The reaction - Zn 2- aq - 2e - - Zn s has a electrode potential of 0-76 V This means-

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Valency and Oxidation State

The reaction ...

Question

The reaction : ${Z n}^{2 +} (a q) + {2 e}^{-} \to Z n (s)$ has a electrode potential of $0.76 V$ This means:

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{Z n}^{+ 2} + {2 e}^{-} ⟶ Z n, E^{o} = - 0.76 V

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

F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V

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

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^{2 +} / Z n

- 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^{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

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