# Redox reactions: identifying what is oxidized vs reduced

$$\ce{SO4^2- + H2O2 + 2H+ -> SO2 + O2 + 2H2O}$$

In the above chemical reaction, I'm having trouble identifying what is oxidized and what is reduced.

My guess is: sulfur has reduced but I can't work out what has oxidised. My guess is the middle oxygen as it has lost two hydrogen atoms to $\ce{2H2O}$.

• The sulphate ion and the hydrogen peroxide are reduced , the hydrogen is oxidised .... – Technetium Jun 12 '16 at 2:24
• I think there would be more steps to the actual reaction tho .... – Technetium Jun 12 '16 at 2:31
• When the species cannot be found in redox tables (or the reaction scheme is just for illustration as in the question, e.g. SO$_2$ does not exist in solution as its a gas), using oxidation numbers is the only simple way. These values can be zero, positive, negative or fractional, e.g S in S$_4$O$_6$ or Fe in Fe$_3$O$_4$. In the question S has a value +6 (6-8*2=-2) in SO$_4$ and +4 in SO$_2$, as the total must be zero in a neutral species. As a decrease in oxidation number corresponds to reduction the sulphate ion is reduced and the peroxide is oxidised. – porphyrin Jul 12 '16 at 19:58

Yes sulfur of $\ce{SO4^2-}$ has undergone reduction as the oxidation number reduces from +6 to +4 and Oxygen of $\ce{H2O2}$ has undergone oxidation as its oxidation number increases from -1 to 0.

$$\ce{\overset{\color{red}{+6}}{S}O4^{2-}\+H2\overset{\color{red}{-1}}{O2}\+2H+->\overset{\color{red}{+4}}{S}O2\+\overset{\color{red}{0}}{O2}\+2H2O}$$

NOTE: Oxidation sate of oxygen in peroxide is -1. This because in the neutral molecule hydrogen peroxide the oxidation state of $\ce{H}$ is +1 and as the sum of oxidation numbers of the two hydrogen atoms and the two oxygen atoms must be zero so each oxygen in $\ce{H2O2}$ has -1 oxidation state.

You can use formal charges, but that's exactly what they are: a formality, not actual charges.

A redox reaction is a redox couple, consisting of an oxidation half reaction and a reduction half reaction. I would consult a table of reduction potentials (e.g.), which lists half-reactions and their standard reduction potentials.

On this table, we find: \begin{align} \ce{O2(g) + 2 H+ (aq) + 2 e– &-> H2O2 (aq)},& E^{\circ} &= \pu{+0.682 V}\\ \ce{SO4^2– (aq) + 4 H+(aq) + 2 e– &-> SO2 (g) + 2 H2O},& E^{\circ} &= \pu{+0.2 V}\\ \end{align}

If I invert the first reaction, I have the following oxidation: \begin{align} \ce{H2O2(aq) &-> O2 (g) + 2 H+ (aq) + 2 e–},& E^{\circ} &= \pu{-0.682 V}\\ \end{align}

Summing this with the second half reaction (reduction) gives the total [redox] reaction provided in your question.

This tells us two things. Hydrogen peroxide is being oxidized (that system is losing electrons), since it is the reactant in the oxidation half reaction. Sulfate and hydrogen ion are participating as reactants in the reduction half reaction; they are reduced.

We didn't actually need to do anything with potentials here, but you can easily see that the total reaction is non-spontaneous.

The oxygen atoms on sulfate ion are not normally redox-active, whereas the oxygen in hydrogen peroxide could be redox-active given that they act that way in other reactions. So I would select the hydrogen peroxide oxygens as the reducing agent here.