Question

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

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

Answer 1

Answer: (C)

$2 Cu ^{+} \longrightarrow Cu ^{+2}+ Cu$
$2 e ^{-}+ Cu ^{+2} \longrightarrow Cu ; E _{1}^{\circ}=0 \cdot 34 \,V ; \ldots . .( i )$
$e ^{-}+ Cu ^{+2} \longrightarrow Cu ^{+} ; E _{2}^{\circ}=0 \cdot 15 \,V ; \ldots( ii )$
$Cu ^{+}+ e ^{-} \longrightarrow Cu ; E _{3}^{\circ}=? \ldots$ (iii)
$Now , \Delta G _{1}^{\circ}=- nFE _{1}^{\circ}=-2 \times 0 \cdot 34 \, F$
$\Delta G _{2}^{\circ}=-1 \times 0 \cdot 15 \, F , \Delta G _{3}^{\circ}=-1 \times E _{3}^{\circ} F$
Again, $\Delta G _{1}^{\circ}=\Delta G _{2}^{\circ}+\Delta G _{3}^{\circ}$
$\Rightarrow -0.68 F=-0.15 F-E_{3}^{\circ} F 0$
$\Rightarrow E_{3}^{\circ}=0.68-0.15=0.53 \, V$
$E _{\text {ell }}^{\circ}= E _{\text {cathode }}^{\circ}\left( Cu ^{+} / Cu \right)- E _{\text {anode }}^{\circ}\left( Cu ^{+2} / Cu ^{+}\right)$
$=0.53-0.15=0.38 \,V .$