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Gibb's Free Energy

Mgs|Mg2+0.1||...

Question

$M g (s) | M g^{2 +} (0.1) | | A g^{+} (1 \times 10^{- 3}) | A g$
$E_{A g^{+} / A g}^{\circ} = + 0.8 volt, E_{M g^{2 +} / M g}^{\circ} = - 2.37 volt$
Calculate the emf of the cell (Round up the answer to two decimal places)
( $\frac{R T}{F} = 0.0591 V$ )

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Solution

$E_{cell}^{\circ} = E_{C a t h o d e}^{\circ} - E_{A n o d e}^{\circ}$
$= 0.80 - (- 2.37) = 3.17$ volt
Cell reaction, $M g + 2 A g^{+} \to 2 A g + M g^{2 +}$
$E_{cell} - \frac{0.0591}{n} log \frac{M g^{2 +}}{[A g^{+}]^{2}}$
$= 3.17 - \frac{0.0591}{2} \times log 10^{2} = 3.17 - 0.0591 = 3.11 V$

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