Volume Strength of H₂O₂

What volume strength means. A sample labelled "$x$ volume" tells you this: 1 mL of this $\ce{H2O2}$ solution, on complete decomposition, releases $x$ mL of $\ce{O2}$ gas measured at STP.

So "10 volume $\ce{H2O2}$" gives 10 mL of $\ce{O2}$ per 1 mL of solution; "20 volume" is twice as strong. The number is a direct read-out of oxygen capacity — no calculation needed just to compare two bottles.

In the lab, $\ce{H2O2}$ is kept in dark bottles because light speeds up its decomposition:

$\ce{2H2O2 ->[\Delta] 2H2O + O2}$

From the reaction to the formula. Start from the decomposition: $\ce{2H2O2 -> 2H2O + O2}$. Here 68 g of $\ce{H2O2}$ produces 22,400 mL of $\ce{O2}$ at STP.

For an $x$-volume solution, 1 mL of solution releases $x$ mL of $\ce{O2}$. So the $\ce{H2O2}$ packed into that 1 mL is

$\frac{68x}{22400} = \frac{17x}{5600}\text{ g}$

Scale up to 1 litre: $\ce{H2O2}$ in 1000 mL $= \dfrac{17x}{5.6}$ g.

Now use Weight = Normality × Equivalent weight. The equivalent weight of $\ce{H2O2}$ is $\frac{34}{2} = 17$ (its n-factor is 2):

$\frac{17x}{5.6} = N \times 17 \;\Rightarrow\; \boxed{x = N \times 5.6}$

And since Normality = n-factor × Molarity = 2M:

$\boxed{x = M \times 11.2}$

The dual role. $\ce{H2O2}$ is unusual — it can act as both an oxidising agent and a reducing agent, in acidic and basic media. In every one of these cases the n-factor is 2, because $\ce{H2O2}$ always moves exactly 2 electrons.

As an oxidising agent ($\ce{H2O2 -> H2O}$; oxygen goes from $-1$ to $-2$):

• Acidic: $\ce{H2O2 + 2H+ + 2e- -> 2H2O}$
• Basic: $\ce{H2O2 + 2e- -> 2OH-}$

As a reducing agent ($\ce{H2O2 -> O2}$; oxygen goes from $-1$ to $0$):

• Acidic: $\ce{H2O2 -> O2 + 2H+ + 2e-}$
• Basic: $\ce{H2O2 + 2OH- -> O2 + 2H2O + 2e-}$

Classic exam fact: with $\ce{KMnO4}$ in acidic medium, $\ce{H2O2}$ behaves as a reducing agent — it reduces $\ce{MnO4-}$ ($\ce{Mn}$ from $+7$ to $+2$) and is itself oxidised to $\ce{O2}$.