Question: $Eº_{cell}$ for the half cell reactions are given below: \[Cu^{2 +} + e^{-} \rightarrow Cu^{+};Eº...

Question

$Eº_{cell}$ for the half cell reactions are given below:

$Cu^{2 +} + e^{-} \rightarrow Cu^{+};Eº = 0.15V$

$Cu^{2 +} + 2e^{-} \rightarrow Cu;Eº = 0.34V$

What will be the $Eº$ of the half cell :

$Cu^{+} + e^{-} \rightarrow Cu?$

Answer 1

$Cu^{2 +} + e^{-} \rightarrow Cu^{+};Eº_{1} = 0.15V,\Delta Gº_{1},n_{1} = 1$

$Cu^{2 +} + 2e^{-} \rightarrow Cu;Eº_{2} = 0.34V,\Delta Gº_{2},n_{2} = 2$

$Cu^{+} + e^{-} \rightarrow Cu;Eº_{3} = ?,\Delta Gº_{3},n_{3} = 1$

$\Delta Gº_{3} = \Delta Gº_{2} - \Delta Gº_{1}$

$- n_{3}FEº_{3} = - n_{2}FEº_{2} + n_{1}FEº_{1}$

$- Eº_{3} = - 2 \times 0.34 + 1 \times 0.15$

$Eº_{3} = 0.68 - 0.15 = + 0.53V$

Question: \(Eº_{cell}\) for the half cell reactions are given below: \[Cu^{2 +} + e^{-} \rightarrow Cu^{+};Eº...