# Balancing Redox Reactions

Balancing oxidation-reduction reactions can be performed via the following steps:

1. Assign oxidation states to all atoms. Identify what is being oxidized and reduced.
2. Separate overall reaction into two half-reactions
3. Balance all atoms except H and O
4. Balance O (by adding H2O)
5. Balance H (by adding H+)
6. Balance charges (by adding e)
7. Make the number of e in both half reactions equal
8. Combine both half-reactions canceling out terms
9. If solution is basic, add enough OH to combine with all H+ (on left hand side) to make H2O. Add same number of OH to right hand side.
10. Combine H+ and OH to make H2O

### Example: Balancing in an acidic solution

Balance the following redox reaction in an acidic solution. $$\require{color}$$

$\mathrm{Fe}^{2+}(aq)+\mathrm{MnO}_{4}^{-}(aq) \longrightarrow \mathrm{Fe}^{3+}(aq)+\mathrm{Mn}^{2+}(aq)$

Step 1: Assign oxidation states to all atoms. Identify what is being oxidized and reduced.

Reactants

• Fe: +2
• Mn: +7
• O: –2

Products

• Fe: +3
• Mn: +2

Fe is oxidized. Mn is reduced.

Step 2: Separate overall reaction into two half-reactions

Oxidation $\mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe^{3+}}(aq)$

Reduction $\mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Mn^{2+}}(aq)$

Step 3: Balance all atoms except H and O

Oxidation $\mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe^{3+}}(aq)$

Reduction $\mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Mn^{2+}}(aq)$

Already balanced. Move to step 4.

Step 4: Balance O (by adding H2O)

Oxidation $\mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe^{3+}}(aq)$

Reduction $\mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Mn^{2+}}(aq) + \textcolor{magenta}{\mathrm{4H_2O}(l)}$

Step 5: Balance H (by adding H+)

Oxidation $\mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe^{3+}}(aq)$

Reduction $\textcolor{magenta}{\mathrm{8H^+}(aq)} + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Mn^{2+}}(aq) +\mathrm{4H_2O}(l)$

Step 6: Balance charges (by adding e)

Oxidation

The oxidation reaction (from Step 5) has a +2 charge on the left and a +3 charge on the right. Balance the charge by adding one electron to the right hand side.

$\mathrm{Fe}^{2+}(aq) \longrightarrow \mathrm{Fe^{3+}}(aq) + \textcolor{magenta}{1e^-}$

Reduction

The reduction reaction (from Step 5) has a +7 charge on the left and +2 charge on the right. Balance the charge by adding 5 electrons to the left hand side.

$\textcolor{magenta}{5e^-} + \mathrm{8H^+}(aq) + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Mn^{2+}}(aq) +\mathrm{4H_2O}(l)$

Step 7: Make the number of e in both half reactions equal

The oxidation reaction has only 1 electron while the reduction reaction has 5 electrons. Multiply the oxidation reaction by 5 to get the number of electrons to equal each other.

Oxidation \begin{align*} \textcolor{magenta}{5}[\mathrm{Fe}^{2+}(aq) &\longrightarrow \mathrm{Fe^{3+}}(aq) + 1e^-] \longrightarrow\\ \textcolor{magenta}{5}\mathrm{Fe}^{2+}(aq) &\longrightarrow \textcolor{magenta}{5}\mathrm{Fe^{3+}}(aq) + \textcolor{magenta}{5}e^- \end{align*}

Reduction $5e^- + \mathrm{8H^+}(aq) + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Mn^{2+}}(aq) +\mathrm{4H_2O}(l)$

Step 8: Combine both half-reactions

\begin{align*} 5\mathrm{Fe}^{2+}(aq) &\longrightarrow 5\mathrm{Fe^{3+}}(aq) + 5e^-\\ 5e^- + \mathrm{8H^+}(aq) + \mathrm{MnO_4^-}(aq) &\longrightarrow \mathrm{Mn^{2+}}(aq) +\mathrm{4H_2O}(l)\\[2ex] \mathrm{5Fe^{2+}}(aq) + \mathrm{8H^+}(aq) + \mathrm{MnO_4^-}(aq) &\longrightarrow \mathrm{5Fe^{3+}}(aq) + \mathrm{Mn^{2+}}(aq) + \mathrm{4H_2O}(l) \end{align*}

We stop here and do not proceed to step 9 since we are balancing this redox reaction for an acidic solution.

### Example: Balancing in a basic solution

Balance the following redox reaction in an acidic solution. $$\require{color}$$

$\mathrm{I}^{-}(aq)+\mathrm{MnO}_{4}^{-}(aq) \longrightarrow \mathrm{I_2}(aq)+\mathrm{MnO_2}(s)$

Step 1: Assign oxidation states to all atoms. Identify what is being oxidized and reduced.

Reactants

• I: –1
• Mn: +7
• O: –2

Products

• I: 0
• Fe: +4
• Mn: –2

I is oxidized. Mn is reduced.

