Question

The standard electrode potential of $Zn^{2+}/Zn$ is $- 0.76\, V$ and that of $Cu^{2+}/Cu$ is $0.34\, V$. The emf $(V)$ and the free energy change $(kJ \,mol^{-1} )$, respectively for a daniell cell will be

• A. - 0.42 and 81
• B. 1.1 and - 213
• C. - 1.1 and 213
• D. 0.42 and -81

Answer 1

Answer: (B)

Given, $E_{ Zn ^{2+} / Zn }^{\circ}=-0.76 \,V$
$E_{ Cu ^{2+} / Cu }^{\circ} =0.34\, V $
$E_{\text {cell }}^{\circ} =0.34-(-0.76) $
$=1.1\, V $
$\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}$
$=-2 \times 96500 \times 11 $
$=-213 \,V $