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Equilibrium Constant and Standard Free Energy Change

The standard ...

Question

The standard electrode potential of a Daniell cell is $1.1 V$ . Find the standard Gibbs energy for the reaction.
$Z n (s) + C u^{2 +} (a q .) \to Z n^{2 +} (a q .) + C u (s)$

- 212.3 kJ/mol

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- 2.123 kJ/mol

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- 21.23 kJ/mol

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- 21230 kJ/mol

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Solution

The correct option is A $- 212.3 kJ/mol$
For the reaction $n = 2$
$F = 96500 C/mol$ and $E = 1.1 V$
$∵ Δ_{r} G^{o} = - 2 \times 1.1 \times 96500 = - 212300 J/mol = - 212.3 kJ/mol$

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Similar questions

1.1

Z n (s) | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u (s)

= 96487

(E^{\circ})

Δ G^{\circ}

Z n (s) + C u^{2 +} (a q) ⟶ Z n^{2 +} (a q) + C u (s)

(1 F = 96500 C m o l^{- 1}) .

(E^{\circ})

+ 1.1 V

Δ G^{\circ}

Z n (s) + C u_{(a q)}^{2 +} \to Z n_{(a q)}^{2 +} + C u (s)

(1 F = 96500 C / m o l)

{Z n}^{2 +} | Z n

- 0.76 V

{C u}^{2 +} | C u

0.34 V

(V)

(k J {m o l}^{- 1})

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

0.350

C u^{2 +} (a q) + 2 e^{-} \to C u

- 0.763

Z n^{2 +} (a q) + 2 e^{-} \to Z n

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