# How to balance equations for reduction/oxidation reaction?

Dec 30, 2013

One way is to use the ion-electron method.

#### Explanation:

EXAMPLE in ACID SOLUTION

Write the balanced chemical equation for the reaction between potassium permanganate and iodide ion in aqueous sulfuric acid to form potassium iodide and manganese(II) sulfate?

1. Write the word equation.

Potassium permanganate + potassium iodide + sulfuric acid → iodine + potassium sulfate

2. Write the skeleton “molecular” equation.

${\text{KMnO"_4 + "KI" + "H"_2"SO"_4 → "I"_2 + "MnSO}}_{4}$

3. Determine the oxidation numbers of each atom on both sides of the equation.

Left hand side: $\text{K = +1; Mn = +7; O = -2; I = 0; H = +1; S = +6}$
Right hand side: $\text{I = 0; Mn = +2, S = +6; O = -2}$

4. Identify the atoms for which the oxidation number changes.

$\text{Mn: +7 → +2; I: +1 → 0}$

5. Write the “skeleton” ionic equation, including only those ions for which an atom has changed oxidation number.

${\text{MnO"_4^(-) + "I"^(-) → "I"_2 + "Mn}}^{2 +}$

6. Separate the skeleton ionic equation into two half-reactions.

${\text{MnO"_4^(-) → "Mn}}^{2 +}$
${\text{I"^(-) → "I}}_{2}$

7. Balance all atoms other than $\text{O}$ and $\text{H}$ in each of the half-reactions.

${\text{MnO"_4^(-) → "Mn}}^{2 +}$
${\text{2I"^(-) → "I}}_{2}$

8. Balance $\text{O}$ by adding $\text{H"_2"O}$ molecules to the appropriate side.

${\text{MnO"_4^(-) → "Mn"^(2+) + "4H}}_{2} O$
${\text{2I"^(-) → "I}}_{2}$

9. Balance $\text{H}$ by adding ${\text{H}}^{+}$ ions to the appropriate side.

${\text{MnO"_4^(-) + "8H"^+ → "Mn"^(2+) + "4H}}_{2} O$
${\text{2I"^(-) → "I}}_{2}$

10. Balance charge by adding electrons to the appropriate side.

The first half-reaction has 7+ on the left and 2+ on the right.
We add ${\text{5e}}^{-}$ to the left to balance charge.

$\text{MnO"_4^(-) + "8H"^+ + "5e"^(-) → "Mn"^(2+) + "4H"_2"O}$

The second half-reaction has 2- on the left and zero on the right.
We add ${\text{2e}}^{-}$ to the right.

"2I"^(-) → "I"_2 + "2e"^⁻

11. Multiply the two half-reactions by a number that gives the lowest common multiple of the electrons transferred in each half-reaction.

In this case, the lowest common multiple of $2$ and $5$ is $10$, so we multiply the top equation by $2$ and the bottom equation by $5$.

2×["MnO"_4^(-) + "8H"^+ + 5e^(-) →" Mn"^(2+) + "4H"_2"O"]
5×["2I"^(⁻) → "I"_2 + "2e"^-]

12. Add the two half-reactions, cancelling species that appear on each side of the overall equation.

$\text{2MnO"_4^(-) + "10I"^(-) + "16H"^+ → "2Mn"^(2+) +"5I"_2 + "8H"_2"O}$

13. Check that atoms are balanced.

Left hand side: $\text{2 Mn; 8 O; 10 I; 16 H}$
Right hand side: $\text{2 Mn; 10 I; 16 H; 8 O}$

14. Check that charges are balanced.

Left hand side:  -2 – 10 + 16 = +4
Right hand side: $+ 4$

The balanced equation is

$\text{2MnO"_4^(-) + "10I"^(-) + "16H"^+→ "2Mn"^(2+) +"5I"_2 + "8H"_2"O}$

15. Add the spectator ions to get the "molecular" equation.

$\text{2KMnO"_4 + "10KI" + "8H"_2"SO"_4 → "2MnSO"_4 +"5I"_2 + "6K"_2"SO"_4 + "8H"_2"O}$

Jan 7, 2016

Here's how you balance in basic solution.

