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Electronic Configuration and Oxidation States

In the follow...

Question

In the following reaction using isotopic O in $H_{2} O_{2}$ , isotopic oxygen goes________

$2 M n O_{4} + 3 H_{2} O_{2}^{18} \to 2 M n O_{2} + 3 O_{2} + 2 H_{2} O + 2 O H^{-}$

with

O_{2}

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with

M n O_{2}

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with OH

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one wiiith

O_{2}

and one with

M n O_{2}

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Open in App

Solution

The correct option is A with $O_{2}$
Oxidation : $3 H_{2} O_{2}^{18} + 6 O H^{-} \to 3 O_{2}^{18} + 6 e^{-} + 6 H_{2} O$

Reduction: $2 M n O_{4}^{-} + 8 H_{2} O + 6 e^{-} \to 2 M n O_{2} + 4 H_{2} O + 8 O H^{-}$

So, option A is correct.

Suggest Corrections

Similar questions

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

2 C u^{+}

\to

C u^{2 +}

C u^{0}

3 M n O_{4}^{2 -}

4 H^{+}

\to

2 M n O_{4}^{-}

M n O_{2}

2 H_{2} O

2 K M n O_{4}

△ - \to

K_{2} M n O_{4}

M n O_{2}

O_{2}

2 M n O_{4}^{-}

3 M n^{2 +}

2 H_{2} O

⟶

5 M n O_{2}

4 H^{+}

C u^{+} \to C u^{2 +} + C u

3 M n O_{4}^{- 2} + 4 H^{+} \to 2 M n O_{4}^{-} + M n O_{2} + 2 H_{2} O

2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}

2 M n O_{4}^{-} + 3 M n^{+ 2} + 2 H_{2} O \to 5 M n O_{2} + 4 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

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