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Redox Reactions and Couple

For which of ...

Question

For which of these redox pairs, the reduction potential vary with pH

A m O_{2}^{2 +} / A m O_{2}^{+}

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A m O_{2}^{2 +} / A m^{4 +}

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M n O_{4}^{-} / M n^{2 +}

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Z n^{2 +} / Z n

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Solution

The correct options are
B $A m O_{2}^{2 +} / A m^{4 +}$
C $M n O_{4}^{-} / M n^{2 +}$
The half reaction involves H $^{+}$ only in
$4 H^{+} + A m O_{2}^{2 +} + 2 e \to A m^{4 +} + 2 H_{2} O$
and $M n O_{4}^{-} + 8 H^{+} + 5 e \to M n^{2 +} + 4 H_{2} O$
(according to oxygen balance)

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Similar questions

p H

A m O_{2}^{2 +} / A m O_{2}^{+}

A m O_{2}^{2 +} / A m^{4 +}

A m^{4 +} / A m^{2 +}

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

Divide the following redox reactions into oxidation and reduction half reactions.

(i) $Z n + P b^{2} \to Z n^{2} + P b$

(ii) $Z n + C u^{2 +} \to Z n^{2} + C u$

(iii) $C l_{2} + 2 B r^{-} \to B r_{2} + 2 C l^{-}$

E^{o}

Z n ⟶ {Z n}^{2 +} + 2 e^{-}

E^{o} = + 0.76 V

F e ⟶ {F e}^{2 +} + 2 e^{-}

E^{o} = + 0.41 V

{F e}^{2 +} + Z n ⟶ {Z n}^{2 +} + F e

E^{o}

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

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

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

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