8.2 Balancing Redox Reactions

Redox reactions involve additional considerations when balancing when compared to balancing regular chemical equations. There exists multiple techniques to balancing redox reactions, one of which is presented below (called the half-reaction method). This method assumes that the redox reaction is taking place in aqueous solution, a common environment for electrochemical reactions.

Steps for Balancing a Redox Reaction in Acidic/Basic Conditions

Assign oxidation numbers to all atoms
Separate reaction into two half-reactions
Balance all atoms except H & O
Balance O (by adding H₂O)
Balance H (by adding H⁺)
Balance the charges in each half-reaction (by adding e^–)
Balance e^– between both half-reactions
Combine both half-reactions and cancel out like terms
- Stop here if balancing for acidic conditions
- Continue to Step 9 if balancing for basic conditions
If the reaction is taking place in a basic environment, add a number of OH^– to both sides of the reaction equal to the number of H⁺ that exists in Step 8. Combine the H⁺ and OH^– on one side of the reaction to make H₂O.

Let us illustrate these steps by balancing the following redox reaction in a basic aqueous environment.

\[\mathrm{Fe^{2+}}(aq) + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{Fe^{3+}}(aq) + \mathrm{Mn^{2+}}(aq)\]

Step 1: Assign oxidation numbers to all atoms.

\[\begin{alignat*}{3} & \mathrm{Fe^{2+}}(aq) ~+~ && \mathrm{MnO_4^-}(aq) ~ \longrightarrow ~ &&\mathrm{Fe^{3+}}(aq) ~+~ && \mathrm{Mn^{2+}}(aq)\\ & ~(2) && ~(7)(-2) && ~(3) && ~(2) \end{alignat*}\]

Fe is oxidized. Mn is reduced.

Step 2: Separate reaction into two half-reactions

I omit the phase labels for clarity. We will add them back in at the end.

Oxidation half-reaction

\[\mathrm{Fe^{2+}} \longrightarrow \mathrm{Fe^{3+}}\]

Reduction half-reaction

\[\mathrm{MnO_4^-} \longrightarrow \mathrm{Mn^{2+}}\]

Step 3: Balance all atoms except H & O

All non-hydrogen/oxygen atoms are already balanced. Move on to Step 4.

\[\mathrm{Fe^{2+}} \longrightarrow \mathrm{Fe^{3+}}\]

\[\mathrm{MnO_4^-} \longrightarrow \mathrm{Mn^{2+}}\]

Step 4: Balance O (by adding H₂O)

\[\mathrm{Fe^{2+}} \longrightarrow \mathrm{Fe^{3+}}\]

\[\mathrm{MnO_4^-} \longrightarrow \mathrm{Mn^{2+}} + \color{magenta}{\mathrm{4H_2O}}\]

Step 5: Balance H (by adding H⁺)

\[\mathrm{Fe^{2+}} \longrightarrow \mathrm{Fe^{3+}}\]

\[\color{magenta}{\mathrm{8H^+}} + \mathrm{MnO_4^-} \longrightarrow \mathrm{Mn^{2+}} + \mathrm{4H_2O}\]

Step 6: Balance the charges in each half-reaction (by adding e^–)

\[\mathrm{Fe^{2+}} \longrightarrow \mathrm{Fe^{3+}} + ~\color{magenta}{e^-}\]

\[\color{magenta}{5e^-} + \mathrm{8H^+} + \mathrm{MnO_4^-} \longrightarrow \mathrm{Mn^{2+}} + \mathrm{4H_2O}\]

Step 7: Balance e^– between both half-reactions

Multiply the oxidation half-reaction by 5. Both half-reactions now have 5 electrons.

\[\begin{align*} \color{magenta}{5}[\mathrm{Fe^{2+}} \longrightarrow \mathrm{Fe^{3+}} + e^-] \rightarrow \\ \mathrm{5Fe^{2+}} \longrightarrow \mathrm{5Fe^{3+}} + 5e^- \end{align*}\]

\[\begin{align*} 5e^- + \mathrm{8H^+} + \mathrm{MnO_4^-} \longrightarrow \mathrm{Mn^{2+}} + \mathrm{4H_2O} \end{align*}\]

Step 8: Combine both half-reactions and cancel out like terms

\[\begin{array}{rl} \require{cancel} \mathrm{5Fe^{2+}} &\longrightarrow \mathrm{5Fe^{3+}} + 5e^- \\[1.5ex] 5e^- + \mathrm{8H^+} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{Mn^{2+}} + \mathrm{4H_2O} \\[0.75ex] \hline \mathrm{5Fe^{2+}} + 5e^- + \mathrm{8H^+} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{5Fe^{3+}} + 5e^- + \mathrm{Mn^{2+}} + \mathrm{4H_2O}\\[3.0ex] \mathrm{5Fe^{2+}} + \cancel{5e^-} + \mathrm{8H^+} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{5Fe^{3+}} + \cancel{5e^-} + \mathrm{Mn^{2+}} + \mathrm{4H_2O}\\[1.5ex] \mathrm{5Fe^{2+}} + \mathrm{8H^+} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{5Fe^{3+}} + \mathrm{Mn^{2+}} + \mathrm{4H_2O} \end{array}\]

If we were balancing the redox reaction in a non-basic environment, we would not continue on to Step 9. We would simply add the phase labels back in and be done.

Step 9: If the reaction is taking place in a basic environment, add a number of OH^– to both sides of the reaction equal to the number of H⁺ that exists in Step 8. Combine the H⁺ and OH^– on one side of the reaction to make H₂O.

Here we add eight OH^–.

\[\begin{align*} \mathrm{5Fe^{2+}} + \mathrm{8H^+} + \color{magenta}{\mathrm{8OH^-}} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{5Fe^{3+}} + \mathrm{Mn^{2+}} + \mathrm{4H_2O} + \color{magenta}{\mathrm{8OH^-}} \end{align*}\]

Combine H⁺ and OH^– on the left hand side to make 8H₂O. Canacel out waters.

\[\begin{align*} \require{cancel} \mathrm{5Fe^{2+}} + \cancelto{4}{8}\!\mathrm{H_2O} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{5Fe^{3+}} + \mathrm{Mn^{2+}} + \cancel{\mathrm{4H_2O}} + \mathrm{8OH^-}\\[1.5ex] \mathrm{5Fe^{2+}} + \mathrm{4H_2O} + \mathrm{MnO_4^-} &\longrightarrow \mathrm{5Fe^{3+}} + \mathrm{Mn^{2+}} + \mathrm{8OH^-} \end{align*}\]

Add the phase labels back into the reaction.

\[\mathrm{5Fe^{2+}}(aq) + \mathrm{4H_2O}(l) + \mathrm{MnO_4^-}(aq) \longrightarrow \mathrm{5Fe^{3+}}(aq) + \mathrm{Mn^{2+}}(aq) +\mathrm{8OH^-}(aq)\]

Practice

Balance the following redox reaction in a basic aqueous environment.

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

Answer

\[6\mathrm{I}^{-}(aq) + \mathrm{4H_2O}(l) + 2\mathrm{MnO}_{4}^{-}(aq) \longrightarrow 3\mathrm{I}_{2}(aq) + 2\mathrm{MnO}_{2}(s) + 8\mathrm{OH^-}(aq)\]