Question

Calculate $ \Delta G^{\circ} $ for the following cell reaction
$ \ce{Zn_{(s)} +Ag2O_{(s)} +H2O_{(l)} ->} $
$ \ce{Zn^{2+}_{n(aq)} +2Ag_{(s)} +2^{-}OH_{(aq)}} $
$ E^{\circ}_{Ag^{+}/Ag}=+0.80V $ and $ E^{\circ}_{Zn^{2+}/Zn}=-0.76\,V $

• A. $ - 305 \,kJ/mol $
• B. $ -301 \,kJ/mol $
• C. $ 305 \,kJ/mol $
• D. $ 301 \,kJ/mol $

Answer 1

Answer: (B)

Given that, $E^{\circ} Ag ^{+} / Ag =+0.80\, V$

$ E^{\circ} Zn ^{2+} / Zn =-0.76 \,V$

$ E_{ Cell }^{\circ} =E_{ R }^{\circ}-E^{\circ} L =0.80-(-0.76)=1.56\, V $

From the equation $\Delta G^{\circ}=-n F E^{\circ}$

(where, $F=96500, n=2$ ) $=-2 \times 96500 \times 1.56$

$=-301080\, J / mol =-301.080 \,kJ / mol$

$=-301 \,kJ / mol$