# How do you balance this redox reaction using the oxidation number method? Fe2+(aq) + MnO4–(aq) --> Fe3+(aq) + Mn2+(aq)

Jul 8, 2014

WARNING: This is a long answer. The balanced equation is $\text{5Fe"^"2+" + "MnO"_4^"-" + "8H"^"+" → "5Fe"^"3+" + "Mn"^"2+" + "4H"_2"O}$.

#### Explanation:

You follow a series of steps in order:

1. Identify the oxidation number of every atom.
2. Determine the change in oxidation number for each atom that changes.
3. Make the total increase in oxidation number equal to the total decrease in oxidation number.
4. Place these numbers as coefficients in front of the formulas containing those atoms.
5. Balance all remaining atoms other than $\text{O}$ and $\text{H}$.
6. Balance $\text{O}$.
7. Balance $\text{H}$.
8. Check that atoms and charges balance.

Here's how it works. Your unbalanced equation is

$\text{Fe"^"2+" + "MnO"_4^"-" → "Fe"^"3+" + "Mn"^"2+}$

1. Identify the oxidation number of every atom.

Left hand side: $\text{Fe = +2; Mn = +7; O = -2}$
Right hand side: $\text{Fe = +3; Mn = +2}$

2. Determine the change in oxidation number for each atom that changes.

$\text{Fe: +2 → +1; Change = +1}$
$\text{Mn: +7 → +2; Change = -5}$

3. Make the total increase in oxidation number equal to the total decrease in oxidation number.

We need 5 atoms of $\text{Fe}$ for every 1 atom of $\text{Mn}$. This gives us total changes of $\text{+5}$ and $\text{-5}$.

4. Place these numbers as coefficients in front of the formulas containing those atoms.

$\textcolor{red}{5} \text{Fe"^"2+" + color(red)(1)"MnO"_4^"-" → color(red)(5)"Fe"^"3+" + color(red)(1)"Mn"^"2+}$

5. Balance all remaining atoms other than $\text{H}$ and $\text{O}$.

Done.

6. Balance $\text{O.}$

Add enough $\text{H"_2"O}$ molecules to the deficient side to balance $\text{O}$.

We have 4 $\text{O}$ atoms on the left, so we need 4 $\text{H"_2"O}$ on the right.

$\textcolor{red}{5} \text{Fe"^"2+" + color(red)(1)"MnO"_4^"-" → color(red)(5)"Fe"^"3+" + color(red)(1)"Mn"^"2+" + color(blue)(4)"H"_2"O}$

7. Balance $\text{H}$.

Add enough $\text{H"^"+}$ ions to the deficient side to balance $\text{H}$.

We have 8 $\text{H}$ atoms on the right, so we need 8 $\text{H"^"+}$ on the left.

$\textcolor{red}{5} \text{Fe"^"2+" + color(red)(1)"MnO"_4^"-" + color(green)(8)"H"^"+" → color(red)(5)"Fe"^"3+" + color(red)(1)"Mn"^"2+" + color(blue)(4)"H"_2"O}$

8. Check that atoms and charges balance.

On the left: : $\text{5 Fe}$; $\text{1 Mn}$; $\text{8 H}$; $\text{4 O}$
On the right: $\text{5 Fe}$; $\text{1 Mn}$; $\text{8H}$; $\text{4 O}$

On the left: $\textcolor{w h i t e}{l l} \text{+10"color(white)(ll) "- 1 + 8 = +17}$
On the right: $\text{+15 + 2"color(white)(mm) = "+17}$

The balanced equation is

$\textcolor{red}{5} \text{Fe"^"2+" + color(red)(1)"MnO"_4^"-" + color(green)(8)"H"^"+" → color(red)(5)"Fe"^"3+" + color(red)(1)"Mn"^"2+" + color(blue)(4)"H"_2"O}$