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Mole Concept and Avogadro's Number

a O3 + 2H+ + ...

Question

$(a) O_{3} + 2 H^{+} + 2 e^{-} ⇌ O_{2} + H_{2} O$
$(b) O_{3} + H_{2} O + 2 e^{-} ⇌ O_{2} + 2 O H^{-}$
what is the equivalent weight of $O_{3}$ in (b)?

48 g

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40 g

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24 g

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20 g

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Solution

The correct option is C $24 g$
equivalent weight of $O_{3}$ in a &b reactions is
$Equivalent weight in (a) = \frac{m o l e c u l a r w e i g h t}{number of electrons involved in reaction}$
$Equivalent weight in (a) = \frac{48}{2} = 24 g e q u i v a l e n t^{- 1} Equivalent weight in (b) = \frac{48}{2} = 24 g e q u i v a l e n t^{- 1}$
Hence option C is correct.

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Similar questions

O_{2}

H_{2} O + \frac{1}{2} O_{2} + 2 e^{-} ⟶ 2 O H^{-}

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

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}

(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

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