Question

$Cu ( s )+ Sn ^{2+}(0.001 M ) \rightarrow Cu ^{2+}(0.01 M )+ Sn ( s )$
The Gibbs free energy change for the above reaction at $298\, K$ is $\times 10^{-1} \,kJ \,mol ^{-1}$;
The value of $x$ is____ . [nearest integer]
$\left[\right.$ Given $\left.: E _{ Cu ^{2+} / Cu }=0.34\, V ; E _{ Sn ^{2+} / Sn }=-0.14\, V ; F =96500 \,C \,mol ^{-1}\right]$

Answer 1

$Cu _{( s )}+ Sn ^{2+}(0.001 M ) \rightarrow Cu ^{2+}(0.01 M )+ Sn _{( s )}$
$E _{\text {cell }}^{\circ}= E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}$
$=-0.14-(0.34)$
$=-0.48 V$
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ Cu ^{2+}\right]}{\left[ Sn ^{2+}\right]}$
$=-0.48-\frac{0.059}{2} \log \frac{0.01}{0.001}$
$=-0.509$
$\Delta G =- nF E _{\text {cell }}$
$=-2 \times 96500 \times(-0.5095)$
$=98333.5\, J / mol$
$=98.335 \, kJ / mol$
$=983.35 \times 10^{-1} \, kJ / mol$
Nearest Integer :$ 983$