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Gibb's Free Energy

For the react...

Question

For the reaction, $Δ G^{0} = - 42927$ joules at $25^{0} C$ .
$H_{2} (g, 1 a t m) + 2 A g C l (s) ⇌ 2 A g (s) + 2 H^{+} (a q, 0.1 M) + 2 C l^{-} (a q, 0.1 M)$
Calculate the emf of the cell of given reaction.

A

0.340 V

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B

0.067 V

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C

1.49 V

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D

- 0.50 V

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Solution

The correct option is A $0.340 V$
We know,
$Δ G^{0} = - n F E^{0}$
$∴$
$E^{0} = - \frac{Δ G^{0}}{n F} = \frac{42927}{2 \times 96500} \approx 0.222 V$

Now, for the cell reaction,
$H_{2} (g, 1 a t m) + 2 A g C l (s) ⇌ 2 A g (s) + 2 H^{+} (a q, 0.1 M) + 2 C l^{-} (a q, 0.1 M)$

By nernst equation,
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g \frac{[H^{+}]^{2} [C l^{-}]^{2}}{p_{H_{2}}}$

$E_{c e l l} = 0.222 - \frac{0.0591}{2} l o g \frac{(0.1)^{2} (0.1)^{2}}{(1)}$

$E_{c e l l} = 0.222 - \frac{0.0591}{2} l o g 10^{- 4}$

$E_{c e l l} = 0.222 + \frac{0.0591}{2} \times 4$

$E_{c e l l} = 0.340 V$

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Similar questions

Q. The standard free energy change for the given reaction is -

10.26 k c a l m o l^{- 1}

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A cell using the above reaction is operated at

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Q. The standard free energy change for the given reaction is -

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Q. The standard free energy change for the reaction,

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Gibb's Free Energy

Standard XII Chemistry

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