Step 2: Separate overall reaction into two half-reactions

Oxidation $\mathrm{I}^{-}(aq) \longrightarrow \mathrm{I_2}(aq)$

Reduction $\mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{MnO_2}(s)$

Step 3: Balance all atoms except H and O

Oxidation $\textcolor{magenta}{2}\mathrm{I}^{-}(aq) \longrightarrow \mathrm{I_2}(aq)$

Reduction $\mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{MnO_2}(s)$

Step 4: Balance O (by adding H2O)

Oxidation $2\mathrm{I}^{-}(aq) \longrightarrow \mathrm{I_2}(aq)$

Reduction $\mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{MnO_2}(s) + \textcolor{magenta}{2\mathrm{H_2O}(l)}$

Step 5: Balance H (by adding H+)

Oxidation $2\mathrm{I}^{-}(aq) \longrightarrow \mathrm{I_2}(aq)$

Reduction $\textcolor{magenta}{4\mathrm{H^+}(aq)} + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{MnO_2}(s) + 2\mathrm{H_2O}(l)$

Step 6: Balance charges (by adding e)

Oxidation

The oxidation reaction (from Step 5) has a –2 charge on the left and 0 charge on the right. Balance the charge by adding two electrons to the right hand side.

$2\mathrm{I}^{-}(aq) \longrightarrow \mathrm{I_2}(aq) + \textcolor{magenta}{2e^-}$

Reduction

The reduction reaction (from Step 5) has a +3 charge on the left and 0 charge on the right. Balance the charge by adding 3 electrons to the left hand side.

$\textcolor{magenta}{3e^-} + 4\mathrm{H^+}(aq) + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{MnO_2}(s) + 2\mathrm{H_2O}(l)$

Step 7: Make the number of e in both half reactions equal

The oxidation reaction has 2 electrons while the reduction reaction has 3 electrons. Multiply the oxidation reaction by 3 and the reduction reaction by 2 to get the number of electrons to equal each other.

Oxidation \begin{align*} \textcolor{magenta}{3}[2\mathrm{I}^{-}(aq) &\longrightarrow \mathrm{I_2}(aq) + 2e^-] \longrightarrow\\ \textcolor{magenta}{6}\mathrm{I}^{-}(aq) &\longrightarrow \textcolor{magenta}{3}\mathrm{I_2}(aq) + \textcolor{magenta}{6}e^- \end{align*}

Reduction \begin{align*} \textcolor{magenta}{2}[3e^- + 4\mathrm{H^+}(aq) + \mathrm{MnO_4^-}(aq) &\longrightarrow \mathrm{MnO_2}(s) + 2\mathrm{H_2O}(l)] \longrightarrow\\ \textcolor{magenta}{6}e^- + \textcolor{magenta}{8}\mathrm{H^+}(aq) + \textcolor{magenta}{2}\mathrm{MnO_4^-}(aq) &\longrightarrow \textcolor{magenta}{2}\mathrm{MnO_2}(s) + \textcolor{magenta}{4}\mathrm{H_2O}(l) \end{align*}

Step 8: Combine both half-reactions

\begin{align*} 6\mathrm{I}^{-}(aq) &\longrightarrow 3\mathrm{I_2}(aq) + 6e^-\\ 6e^- + 8\mathrm{H^+}(aq) + 2\mathrm{MnO_4^-}(aq) &\longrightarrow 2\mathrm{MnO_2}(s) +4\mathrm{H_2O}(l)\\[2ex] 6\mathrm{I}^{-}(\mathrm{aq})+ 8\mathrm{H}^{+}(aq) + 2\mathrm{MnO}_{4}^{-}(aq) &\longrightarrow 3\mathrm{I}_{2}(aq) + 2\mathrm{MnO}_{2}(s) + 4\mathrm{H}_{2}\mathrm{O}(l) \end{align*}

Step 9: If solution is basic, add enough OH to combine with all H+ (on left hand side) to make H2O. Add same number of OH to right hand side.

$5\mathrm{I}^{-}(\mathrm{aq})+ 8\mathrm{H}^{+}(aq) + \textcolor{magenta}{8\mathrm{OH^-}(aq)} + 2\mathrm{MnO}_{4}^{-}(aq) \longrightarrow 3\mathrm{I}_{2}(aq) + 2\mathrm{MnO}_{2}(s) + 4\mathrm{H}_{2}\mathrm{O}(l) + \textcolor{magenta}{8\mathrm{OH^-}(aq)}$

Step 10: Combine H+ and OH to make H2O

\begin{align*} 6\mathrm{I}^{-}(\mathrm{aq}) + \textcolor{red}{8\mathrm{H}^{+}(aq) + 8\mathrm{OH^-}(aq)} + 2\mathrm{MnO}_{4}^{-}(aq) &\longrightarrow 3\mathrm{I}_{2}(aq) + 2\mathrm{MnO}_{2}(s) + 4\mathrm{H}_{2}\mathrm{O}(l) + 8\mathrm{OH^-}(aq)\\[2ex] 6\mathrm{I}^{-}(\mathrm{aq}) + \textcolor{magenta}{\mathrm{8H_2O}(l)} + 2\mathrm{MnO}_{4}^{-}(aq) &\longrightarrow 3\mathrm{I}_{2}(aq) + 2\mathrm{MnO}_{2}(s) + 4\mathrm{H}_{2}\mathrm{O}(l) + 8\mathrm{OH^-}(aq) \end{align*}

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