#### Explanation:

You can first balance the equation as if it were in acid and then make a final change to convert it to basic conditions.

Balance the following equation in basic solution:

${\text{MnO"_4^(-) + "CN"^(-) → "MnO"_2 + "CNO}}^{-}$

1. Determine the oxidation numbers of each atom.

Left hand side: $\text{Mn = +7; O = -2}$; we don’t know $\text{C}$ or $\text{N}$, but we can take $\text{CN}$ as a group and say $\text{CN = -1}$
Right hand side: $\text{Mn = +4; O = -2; CN = +1}$

2. Identify the atoms for which the oxidation number changes.

"Mn": +7 → +4
"CN": -1 → +1

3. Separate the skeleton ionic equation into two half-reactions.

${\text{MnO"_4^(-) → "MnO}}_{2}$
${\text{CN"^(-) → "CNO}}^{-}$

4. Balance all atoms other than $\text{O}$ and $\text{H}$ in each half-reaction.
Done.

5. Balance $\text{O}$: add $\text{H"_2"O}$ molecules to the appropriate side.

$\text{MnO"_4^(-) → "MnO"_2+ "2H"_2"O}$
${\text{CN"^(-) + "H"_2"O" → "CNO}}^{-}$

9. Balance $\text{H}$: add ${\text{H}}^{+}$ ions to the appropriate side.

$\text{MnO"_4^(-) + 4"H"^(+) → "MnO"_2+ "2H"_2"O}$
${\text{CN"^(-) + "H"_2"O" → "CNO"^(-)+ 2"H}}^{+}$

10. Balance charge: add electrons to the appropriate side.

$\text{MnO"_4^(-) + 4"H"^(+) + 3"e"^(-) → "MnO"_2+ "2H"_2"O}$
${\text{CN"^(-) + "H"_2"O" → "CNO"^(-)+ 2"H"^+ + 2"e}}^{-}$

11. Multiply the two half-reactions by a number that gives the lowest common multiple of the electrons transferred in each half-reaction.

The lowest common multiple of $2$ and $3$ is $6$, so we multiply the top equation by $2$ and the bottom equation by $3$.

2 × ["MnO"_4^(-) + 4"H"^(+) + 3e^(-) → "MnO"_2+ "2H"_2"O"]
3×["CN"^(-) + "H"_2"O" → "CNO"^(-)+ 2"H"^+ + 2e^-]

12. Add the two half-reactions, cancelling species that appear on each side of the overall equation.

$\text{2MnO"_4^(-) + "3CN"^(-) + "2H"^(+) → "2MnO"_2 +"3CNO"^(-) + "H"_2"O}$

13. Convert to base: add enough multiples of the equation $\text{H"^+ + "OH"^(-) → "H"_2"O}$ or ${\text{H"_2"O" → "H"^+ + "OH}}^{-}$ to cancel the ${\text{H}}^{+}$ ions.

Our equation has ${\text{2H}}^{+}$ on the left hand side, so we must add two multiples of ${\text{H"_2"O" → "H"^+ + "OH}}^{-}$ and cancel species that appear on each side.

"2MnO"_4^(-) + "3CN"^(-) + color(red)(cancel(color(black)("2H"^(+)))) → "2MnO"_2 +"3CNO"^(-) + color(red)(cancel(color(black)("H"_2"O")))
stackrel(1)color(red)(cancel(2))color(black)( "H"_2"O") → color(red)(cancel(color(black)("2H"^+))) + "2OH"^-
stackrel(———————————————————)("2MnO"_4^(-) + "3CN"^(-) + "H"_2"O" → "2MnO"_2 +"3CNO"^(-) + "2OH"^-)

14. Check that atoms are balanced.

Left hand side: $\text{2 Mn; 9 O; 3 C; 3 N; 2 H}$
Right hand side: $\text{2 Mn; 9 O; 3 C;3 N; 2 H}$

15. Check that charges are balanced.

Left hand side: -2 – 3 = -5
Right hand side: -3 – 2 = -5

The balanced equation in basic solution is

${\text{2MnO"_4^(-) + "3CN"^(-) + "H"_2"O" → "2MnO"_2 +"3CNO"^(-) + "2OH}}^{-}$