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

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 -

#MnO_4^(-) + H^(+) + HSO_3^(-) -> Mn^(2+) + SO_4^(2-) + H_2O#

Let's assign oxidation numbers for each atom involved.

For #"MnO"_4^(-) -># #"O"#: -2, #"Mn"#: +7 - remember that the sum of the oxidation numbers for each atom must equal the charge of the molecule. In this case, #4*(-2) + 7 =-1#, which matches the negative charge on the molecule.
For #"H"^(+)-># +1;
For #"HSO"_3^(-) -># #"H"#: +1, #"O"#: -2, #"S"#: +4;
For #"Mn"^(2+) -># +2;
For #"SO"_4^(2-) -># #"O"#: -2, #"S"#: +6;
For #"H"_2"O" -># #"H"#: +1, #"O"#: -2.

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

The half-reactions are as follows:

#MnO_4^(-) + 5e^(-) -> Mn^(2+)# (1)
#HSO_3^(-) -> SO_4^(2-) + 2e^(-)# (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_2O# on that side, and #2H^(+)# on the other side of the equation. So (1) becomes

#8H^(+) + MnO_4^(-) + 5e^(-) -> Mn^(2+) + 4H_2O# (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_2O + HSO_3^(-) -> SO_4^(2-) + 2e^(-) + H^(+) + 2H^(+)#, which is equivalent to

#H_2O + HSO_3^(-) -> SO_4^(2-) + 2e^(-) + 3H^(+)# (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

#16H^(+) + 2MnO_4^(-) + 10e^(-) -> 2Mn^(2+) + 8H_2O#
#5H_2O + 5HSO_3^(-) -> 5SO_4^(2-) + 10e^(-) + 15H^(+)#

Add these last two equations and you'll get

#H^(+) + 2MnO_4^(-) + 5SO_3^(-) -> 2Mn^(2+) + 5SO_4^(2-) + 3H_2O#

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:

http://redoxanswers.weebly.com/balancing-redox-reactions.html

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

Your unbalanced equation is

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