For the following cell reaction

$\mathrm{Ag}|\mathrm{Ag}+|\mathrm{AgCl}\left|{\mathrm{Cl}}^{-}\right|{\mathrm{Cl}}_2,$Pt;

$∆{\mathrm{G}}_{\mathrm{f}}^0 \left(\mathrm{AgCl}\right) = – 109\mathrm{kJ}/\mathrm{mol},$

$∆{\mathrm{G}}_{\mathrm{f}}^0 ({\mathrm{Cl}}^{-} ) 129\mathrm{KJ} / \mathrm{mol} \mathrm{and}$

$∆{\mathrm{G}}_{\mathrm{f}}^0(\mathrm{Ag} ) 78\mathrm{KJ} / \mathrm{mol}.$${\mathrm{E}}^0$ of the cell is:

Here we have,
$∆{\mathrm{G}}_{\mathrm{f}}^0 \left(\mathrm{AgCl}\right)=– 109\mathrm{kJ}/\mathrm{mol}$,
$∆{\mathrm{G}}_{\mathrm{f}}^0 ({\mathrm{Cl}}^{-} )=129\mathrm{KJ} / \mathrm{mol}$  and
$∆{\mathrm{G}}_{\mathrm{f}}^0\left({\mathrm{Ag}}^{+}\right) 78\mathrm{KJ} / \mathrm{mol}.$
We have to find out Ecell0=?
At anode reaction will be,    
$\mathrm{Ag}⟶{\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}$
At cathode reaction will be,
$\mathrm{AgCl}+{\mathrm{e}}^{-} ⟶{\mathrm{Ag}}_{\left(\mathrm{S}\right)}+{\mathrm{Cl}}^{-}$
Therefore, The net cell reaction will be,
$\mathrm{AgCl}⟶{\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}$
We know that ,Free energy is given by
$\therefore ∆{\mathrm{G}}_{\mathrm{reaction}}^0=\mathrm{\Sigma \Delta }{\mathrm{G} }_{\mathrm{product}}^0-{\mathrm{\Sigma \Delta G}}_{\mathrm{reactant}}^0$
Product of Ag and Cl is 78 and 129 and reactant of AgCl is -109
By putting above values in equation we get,

$∆G_{reaction}^0=(78-129)-(-109)$
$∆{\mathrm{G}}_{\mathrm{reaction}}^0=+58\mathrm{kJ}/\mathrm{mol}$
We know that,
${\mathrm{\Delta G}}^0=-{\mathrm{nFE}}^0$
By putting values we get
$58\times {10}^3=-1\times 96500\times {\mathrm{E}}_{\mathrm{cell}}^0$
${\mathrm{E}}_{\mathrm{cell}}^0=-58\times 1000/ 965000=-0.60\mathrm{V}$
Therefore the required option is option A.