\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{at} 298\mathrm{K} )$