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We have cell ...

Question

We have cell
$C u / C u^{2 +} / / A g^{+} / A g$
Reactions are given as below
$C u^{2 +} + e \to C u^{+} {E_{1}}^{\circ} = 0.15 V$
$C u^{+} + e \to C u {E_{2}}^{\circ} = 0.5 V$
$A g^{+} + e \to A g {E_{3}}^{\circ} = 0.799 V$
Calculate Gibbs free Energy.

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Solution

we have cell
$C u / C u^{2 +} / / A g^{+} / A g$
oxidation half reaction
$C u \to C u^{2 +} + 2 e^{-} . . . . (1)$
Reduction half reaction
$\begin{matrix} A g^{+} + e^{-} \to A g 2 A g + 2 e^{-} \to 2 A g . . . (2) \end{matrix}$
So, cell reactions
$C u + 2 A g^{+} \to C u^{2 +} + 2 A g . . . (3)$
we have
$C u^{2 +} + e \to C u^{+} E_{1}^{\circ} = 0.15 V . . . (4)$
$\begin{matrix} Δ G_{4} = - n E F = - 1 E^{\circ} F = - 0.15 F C u^{+} + e \to C u E_{2}^{\circ} = 0 .5 V . . . . (5) \end{matrix}$
$\begin{matrix} Δ G_{5} = - n E^{\circ} F = (- 1) E^{\circ} F Δ G_{5} = - 0.5 F \end{matrix}$
$\begin{matrix} S o a d d i n g e q u a t i o n s (4) & (5) w e g e t C u^{2 +} + 2 e^{-} \to C u . . . . (6) \end{matrix}$
$\begin{matrix} Δ G_{6} = Δ G_{4} + Δ G_{5} = - 0.15 F + (- 0.5 F) (- 2) E^{\circ} F = - 0.65 F E^{\circ} = 0.325 V \end{matrix}$
We have $E^{\circ} = 0.325 V$
equation
$C u^{2 +} + 2 e^{-} \to C u E^{\circ} = 0.325 V$
which is reduction potention of electrode on left side
we have $\begin{matrix} A g^{+} + e^{-} \to A g E_{3}^{\circ} = 0.799 V S o, 2 A g^{+} + 2 e^{-} \to 2 A g \end{matrix}$
$E_{3}^{\circ}$ remains as it is $E_{3}^{\circ} = 0.799 V$
which is reduction potential of right electrodes
so, we have $\begin{matrix} E c e l l = E_{R}^{\circ} - E_{L}^{\circ} = 0.799 - 0.325 E c e l l = 0.474 V \end{matrix}$
which is potential of cell that is for cell reaction equation..(3)
so $\begin{matrix} Δ G_{3} = (- 2) E^{\circ} F = - 2 \times 0.474 \times 96500 Δ G_{3} = - 91482 J / m o l e \end{matrix}$

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