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Gibbs Free Energy & Spontaneity

Calculate G...

Question

Calculate $△ G^{\circ}$ for the following cell reaction:
$Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ Z n^{2 +} (a q) + 2 A g (s) + 2 O H^{-} (a q)$
$E_{A g^{+} / A g}^{\circ} = + 0.80 V$ and $E_{Z n^{2 +} / Z N}^{\circ} = - 0.76 V$ .

- 305 k J / m o l

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- 301 k J / m o l

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+ 305 k J / m o l

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+ 301 k J / m o l

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Solution

The correct option is B $- 301 k J / m o l$

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Similar questions

Δ G

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}^{+} / A g}^{0} = + 0.80 V

E_{{Z n}^{+ 2} / Z n}^{0} = - 0.76 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 (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

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

E^{\circ}

2 A g + Z n^{2 +} \to 2 A g^{+} + Z n

E_{A g^{+} / A g}^{\circ} = 0.80 V, E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

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