1. Tardigrade
  2. Question
  3. Chemistry
  4. E° values for the half cell reactions are given below: Cu2++e- → Cu+ ; E°=0.15 V Cu2++2e- → Cu ; E°=0.34 V What will be the E° of the half-cell: Cu+ +e- → Cu ?

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Q. $E^{\circ}$ values for the half cell reactions are given below :
$Cu^{2+}+e^{-} \to Cu^{+}$ ; $\,E^{\circ}=0.15\,V$
$ Cu^{2+}+2e^{-} \to Cu$ ; $\,E^{\circ}=0.34\,V$
What will be the $E^{\circ}$ of the half-cell : $Cu^+ +e^- \to Cu$ ?

Electrochemistry

Solution:

$Cu^{2+}+e^{-} \to Cu^{+}$ ;
$Cu^{2+}+2e^{-} \to Cu$ ;
$Cu^{+}+e^{-} \to Cu$ ;
$E^{\circ}_{1}=0.15\,V, \Delta G^{\circ}_{1}, n_{1}=1$
$E^{\circ}_{2}=0.34\,V, \Delta G^{\circ}_{2}, n_{2}=2$
$E^{\circ}_{3}=?$, $\Delta G^{\circ}_{3}$, $n_{3}=1$
$\Delta G^{\circ}_{3}=\Delta G^{\circ}_{2}-\Delta G^{\circ}_{1}$
$\Rightarrow -n_{3}FE^{\circ}_{3}=-n_{2}FE^{\circ}_{2}+n_{1}FE^{\circ}_{1}$
$-E^{\circ}_{3}=-2\times0.34+1\times0.15$
$E^{\circ}_{3}=0.68-0.15=+0.53\,V$