Question

$C{u}_{aq}^{+}$is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction:
$2C{u}^{+}(aq)\rightleftharpoons C{u}^{2+}(aq)+Cu(s)$
choose correct $E^{\circ }$ for above reaction if
$E^{\circ }{}_{Cu}^{2+}=0\cdot 34V$ and $E^{\circ }{}_{Cu}^{2+}=0\cdot 15V$

• A. -0.38 V
• B. +0.49 V
• C. +0.38 V
• D. -0.19 V

Answer 1

Answer: (C)

$2C{u}^{+}⟶C{u}^{+2}+Cu$
$2e^{-}+C{u}^{+2}⟶Cu;E_1^{\circ }=0\cdot 34V;\dots ..(i)$
$e^{-}+C{u}^{+2}⟶C{u}^{+};E_2^{\circ }=0\cdot 15V;\dots (ii)$
$C{u}^{+}+e^{-}⟶Cu;E_3^{\circ }=?\dots$ (iii)
$Now,\Delta G_1^{\circ }=-nF{E}_1^{\circ }=-2\times 0\cdot 34F$
$\Delta G_2^{\circ }=-1\times 0\cdot 15F,\Delta G_3^{\circ }=-1\times E_3^{\circ }F$
Again, $\Delta G_1^{\circ }=\Delta G_2^{\circ }+\Delta G_3^{\circ }$
$\Rightarrow -0.68F=-0.15F-E_3^{\circ }F0$
$\Rightarrow E_3^{\circ }=0.68-0.15=0.53V$
$E_{\text{ell }}^{\circ }=E_{\text{cathode }}^{\circ }\left(C{u}^{+}/Cu\right)-E_{\text{anode }}^{\circ }\left(C{u}^{+2}/C{u}^{+}\right)$
$=0.53-0.15=0.38V.$