Question

Given the standard potential for the following half-cell reaction at 298 K,
$C{u}^{+}\left(aq\right)+e^{-}\to Cu\left(s\right);$ $E^{\circ }=0.52V$
$C{u}^{2+}\left(aq\right)+e^{-}\to C{u}^{+}\left(aq\right);$ $E^{\circ }=0.16V$
Calculate the $\Delta G^{\circ }$ (kJ) for the reaction, $\left[2C{u}^{+}\left(aq\right)\to Cu\left(s\right)+C{u}^{2+}\right]$

• A. -34.740
• B. -65.720
• C. -69.480
• D. -131.440

Answer 1

Answer: (A)

$C{u}^{+}+e^{-}\to Cu;E^{\circ }=0.52$ V
$\Delta G_1=-nF{E}^{\circ }=-1\times 96500\times 0.52$ ...(i)
$C{u}^{2+}+e^{-}\to C{u}^{+};E^{\circ }=0.16$ V
or $C{u}^{+}\to C{u}^{2+}+e^{-};E^{\circ }=-0.16$ V
$\Delta G_2=-1\times 96500\times \left(-0.16\right)$ ...(ii)
On adding equations (i) and (ii), we get
$2C{u}^{+}\to Cu+C{u}^{2+}$ ,
$\Delta G=\Delta G_1+\Delta G_2$
$=\left(-96500\times 0.52\right)+\left(96500\times 0.16\right)$
$=96500\left(-0.52+0.16\right)$
$=-\left(96500\times 0.36\right)$
$\text{=}-\text{34740 J}\text{=}-\text{34}\text{.740}\text{kJ}$