A few drops of potassium iodide solution are added to a concentrated solution of hydrogen peroxide, where the iodide ion acts as a catalyst. The mixture rapidly decomposes with vigorous fizzing, releasing a stream of oxygen gas: 2H2O2(aq) → 2H2O(l) + O2(g) What is the oxidation…

AS Chemistry 9701: Oxidation Number of Oxygen in Hydrogen Peroxide (MCQ)

Step 1: Find the oxidation number of oxygen in $\text{H}_2\text{O}_2$

Rules used: the sum of oxidation numbers in a neutral molecule is $0$ , and hydrogen is $+1$ here (not bonded to a metal). Let $x$ be the oxidation number of oxygen:

$2(+1) + 2x = 0 \implies 2x = -2 \implies x = -1$

So oxygen has an oxidation number of $-1$ in $\text{H}_2\text{O}_2(aq)$ . This is the characteristic “peroxide” oxidation state, different from the more familiar $-2$ seen in water or most oxides, because the two oxygen atoms are joined directly by an $\text{O}-\text{O}$ bond between atoms of equal electronegativity, which contributes nothing to either atom’s oxidation number.

Step 2: Track oxygen’s oxidation number into each product

In $\text{H}_2\text{O}(l)$ : with hydrogen at $+1$ (two H atoms give $+2$ ) and the molecule neutral, oxygen must be $-2$ . Going from $-1$ (in $\text{H}_2\text{O}_2$ ) to $-2$ (in $\text{H}_2\text{O}$ ), oxygen gains an electron: this is reduction.

In $\text{O}_2(g)$ : oxygen is in its elemental form, so its oxidation number is $0$ . Going from $-1$ (in $\text{H}_2\text{O}_2$ ) to $0$ (in $\text{O}_2$ ), oxygen loses an electron: this is oxidation.

Step 3: Classify the overall redox process

Oxygen atoms that all started in the same reactant ( $\text{H}_2\text{O}_2$ , oxidation number $-1$ ) end up split between two different products: some are reduced (to $-2$ in $\text{H}_2\text{O}$ ) and others are oxidised (to $0$ in $\text{O}_2$ ). This simultaneous oxidation and reduction of the same element, from the same starting species, is called disproportionation.

Why the other options are wrong

A ( $-2$ ; oxidation only): gets the oxidation number wrong (it is $-1$ , not $-2$ , in a peroxide) and also misses that some of the oxygen is simultaneously reduced, not just oxidised.
C ( $-1$ ; reduction only): gets the oxidation number right, but only accounts for the oxygen ending up in $\text{H}_2\text{O}$ . It ignores the oxygen that is oxidised to $\text{O}_2$ .
D ( $-2$ ; disproportionation): correctly identifies disproportionation as the process, but starts from the wrong oxidation number for oxygen in $\text{H}_2\text{O}_2$ .

Final answer

Oxidation number of oxygen in $\text{H}_2\text{O}_2(aq)$ is $\boxed{-1}$ , and it undergoes disproportionation, option B.