# How do you balance the acid equation "MnO"_4^"-" + "H"^"+" + "HSO"_3^"-" → "Mn"^"2+" + "SO"_4^"2-" + "H"_2"O"?

Jan 16, 2015

I assume the "(acid)" means you must balance this equation in an acid aqueous solution. The answer will be quite long, but I think it'll be worth it.

Let's start with the equation given to you - all the species are in aqueous solution, except water, which is in liquid state, so I won't write (aq) for each of them -

$M n {O}_{4}^{-} + {H}^{+} + H S {O}_{3}^{-} \to M {n}^{2 +} + S {O}_{4}^{2 -} + {H}_{2} O$

Let's assign oxidation numbers for each atom involved.

For ${\text{MnO}}_{4}^{-} \to$ $\text{O}$: -2, $\text{Mn}$: +7 - remember that the sum of the oxidation numbers for each atom must equal the charge of the molecule. In this case, $4 \cdot \left(- 2\right) + 7 = - 1$, which matches the negative charge on the molecule.
For ${\text{H}}^{+} \to$ +1;
For ${\text{HSO}}_{3}^{-} \to$ $\text{H}$: +1, $\text{O}$: -2, $\text{S}$: +4;
For ${\text{Mn}}^{2 +} \to$ +2;
For ${\text{SO}}_{4}^{2 -} \to$ $\text{O}$: -2, $\text{S}$: +6;
For $\text{H"_2"O} \to$ $\text{H}$: +1, $\text{O}$: -2.

Notice that $\text{Mn}$ goes from +7 on the reactants' side, to +2 on the products side, which means it has been reduced, while $\text{S}$ goes from +4 on the reactants' side, to +6 on the products' side - it has been oxidized.

The half-reactions are as follows:

$M n {O}_{4}^{-} + 5 {e}^{-} \to M {n}^{2 +}$ (1)
$H S {O}_{3}^{-} \to S {O}_{4}^{2 -} + 2 {e}^{-}$ (2)

Notice that in (1), you are 4 oxygen atoms short on the products' side. In acid aqueous solution, for every oxygen atom you need, you must add ${H}_{2} O$ on that side, and $2 {H}^{+}$ on the other side of the equation. So (1) becomes

$8 {H}^{+} + M n {O}_{4}^{-} + 5 {e}^{-} \to M {n}^{2 +} + 4 {H}_{2} O$ (1.1)

In (2), you need an oxygen on the reactants' side. You also need a hydrogen on the products' side, which means that you must add an ${H}^{+}$ for every missing hydrogen atom on that respective side of the equation. Thus, (2) becomes

${H}_{2} O + H S {O}_{3}^{-} \to S {O}_{4}^{2 -} + 2 {e}^{-} + {H}^{+} + 2 {H}^{+}$, which is equivalent to

${H}_{2} O + H S {O}_{3}^{-} \to S {O}_{4}^{2 -} + 2 {e}^{-} + 3 {H}^{+}$ (2.1)

Now you can focus on balancing the electrons lost and gained. In order to balance out these electrons, you must multiply equation (1.1) by 2 and equation (2.1) by 5. This will get you

$16 {H}^{+} + 2 M n {O}_{4}^{-} + 10 {e}^{-} \to 2 M {n}^{2 +} + 8 {H}_{2} O$
$5 {H}_{2} O + 5 H S {O}_{3}^{-} \to 5 S {O}_{4}^{2 -} + 10 {e}^{-} + 15 {H}^{+}$

Add these last two equations and you'll get

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

That represents your balanced chemical equation.

As a conclusion, you must remember to balance oxygen and hydrogen atoms first when in aqueous solution, then proceed to the electrons lost and gained.

Here's a link to a whole bunch of other examples:

Jan 17, 2015

For one method, see How do you balance redox equations by oxidation number method?

MnO₄⁻ + H⁺ + HSO₃⁻ → Mn²⁺ + SO₄²⁻+ H₂O

Step 1. The oxidation numbers are:

Left hand side: Mn =+7;O = -2; H= +1; S = +4
Right hand side: Mn = +2; S = +6; O = -2; H = +1

Step 2. The changes in oxidation number are:

Mn: +7 → +2; Change = -5
S: +4 → +6; Change = +2

Step 3. Equalize the changes in oxidation number.

You need 5 atoms of S for every 2 atoms of Mn. This gives us total changes of +10 and -10.

Step 4. Insert coefficients to get these numbers.

2 MnO₄⁻ + H⁺ + 5 HSO₃⁻ → 2 Mn²⁺ + 5 SO₄²⁻+ H₂O

Step 5. Balance O by adding H₂O molecules to the appropriate side.

2 MnO₄⁻ + H⁺ + 5 HSO₃⁻ → 2 Mn²⁺ + 5 SO₄²⁻+ 3 H₂O

Step 6. Balance H by adding H⁺ ions to the appropriate side.

2 MnO₄⁻ +1 H⁺ + 5 HSO₃⁻ → 2 Mn²⁺ + 5 SO₄²⁻+ 3 H₂O

Step 7. Check that all atoms balance.

Left hand side: 2Mn; 23 O; 6 H; 5 S
Right hand side: 2 Mn; 5 S; 23 O; 6H

Step 8. Check that all charges balance.

Left hand side: 2- + 1+ + 5- = 6-
Right hand side: 4+ + 10- = 6-

The balanced equation is

2MnO₄⁻ +H⁺ + 5HSO₃⁻ → 2Mn²⁺ + 5SO₄²⁻+ 3H₂O