Calculate $E^{\circ }$ values of the following reactions.

$2Ag+Z{n}^{2+}\to 2A{g}^{+}+Zn$

Given that: $E_{A{g}^{+}/Ag}^{\circ }=0.80V,E_{Z{n}^{2+}/Zn}^{\circ }=-0.76V$

The correct option is C $-1.56V$
$E_{A{g}^{+}/Ag}^{\circ }=-0.80V$

$A{g}^{+}+e^{-}\to Ag;E^{\circ }=+0.80V$ - (1)

$Z{n}^{2+}+2e^{-}\to Zn;E^{\circ }=-0.76V$ - (2)

Inverting (1) and multiply with 2,

$2Ag\to 2A{g}^{+}+2e^{-};E^{\circ }=-0.80V$ - (3)

$\Delta G_2^{\circ }=-nF{E}_2^{\circ }=2F\left(0.76V\right)$

$\Delta G_3^{\circ }=-nF{E}_3^{\circ }=2F\left(0.80V\right)$

$\Delta G_r^{\circ }=\Delta G_2^{\circ }+\Delta G_3^{\circ }=2F\left(1.56V\right)=-2F{E}_r^{\circ }$

$E_r^{\circ }=-1.56V$