The potential of the given following cell is $0.092 v o l t$ . at $298 K$ temperature.
Calculate the $p H$ of $H C l$ soluton $(E_{S n / S n^{2 +}}^{\circ} = + 0.14 v o l t)$
$^{⊖} S n | S n^{2} (0.05 M) | | H^{+} (x M) | H_{2 (g)} (1 b a r) | P t^{\oplus}$ .

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Solution

Using Nernst equation for the cell reaction:
Anode: $S n \to S n^{2 +} + 2 e$ ; $E_{S n^{2 +} / S n}^{0} = - 0.14 V$
Cathode: $2 H^{+} + 2 e \to H_{2}$ ; $E_{H^{+} / H_{2}}^{0} = 0 V$
cell reaction:
$S n + 2 H^{+} \to S n^{2 +} + H_{2}$
$E_{c e l l} = E^{0} - \frac{R T}{n F} ln (\frac{[{S n}^{2 +}]}{{[H^{+}]}^{2}})$
Given: $E_{c e l l} = 0.092 V$ and $E^{0} = E_{H^{+} / H_{2}}^{0} - E_{S n^{2 +} / S n}^{0}$
$E^{0} = 0 - (- 0.14) = 0.14 V$
$E_{c e l l} = E^{0} - \frac{R T}{n F} ln (\frac{[{S n}^{2 +}]}{{[H^{+}]}^{2}})$
$0.092 = 0.14 - \frac{0.059}{2} ln (\frac{0.05}{x^{2}}) x = 0.1$
Calculation of pH:
$p H = - log ([H^{+}])$
$p H = - log (0.1) p H = 1$

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