Hydrogen peroxide ( $H_{2} O_{2}$ $H_2O_2$ ) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?

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Hydrogen peroxide ( $H_{2} O_{2}$ $H_2O_2$ ) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?

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Answer 1 · 2015-12-20T20:18:14

$2 H_{2} O_{2} (l) ⇌ 2 H_{2} O (l) + O_{2} (g)$ $2H_2O_2(l) rightleftharpoons 2H_2O(l) + O_2(g)$

Explanation:

This is in fact a disproportionation reaction in that oxygen in peroxide ( $- I$ $-I$ oxidation state) has given water ( $- I I$ $-II$ oxidation state for oxygen) and zerovalent oxygen gas ( $0$ $0$ oxidation state for oxygen).

Often a $M n^{2 +}$ $Mn^(2+)$ salt is added to catalyze this reaction.

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Hydrogen peroxide ( $H_{2} O_{2}$ $H_2O_2$ ) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?

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Explanation:

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Hydrogen peroxide (H2O2H_2O_2) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?

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Hydrogen peroxide ( $H_{2} O_{2}$ $H_2O_2$ ) decomposes to form water and oxygen. How do you write the balanced equation for this reaction?