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Electrode Potential

Calculate Δ...

Question

Calculate $Δ_{r} G$ for the reaction at $27^{o} C$ .
$H_{2} (g) + 2 {A g}^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)$
Given: $P_{H_{2}} = 0.5$ bar; $[{A g}^{+}] = 10^{- 5} M$ ;
$[H^{+}] = 10^{- 3} M; Δ_{r} G^{o} [{A g}^{+} (a q)] = 77.1 k J / m o l$

A

- 154.2 k J / m o l

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B

- 178.9 k J / m o l

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C

- 129.5 k J / m o l

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D

None of these

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Solution

The correct option is C $- 129.5 k J / m o l$
$Δ_{r} G^{o} = 0 - 77.1 = - 154.2 k J / m o l$
$Q = \frac{[H^{+}]^{2}}{P_{H_{2}} [A g^{+}]^{2}}$
$= \frac{10^{- 6}}{0.5 \times (10^{- 19})} = 2 \times 10^{4}$
$Δ G = Δ_{r} G^{o} + R T l n Q$
$Δ_{r} G = - 154.2 + \frac{8.314 \times 300 l n (2 \times 10^{4})}{1000}$
$= - 129.5 k J / m o l$

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Similar questions

Δ_{r}

G for the reaction at

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

p_{H_{2}}

0.5 bar; [A g^{+}] = 10^{- 5} M; log 2 =0.3 log 2= 0.7

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 kJ/mol

Δ_{r}

G for the reaction at

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

p_{H_{2}}

0.5 bar; [A g^{+}] = 10^{- 5} M; log 2 =0.3 log 2= 0.7

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 kJ/mol

Δ_{r} G

for the reaction at

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

P_{H_{2}} = 0.5 b a r; [A g^{+}] = 10^{- 5} M

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 k J m o l^{- 1}

(G i v e n, l n (2 \times 10^{4}) = 9.90)

Δ_{r} G

for the reaction at

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

P_{H_{2}} = 0.5 b a r; [A g^{+}] = 10^{- 5} M

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 k J m o l^{- 1}

(G i v e n, l n (2 \times 10^{4}) = 9.90)

Q. Calculate the equilibrium constant for the following reaction:

C u (s) + 2 {A g}^{+} (a q .) ⇌ {C u}^{2 +} (a q) + 2 A g (s)

25^{o} C

E_{c e l l}^{o} = 0.47 v o l t, R = 8.314 J K^{- 1} {m o l}^{- 1}, F = 96500 c o u l o m b

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