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Volume of Gases and Number of Moles

What is the e...

Question

What is the equivalent weight of $O_{2}$ in the following reaction?
$H_{2} O + \frac{1}{2} O_{2} + 2 e^{-} ⟶ 2 O H^{-}$

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Solution

The given reaction is $H_{2} O + \frac{1}{2} O_{2} + 2 e^{-} ⟶ 2 O H^{-}$
Multiply this reaction with 2.
$2 H_{2} O + O_{2} + 4 e^{-} ⟶ 4 O H^{-}$
Thus, 1 mole of oxygen accepts 4 moles of electrons.
Hence, the equivalent weight of $O_{2}$ is one-fourth of molecular weight.
$Equivalent mass = \frac{Molecular mass}{Number of electrons transferred} = \frac{32}{4} = 8$
Hence, the equivalent weight of $O_{2}$ in the given reaction is 8.

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Similar questions

(a) O_{3} + 2 H^{+} + 2 e^{-} ⇌ O_{2} + H_{2} O

(b) O_{3} + H_{2} O + 2 e^{-} ⇌ O_{2} + 2 O H^{-}

O_{3}

H_{2} O_{2}

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

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

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

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

H_{2} O_{2}

H_{2} O_{2}

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

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

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

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

H_{2} O_{2}

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

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

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

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

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