Balancing Redox Reactions

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

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