# How do you balance the redox reaction? "KMnO"_4 + "H"_2"O"_2 + "H"_2"SO"_4 -> "MnSO"_4 + "K"_2"SO"_4 + "O"_2 + "H"_2"O" ?

Aug 10, 2015

$2 K M n {O}_{4} + 5 {H}_{2} {O}_{2} + 3 {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + {K}_{2} S {O}_{4} + 5 {O}_{2} + 8 {H}_{2} O$

#### Explanation:

Before you start balancing this equation, it's important to recognize that you're in acidic medium because the reaction take place in the presence of sulfuric acid.

This means that any protons that will remain unbalanced on the reactants' side will come from the sulfuric acid.

Moreover, not all the species are relevant to the reaction. More precisely, the potassium cation and the sulfate anion are spectator ions. The net ionic equation actually looks like this

$M n {O}_{4}^{-} + {H}_{2} {O}_{2} + 2 {H}^{+} \to M {n}^{2 +} + {O}_{2} + {H}_{2} O$

So, start by assigning oxidation numbers to all the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 1}}{K} \stackrel{\textcolor{b l u e}{+ 7}}{M} n {\stackrel{\textcolor{b l u e}{- 2}}{O}}_{4} + {\stackrel{\textcolor{b l u e}{+ 1}}{H}}_{2} \stackrel{\textcolor{b l u e}{+ 6}}{S} {\stackrel{\textcolor{b l u e}{- 2}}{O}}_{4} + {\stackrel{\textcolor{b l u e}{+ 1}}{H}}_{2} {\stackrel{\textcolor{b l u e}{- 1}}{O}}_{2} \to \stackrel{\textcolor{b l u e}{+ 2}}{M} n \stackrel{\textcolor{b l u e}{+ 6}}{S} {\stackrel{\textcolor{b l u e}{- 2}}{O}}_{4} + {\stackrel{\textcolor{b l u e}{+ 1}}{K}}_{2} \stackrel{\textcolor{b l u e}{+ 6}}{S} {\stackrel{\textcolor{b l u e}{- 2}}{O}}_{4} + {\stackrel{\textcolor{b l u e}{0}}{O}}_{2} + {\stackrel{\textcolor{b l u e}{+ 1}}{H}}_{2} \stackrel{\textcolor{b l u e}{- 2}}{O}$

Notice that the oxidation number of mangenese changes from $\text{+7}$ on the reactants' side to $\text{+2}$ on the products' side, which means that mangenese is being reduced.

On the other hand, the oxidation number of oxygen in hydrogen peroxide, ${H}_{2} {O}_{2}$, changes from $\text{-1}$ on the reactants' side to $\text{0}$ in oxygen gas, ${O}_{2}$, on the products' side.

This means that oxygen is being oxidized. The two half-reactions will look like this

• reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 7}}{M} n {O}_{4}^{-} + 5 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{M} {n}^{2 +}$

Here, each manganese atom gains 5 electrons.

Since you're in acidic medium, you can balance the oxygen atoms by adding water to the side that needs oxygen and the hydrogen atoms by adding protons, ${H}^{+}$, to the side that needs hydrogen.

The reduction half-reaction will thus be

$8 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 7}}{M} n {O}_{4}^{-} + 5 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{M} {n}^{2 +} + 4 {H}_{2} O$

• oxidation half-reaction

${H}_{2} {\stackrel{\textcolor{b l u e}{- 1}}{O}}_{2} \to {\stackrel{\textcolor{b l u e}{0}}{O}}_{2} + 2 {e}^{-}$

Here, each oxygen atom loses one electron, so two oxygen atoms will lose 2 electrons.

Once again, balance the oxygen and hydrogen accordingly

${H}_{2} {\stackrel{\textcolor{b l u e}{- 1}}{O}}_{2} \to {\stackrel{\textcolor{b l u e}{0}}{O}}_{2} + 2 {e}^{-} + 2 {H}^{+}$

Now, in a redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.

This means that you need to multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 to get a total of $10 {e}^{-}$ transferred in the reaction.

$\left\{\begin{matrix}16 {H}^{+} + 2 M n {O}_{4}^{-} + 10 {e}^{-} \to 2 M {n}^{2 +} + 8 {H}_{2} O \\ 5 {H}_{2} {O}_{2} \to 5 {O}_{2} + 10 {e}^{-} + 10 {H}^{+}\end{matrix}\right.$

Add these two half-reactions to get

$16 {H}^{+} + 2 M n {O}_{4}^{-} + \textcolor{red}{\cancel{\textcolor{b l a c k}{10 {e}^{-}}}} + 5 {H}_{2} {O}_{2} \to 2 M {n}^{2 +} + \textcolor{red}{\cancel{\textcolor{b l a c k}{10 {e}^{-}}}} + 5 {O}_{2} + 10 {H}^{+} + 8 {H}_{2} O$

Finally, reduce the number of protons to get

$6 {H}^{+} + 2 M n {O}_{4}^{-} + 5 {H}_{2} {O}_{2} \to 2 M {n}^{2 +} + 5 {O}_{2} + 8 {H}_{2} O$

Now, all those protons are coming from the sulfuric acid, which means that the balanced chemical equations will be - add the spectator ions, too

$2 K M n {O}_{4} + 5 {H}_{2} {O}_{2} + \textcolor{red}{3} {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + {K}_{2} S {O}_{4} + 5 {O}_{2} + 8 {H}_{2} O$

SIDE NOTE This is actually the half-reaction method, not the oxidation numbers method. The oxidation numbers method does not use water and protons (for reactions that take place in acidic medium) to balance hydrogen and oxygen.

To balance this equation using the oxidation number method, figure out the number of electrons transferred, which is of course equal to 10, multiply the compounds so that these electrons are balanced, then balance the rest of the atoms by inspection.

So, you know that you need to multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$. This means that you have

$2 K M n {O}_{4} + 5 {H}_{2} {O}_{2} + {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + {K}_{2} S {O}_{4} + 5 {O}_{2} + {H}_{2} O$

Now balance the equation by balancing the number of sulfate anions present on both sides of the equation. Notice that you have one sulfate anion on the reactants' side and three on the products side, so multiply sulfuric acid by $3$ to get

$2 K M n {O}_{4} + 5 {H}_{2} {O}_{2} + 3 {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + {K}_{2} S {O}_{4} + 5 {O}_{2} + {H}_{2} O$

Finally, balance hydrogen and oxygen by multiplying the water by $8$

$2 K M n {O}_{4} + 5 {H}_{2} {O}_{2} + 3 {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + {K}_{2} S {O}_{4} + 5 {O}_{2} + 8 {H}_{2} O$

And there you have it.