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 the correct $E^0$ for above reaction if $E_{C{u}^{2+}}^0/C{u}^{+}=0.34V$ and $E_{C{u}^{2+}/C{u}^{+}}^0=0.15V$

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

Answer 1

Answer: (C)

From given data (from $\Delta G^0=-n{E}^{\circ }F)$
(i) $Cu(s)⟶C{u}^{2+}(aq)+2e^{-}\cdot \Delta G_1^0=-2\times (-0.34)\times F$
(ii) $C{u}^{2+}(aq)+e^{-}⟶C{u}^{+}(aq)\cdot \Delta G_2^0=-1\times (0.15)F$
On adition
$Cu(s)⟶C{u}^{+}(aq)+e^{-}\cdot \Delta G_3^0=-1\times E^0\times E$
$\Delta G_3^0=\Delta G_1^0+\Delta G_2^0$
$=(-2\times -0.34\times F)+(-1\times 0.15\times F)$
$=+0.68F-0.15F-0.53F$
or $E^{\circ }=-0.53V$
Reaction,
$2C{u}^{+}(aq)\rightleftharpoons C{u}^{2+}(aq)+Cu(s)E^{\circ }=\text{ ? }$
So, $C{u}^{+}(aq)+e^{-}\rightleftharpoons Cu(s),E^0=+0.53V$
$C{u}^{+}(aq)\rightleftharpoons C{u}^{2+}(aq)+e^{-};E^0=-0.15V$
$3C{u}^{+}(aq)\rightleftharpoons C{u}^{2+}(aq)+Cu(s),E^0=+0.38V$