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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.

0.340 V

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0.067 V

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1.49 V

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- 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

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

25^{\circ} C .

H_{2} (g) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q)

25^{\circ} C

P_{H_{2}} = 1 a t m

[H^{+}] and [C l^{-}] = 0.1 M

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

25^{\circ} C .

H_{2} (g) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q)

25^{\circ} C

P_{H_{2}} = 1 a t m

[H^{+}] and [C l^{-}] = 0.1 M

\frac{2.303 R T}{F} = 0.059, R = 8.314 J K^{- 1}, F = 96500 C

H_{2} (g) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (a q .) + 2 C l^{-} (a q .)

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

25^{\circ} C

25^{\circ} C

P_{H_{2}} = 1 a t m . [H^{+}]

[C l^{-}] = 0.1

2 A g C l (s) + H_{2} (g) (1 a t m) \to 2 A g (s) + 2 H^{+} (0.1 M) + 2 C l^{-} (0.1 M),

Δ G^{o} = - 43600

25^{o}

[l o g 10^{- n} = - n]

C d_{(s)} | C d C l_{2 (a q)} (0.1 M) | | A g C l_{(a q)} | A g_{(s)}

C d_{(s)} + 2 A g C l_{(s)} \Rightarrow 2 A g_{(s)} + C d_{(a q)}^{2 +} + 2 C l_{(a q)}^{-}

0.6915 V

0^{o} C

0.6573

V

25^{o} C

25^{o} C

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