# 22.10: Balancing Redox Reactions- Half-Reaction Method


The picture below shows one of the two Thunder Dolphin amusement ride trains. This train has an orange stripe, while its companion has a yellow stripe. Pigments of these colors are often made with a dichromate salt (usually sodium or potassium dichromate). These brightly colored compounds serve as strong oxidizing agents in chemical reactions.

## Balancing Redox Equations: Half-Reaction Method

Another method for balancing redox reactions uses half-reactions. Recall that a half-reaction is either the oxidation or reduction that occurs, treated separately. The half-reaction method works better than the oxidation-number method when the substances in the reaction are in aqueous solution. The aqueous solution is typically either acidic or basic, so hydrogen ions or hydroxide ions are present.

In general, the half-reactions are first balanced by atoms separately. Electrons are included in the half-reactions. These are then balanced so that the number of electrons lost is equal to the number of electrons gained. Finally, the two half-reactions are added back together. The example is the oxidation of $$\ce{Fe^{2+}}$$ ions to $$\ce{Fe^{3+}}$$ ions by dichromate $$\left( \ce{Cr_2O_7^{2-}} \right)$$ in acidic solution. The dichromate ions are reduced to $$\ce{Cr^{3-}}$$ ions.

Step 1: Write the unbalanced ionic equation.

$\ce{Fe^{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow \ce{Fe^{3+}} \left( aq \right) + \ce{Cr^{3+}} \left( aq \right)\nonumber$

Notice that the equation is far from balanced, as there are no oxygen atoms on the right side. This will be resolved by the balancing method.

Step 2: Write separate half-reactions for the oxidation and the reduction processes. Determine the oxidation numbers first, if necessary.

\begin{align*} &\text{Oxidation:} \: \ce{Fe^{2+}} \left( aq \right) \rightarrow \ce{Fe^{3+}} \left( aq \right) \\ &\text{Reduction:} \: \overset{+6}{\ce{Cr_2}} \ce{O_7^{2-}} \left( aq \right) \rightarrow \ce{Cr^{3+}} \left( aq \right) \end{align*}\nonumber

Step 3: Balance the atoms in the half-reactions other than hydrogen and oxygen. In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced with the chromium atoms.

$\ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right)\nonumber$

Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction half-reaction above, seven $$\ce{H_2O}$$ molecules will be added to the product side.

$\ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right)\nonumber$

Now the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance. In this example, fourteen $$\ce{H^+}$$ ions will be added to the reactant side.

$14 \ce{H^+} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right)\nonumber$

Step 5: Balance the charges by adding electrons to each half-reaction. For the oxidation half-reaction, the electrons will need to be added to the product side. For the reduction half-reaction, the electrons will be added to the reactant side. By adding one electron to the product side of the oxidation half-reaction, there is a $$2+$$ total charge on both sides.

$\ce{Fe^{2+}} \left( aq \right) \rightarrow \ce{Fe^{3+}} \left( aq \right) + \ce{e^-}\nonumber$

There is a total charge of $$12+$$ on the reactant side of the reduction half-reaction $$\left( 14 - 2 \right)$$. The product side has a total charge of $$6+$$ due to the two chromium ions $$\left( 2 \times 3 \right)$$. To balance the charge, six electrons need to be added to the reactant side.

$6 \ce{e^-} + 14 \ce{H^+} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) \rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right)\nonumber$

Now equalize the electrons by multiplying everything in one or both equations by a coefficient. In this example, the oxidation half-reaction will be multiplied by six.

$6 \ce{Fe^{2+}} \left( aq \right) \rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 6 \ce{e^-}\nonumber$

Step 6: Add the two half-reactions together. The electrons must cancel. Balance any remaining substances by inspection. If necessary, cancel out $$\ce{H_2O}$$ or $$\ce{H^+}$$ that appear on both sides.

\begin{align*} 6 \ce{Fe^{2+}} \left( aq \right) &\rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + \cancel{ 6 \ce{e^-}} \\ \cancel{6 \ce{e^-}} + 14 \ce{H^+} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) &\rightarrow 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right) \\ \hline 14 \ce{H^+} \left( aq \right) + 6 \ce{Fe^{2+}} \left( aq \right) + \ce{Cr_2O_7^{2-}} \left( aq \right) &\rightarrow 6 \ce{Fe^{3+}} \left( aq \right) + 2 \ce{Cr^{3+}} \left( aq \right) + 7 \ce{H_2O} \left( l \right) \end{align*}\nonumber

Step 7: Check the balancing. In the above equation, there are $$14 \: \ce{H}$$, $$6 \: \ce{Fe}$$, $$2 \: \ce{Cr}$$, and $$7 \: \ce{O}$$ on both sides. The net charge is $$24+$$ on both sides. The equation is balanced.

## Summary

22.10: Balancing Redox Reactions- Half-Reaction Method is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.