How do you balance the acid equation #"MnO"_4^"-" + "H"^"+" + "HSO"_3^"-" → "Mn"^"2+" + "SO"_4^"2-" + "H"_2"O"#?
2 Answers
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 -
Let's assign oxidation numbers for each atom involved.
For
For
For
For
For
For
Notice that
The half-reactions are as follows:
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
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
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
Add these last two equations and you'll get
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