Question

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

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

Answer 1

Answer: (A)

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