Question

Standard electrode potential for the cell with cell reaction
$Zn ( s )+ Cu ^{2+}( aq ) \rightarrow Zn ^{2+}( aq )+ Cu ( s )$ is $1.1 V$.
Calculate the standard gibbs energy change for the cell reaction.
(Given $F =96487 \,C \,mol ^{-1}$ )

• A. $-200.27\, kJ\, mol ^{-1}$
• B. $-212.27 \,kJ\, mol ^{-1}$
• C. $-212.27 \,J\, mol ^{-1}$
• D. $-200.27 \,J \,mol ^{-1}$

Answer 1

Answer: (B)

$\Delta G ^{\circ}=- nF \cdot E _{\text {cell }}^{\circ}$
$\Delta G ^{\circ}=\frac{-2 \times 96487 \times 1.1}{1000}$
$\Delta G ^{\circ}=-212.27 \,kj\, mol ^{-1}$