Question

$Cu(s)+S{n}^{2+}(0.001M)\to C{u}^{2+}(0.01M)+Sn(s)$
The Gibbs free energy change for the above reaction at $298K$ is $\times 10^{-1}kJmo{l}^{-1}$;
The value of $x$ is____ . [nearest integer]
$[$ Given $:E_{C{u}^{2+}/Cu}=0.34V;E_{S{n}^{2+}/Sn}=-0.14V;F=96500Cmo{l}^{-1}]$

Answer 1

Answer: 983

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