Calculate $\Delta G^{\circ}$ for the following cell reaction. $\begin{aligned} &…

Calculate $\Delta G^{\circ}$ for the following cell reaction. $\begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \\ & \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2^{-} \mathrm{OH}(\mathrm{aq}) \\ & E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V} \text { and } E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\end{aligned}$
  1. $-305 \mathrm{~kJ} / \mathrm{mol}$
  2. $-301 \mathrm{~kJ} / \mathrm{mol}$
  3. $305 \mathrm{~kJ} / \mathrm{mol}$
  4. $301 \mathrm{~kJ} / \mathrm{mol}$

Solution

$\quad E_{\text {cell }}^{\circ}=E^{\circ}{ }_C-E_A^{\circ}=0.80-(-0.76)=1.56 \mathrm{~V}$ From the equation $\Delta G^{\circ}=-n F E^{\circ}$ $\begin{aligned} & =-2 \times 96500 \times 1.56(n=2) \\ & =-301080 \mathrm{~J} / \mathrm{mol} \\ & =-301 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$

Asked in: AP EAMCET 2015

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