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Gibb's Free Energy

The zinc/silv...

Question

The zinc/silver oxide cell is used in electric watches. The reaction is as follows
$Z n^{+ 2} + 2 e ⟶ Z n; E^{0} = - 0.76 V$
$A g_{2} O + H_{2} O ⟶ 2 A g + 2 O H^{-}; E^{0} = 0.344 V$
If $F$ is $96500 m o l^{- 1}$ , then $Δ G^{0}$ of the cell is:

413 k J m o l^{- 1}

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113 k J m o l^{- 1}

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213 k J m o l^{- 1}

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313 k J m o l^{- 1}

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Solution

The correct option is D $213 k J m o l^{- 1}$
$Δ G^{o} = - n F E_{c e l l}^{o}$
$n$ = number of $e^{-}$ transferred= $2$
$E_{c e l l}^{o} = E_{c a t h o d e} - E_{a n o d e}$
$= 0.344 - (- 0.76)$
$= 1.104 V$
$Δ G^{o} = - 2 \times 96500 \times 1.104 = 213 k J m o l^{- 1}$

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Similar questions

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

Z n^{2 +} (a q) + 2 e \to Z n (s); E^{0} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) 2 e \to 2 A g (s) + 2 O H^{-} (a q), E^{0} = 0.34 V

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{\circ} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) + 2 e^{-} \to 2 A g (s) + 2 O H^{-} (a q), E^{\circ} = 0.34 V

Δ_{r} G^{o}

Z n_{(s)} + A g_{2} O_{(s)} + H_{2} O_{(l)} \to Z n_{(a q)}^{2 +} + 2 A g_{(s)} + 2 O H_{(a q)}^{-}

[E_{A g_{2} O / A g}^{o} = 0.344 V, E_{Z n^{2 +} / Z n}^{o} = - 0.76 V]

Z n \to Z n^{+ +} + 2 e^{-}, E^{0} = 0.76 V

C u \to C u^{+ +} + 2 e^{-}, E^{0} = 0.34 V

Z n \to Z n^{2 +} + 2 e^{-} (E_{(Z n^{2 +} / Z n)}^{0} = - 0.76 V)

F e \to F e^{2 +} + 2 e^{-} E^{0} = 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

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