Half-Reaction Method — Balancing Redox Reactions

Importance of Half-Reactions

Redox reactions include both oxidation (lose of electrons) and reduction (gain of electrons). The compounds in a redox reaction can be described as:

• Oxidized species/Reducing agent

• Reduced species/Oxidizing agent

A full redox reaction equation can be broken down into half-reactions. Half reactions represent the movement of electrons within the oxidation or reduction step of the reaction. Looking at each redox reaction part separately is useful for examining electron movement.

Uses for Half-Reactions

There are a two main ways half-reactions can be used:

• They are used to describe reactions occurring in an electrochemical cell. Each half-reaction represents the reaction occurring at the two electrodes.

• They can be used to balance redox reactions. In basic or acidic reaction conditions, $OH^-$, $H^+$, or $H_2O$ are added to balance the redox equation.

Balancing with Half-Reactions Example

The half-reaction method can be used to balance redox reactions. The redox reaction is separated into two parts: oxidation and reduction. Each part of the reaction is balanced separately and then added together.

The example reaction is between potassium permanganate ($KMnO_4$) and sodium sulfite ($Na_2SO_3$) in a basic aqueous solution.

The unbalanced net ionic equation is: $$MnO_4^-(aq)\ +\ SO_3^{2-}(aq)\ \rightarrow\ MnO_4^{2-}(aq)\ +\ SO_4^{2-}(aq)$$ First, identify the the oxidized and reduced species and write an unbalanced half-reaction for each.

• Manganese changes from +7 to +6 (reduction) and sulfur changes from +4 to +6 (oxidation). This makes $MnO_4^-$ the oxidizing agent and $SO_3^{2-}$ the reducing agent: $$Oxidation: MnO_4^-(aq)\ \rightarrow\ MnO_4^{2-}(aq)\\ Reduction: SO_3^{2-}(aq)\ \rightarrow\ SO_4^{2-}(aq)$$

Now balance each half-reaction by starting with all atoms except H and O. Then balance for O atoms and add $H_2O$ to the side of the reaction missing O. Then balance for H by adding $H^+$ to the side missing H.

• Note in this example all atoms are already balanced, so only H and O atoms are balanced: $$Oxidation: MnO_4^-(aq)\ \rightarrow\ MnO_4^{2-}(aq)\\ Reduction: SO_3^{2-}(aq)\ +\ H_2O(l)\ \rightarrow\ SO_4^{2-}(aq)\ +\ 2H^+(aq)$$

Now we add electrons to the most positively charged side of the reaction to balance the charges on each side. The coefficients and the charge play a role. For example, if we had $2 Cr^{3+}$, the charge would be +6.

• In the ox step, an electrons is added on the left to match the -2 charge on the right. In the red step, electrons are added on the right to match the -2 charge on the left. $$Oxidation: MnO_4^-(aq)\ +\ 1e^- \rightarrow\ MnO_4^{2-}(aq)\\ Reduction: SO_3^{2-}(aq)\ +\ H_2O(l)\ \rightarrow\ SO_4^{2-}(aq)\ +\ 2H^+\ +\ 2e^-$$

Next multiply each half-reaction so that the number of electrons in each are equal.

• The oxidation half-reaction will need to be multiplied by 2 so that it has the same number of electrons as the reduction reaction. $$Oxidation: 2MnO_4^-(aq)\ +\ 2e^- \rightarrow\ 2MnO_4^{2-}(aq)\\$$

Now we can add the two half-reactions together. Any species that occur on both sides of the reactions are canceled. $$2MnO_4^-(aq)\ +\ 2e^- \rightarrow\ 2MnO4^{2-}(aq)\\ \> \underline {SO_3^{2-}(aq)\ +\ H_2O(l)\ \rightarrow\ SO_4^{2-}(aq)\ +\ 2H^+\ +\ 2e^-}\\ 2MnO_4^-(aq)\ + SO_3^{2-}(aq)\ + H_2O(l)\ \rightarrow\ 2MnO_4^{2-}(aq)\ + SO_4^{2-}(aq)\ + 2H^+$$

If a reaction occurs in a basic or acidic solution. The last step is to add $H^+$ or $OH^-$ to the reactants to account for those conditions.

• This reaction occurs in a basic solution. $OH^-$ is added to both sides to neutralize the $H^+$ ions. This creates 2 water molecules on the right. Since, the left side also had a water molecule, one water molecule from each side cancels out leaving water only on the right side.

$$2MnO_4^-(aq) + SO_3^{2-}(aq) +2OH^-(aq) \rightarrow 2MnO_4^{2-}(aq) + SO_4^{2-}(aq) + H_2O(l)$$

(Video) 19.2 How to Balance Oxidation Reduction Reactions | General Chemistry

by Chad's Prep

In this video, Chad explains the half-reaction method. He starts with a quick refresher on oxidation and reductions. Then, he shows you two examples of balancing redox equations. In the end, he converts one of the equations from acidic to basic conditions. This is an excellent video to help you master the half-reaction method.