# What is equivalent weight of H2O2 in decomposition reaction?

##### 1 Answer
Jun 12, 2018

The equivalent weight of ${\text{H"_2"O}}_{2}$ is 34.01 g.

#### Explanation:

The equivalent weight is the molecular weight ($\text{MM}$) divided by the number of electrons transferred per mole of reactant ($n$).

color(blue)(bar(ul(|color(white)(a/a)"Equivalent weight" = "MM"/ncolor(white)(a/a)|)))" "

We need the moles of electrons, so let's balance the equation by the ion-electron method.

Unbalanced equation: ${\text{H"_2"O"_2 → "H"_2"O" + "O}}_{2}$

Reduction: 1 × ["H"_2"O"_2 + "2H"^"+" + 2"e"^"-"→ "2H"_2"O" ]
Oxidation: 1 × ul(["H"_2"O"_2 → "O"_2 + "2H"^"+" + "2e"^"-"]color(white)(mm))
Overall: $\textcolor{w h i t e}{m m m m} 2 \text{H"_2"O"_2 → "O"_2 + "2H"_2"O}$

We see that the reaction involves the transfer of 2 mol of electrons.

Since the reaction also involves 2 mol of ${\text{H"_2"O}}_{2}$, the number of electrons transferred per mole of ${\text{H"_2"O}}_{2}$ is

n = ("2 mol e"^"-")/("2 mol H"_2"O"_2) = 1color(white)(l) "mol electrons per mole of H"_2"O"_2"

$\text{Equivalent weight" = "MM"/n = "34.01 g"/1 = "34.01 g}$

The equivalent weight of ${\text{H"_2"O}}_{2}$ in this reaction is 34.01 g.