The emf of the cell reaction,
$Z n (s) + {C u}^{2 +} (a q .) \to {Z n}^{2 +} (a q .) + C u (s)$
is $1.1 V$ . Calculate free energy change for the reaction. If enthalpy of the reaction is $- 216.7 k J {m o l}^{- 1}$ , calculate the entropy change for the reaction.

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Solution

$- Δ G^{o} = n \times F \times E^{o} = 2 \times 96500 \times 1.1 = 212.3 k J$
$Δ G^{o} = - 212.3 k J {m o l}^{- 1}$
$Δ G^{o} = Δ H^{o} - T Δ S^{o}$
$Δ S^{o} = \frac{Δ H^{o} - Δ G^{o}}{T}$
$= \frac{- 216.7 - (- 212.3)}{298}$
$- 0.01476 k J K^{- 1} {m o l}^{- 1}$
$- 1476 J K^{- 1} {m o l}^{- 1}$

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Similar questions

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

1.1 V

- 216.7 k J {m o l}^{- 1}

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

1.1 V

- 216.7 k J {m o l}^{- 1}

Z n_{(s)} + C u_{(a q)}^{2 +} ⇌ Z n_{(a q)}^{2 +} + C u_{(s)}

= 1.1 v o l t

F = 96500

C u^{2 +} (C_{1}) (a q .) + Z n (s) \to Z n^{2 +} (C_{2}) (a q .) + C u (s)

(△ G)

(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}) .

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