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Balancing Redox Reaction

The half reac...

Question

The half reaction which correctly represents the change that takes place in the reaction between $H_{2} O_{2}$ and alkaline potassium ferricyanide is?

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

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H_{2} O_{2} \to 2 H^{-} + O_{2} + 2 e^{-}

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H_{2} O_{2} \to 2 H^{-} + O_{2} + 2 H_{2} O + 2 e^{-}

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H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

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Solution

The correct option is A $H_{2} O_{2} \to 2 H^{-} + O_{2} + 2 e^{-}$
$H_{2} O_{2}$ should get oxidised as it acts as a reducing agent. Hence, $O_{2}$ should formed as one of the products of this half reaction $($ as it would then be an oxidation half reaction as oxygen gets oxidised from $- 1$ to $0)$ . Therefore, the reaction is: $H_{2} O_{2} \to 2 H^{-} + O_{2} + 2 e^{-}$ .

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H_{2} O_{2}

(a)

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

(b)

H_{2} O_{2} - 2 e^{-} \to O_{2} + 2 H^{+}

(c)

H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

(d)

H_{2} O_{2} + 2 O H^{-} - 2 e^{-} \to O_{2} + 2 H_{2} O

H_{2} O_{2}

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