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Gibb's Energy and Nernst Equation

Calculate the...

Question

Calculate the pressure of hydrogen passed through cathode half cell at $298 K$ .
$P t (s) | H_{2} (g, 800 t o r r) | H C l (a q) | H_{2} (g, p_{2}) | P t (s)$
Given:
$E_{c e l l} = 0.018 V 10^{2.3} \approx 200$

p_{2} = 200 t o r r

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p_{2} = 400 t o r r

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p_{2} = 100 t o r r

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p_{2} = 300 t o r r

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Solution

The correct option is A $p_{2} = 200 t o r r$
From the cell representation,
Anode reaction:
$H_{2} (800 t o r r) \to 2 H^{+} + 2 e^{-}$
Cathode reaction:
$2 H^{+} + 2 e^{-} \to H_{2} (p_{2})$
Thus, the cell reaction is
$H_{2} (800 t o r r) \to H_{2} (p_{2})$

By Nernst equation,
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g \frac{p_{2}}{800}$
$E_{c e l l}^{0} = 0$
$E_{c e l l} = - \frac{0.0591}{2} l o g \frac{p_{2}}{800}$

$0.018 = \frac{0.0591}{2} l o g \frac{800}{p_{2}}$

$0.61 = l o g \frac{800}{p_{2}}$

$0.61 = l o g 800 - l o g p_{2}$

$0.61 = 2.90 - l o g p_{2}$

$2.3 \approx l o g p_{2}$

$p_{2} \approx 200 t o r r$

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