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Gibb's Energy and Nernst Equation

At 25∘ C, the...

Question

At $25^{o} C$ , the free energy of formation of $H_{2} O (l)$ is $- 61076 c a l / m o l$ . The free energy of ionisation of water to $H^{+}$ and $O H^{-}$ is $19050 c a l / m o l$ . The reversible EMF (in V) of the following cell at $25^{o} C$ is:
$H_{2} (g) (1 a t m) | H^{+} | | O_{2} (g) (1 a t m) | O H^{-}$

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Solution

Given:
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⇋ H_{2} O (l)$ --(i)
$H_{2} O ⇋ H^{+} + O H^{-}$ --(ii)

$△ G_{1} = - 61076 c a l / m o l$ , $△ G_{2} = 19050 c a l / m o l$
Cell reaction will be:
at anode: $H_{2} ⇋ 2 H^{+} + 2 e^{-}$
at cathode: $\frac{1}{2} O_{2} + H_{2} O + 2 e^{-} ⇋ 2 O H^{-}$
so overall reaction is: $H_{2} + \frac{1}{2} O_{2} + H_{2} O + ⇋ 2 H^{+} + 2 O H^{-}$

multiply Eq.(ii) by 2 and adding eq.(i), we get:
$H_{2} + \frac{1}{2} O_{2} + H_{2} O + ⇋ 2 H^{+} + 2 O H^{-}$
$∴ △ G = △ G_{1} + 2 △ G_{2}$
$= - 61076 + 2 \times 19050$
$= - 22976 c a l / m o l$
$= - 22976 \times 4.2 = - 96499.2 J / m o l$ or $= 96500 J / m o l$

Now, $△ G = - n F E_{c e l l}$
$- 96500 = - 2 \times 96500 \times E_{c e l l}$
$∴ E_{c e l l} = 0.5 V$

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