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Electrochemical Series

For Zn 2+ /...

Question

For ${Z n}^{2 +} / Z n, E^{o} = - 0.76 V, f o r {A g}^{+} / A g, E^{o} = 0.799 V$ The correct statement is :

the reaction

Z n

getting reduced

A g

getting oxidized is spontaneous

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Z n

undergoes reduction and

A g

is oxidized

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Z n

undergoes oxidation and

{A g}^{+}

gets reduced

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no suitable answer

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Solution

The correct option is C $Z n$ undergoes oxidation and ${A g}^{+}$ gets reduced
The standard electrode potential for silver is higher than the standard electrode potential for zinc. This shows that silver ions have higher tendency to gain an electron than zinc ions. Hence, zinc is oxidized and silver ions are reduced.

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Similar questions

Z n^{+ 2} + 2 e^{-} ⇌ Z n; E^{o} = - 0.76 V

A g^{+} + e^{-} ⇌ A g; E^{o} = + 0.80 V

{Z n}^{2 +} / Z n

E^{\circ} = - 0.76 V

{A g}^{+} / A g E^{\circ} = 0.799 V

Z n^{2 +} / Z n, E^{\circ} = - 0.76 V

A g^{+} / A g, E^{\circ} = 0.799 V

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{o} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{o} = 1.10 V

Z n | Z n^{2 +} | | A g^{+} (1 M) | A g

25^{o} C

Z n

A g

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

A g (s) \to A g^{+} (a q) + e^{-}, E^{0} = - 0.80 V

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