Question

$\Delta G ^{0}$ for the reaction, $Cu ^{2+}+ Fe \rightarrow Fe ^{2+}+ Cu$ is: [given: $\left. E _{ Cu ^{2+} / Cu }^{0}=+0.34\, V , E _{ Fe ^{2+} / Fe }^{0}=-0.44 \,V \right]$

• A. 11.44 KJ
• B. 180.8 KJ
• C. -150.5 KJ
• D. 28.5 KJ

Answer 1

Answer: (C)

$E _{\text {cell }}^{0}= E _{ Cu ^{2+} / Cu }^{0}- E _{ Fe ^{2+} / Fe }^{0}$
$=0.34-(-0.44)=0.78$
$ \Delta G ^{0}=- n FE _{\text {cell }}^{0}$
$=-2 \times 96500 \times 0.78$
$ =150540 \,J$
$ =-150.5 \,KJ$