After explaining what a redox reaction is on our previous notes, today we will be talking about balancing redox reactions.

We saw that a redox reaction is a reaction in which oxidation and reduction are occurring simultaneously. In other words, it is a reaction one species undergoes reduction whereas the other undergoes oxidation.^[1]

During a redox reaction, equilibrium is usually established between the reactants and products of a redox reaction. And every redox reaction is made up of two half-reactions called the redox half equation.

The two redox half equations show how electrons are being transferred (oxidation) and how they transferred electrons are gained by the other species (reduction).

 

Rules for balancing redox reactions

The method used to balance redox reactions is called the Half Equation Method. In this method, the equation is separated into two half-equations; one for oxidation and one for reduction.

The half reaction method provides a systematic approach for balancing redox equation. In this method, the net reaction is broken down into two half reactions which will give us two half equations as earlier menioned.

Each half equation is then balanced separately and the two half equations added to give the overall balanced redox equation for the given reaction.

Steps to follow when balancing redox reactions by half equation method in acidic medium

1. Start by writing down the unbalanced redox equation for the reaction
2. Next, separate the redox equation into two half equations representing the oxidation and reduction half equations.
3. Balance each half-reaction equation with respect to all elements other than O and H.
4. Balance each half-reaction equation with respect to oxygen by adding the appropriate number of water (H₂O) molecules to the side deficient in oxygen.
5. Balance the hydrogen atoms (including those added in step 4 to balance the oxygen atom) by adding H⁺ ions to the opposite side of the equation.
6. Add up the charges on each side. Make them equal by adding enough electrons (e^–) to the more positive side. 
7. The e^– on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.
8. Obtain a complete balanced equation by adding the two half equations and cancel any common terms.
9. Finally, indicate the phase of the elements and ions.

 

In a basic medium

Chemical equations for reactions in basic solutions are balanced slightly different from reactions in acidic solutions.

In basic solutions, Step 5 above changes to:

Balance each half-reaction equation with respect to hydrogen by adding the appropriate numbers of OH^–.

However, if we simply add OH^– to balance for H, then each OH ion added adds an atom of oxygen (O) in addition to H. This will then affect the balance for Oxygen that was obtained in step 4.

To preserve the Oxygen balance, we add one molecule of H2O to the H deficient side and OH^– to the other side for each deficient H.

 

Neutral conditions

The first step to balance any redox reaction is to separate the reaction into half-reactions. The substance being reduced will have electrons as reactants, and the oxidised substance will have electrons as products.

You can then balance the half equations with respect to the different products and electrons and then add up the two to form the over all equation.

Balancing redox reactions: Examples

Now, let’s try balancing some sample redox reactions using the half equation method by applying the steps above in both an acidic, a basic and a neural dedium

Example in acidic medium

Balance the following redox reaction in acidic conditions.

Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

Step 3: Add H₂O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H₂O molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:

Step 4: Balance hydrogen by adding protons (H⁺). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.

Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e^– and the other reaction has 2e^–, so it should be multiplied by 3. This gives:

Step 7: Add the reactions and cancel out common terms.

Balancing redox reactions in basic solutions example

Let’s balance the following redox reaction in basic conditions.

Solution

Step 1: Separate the half-reactions.

Step 2: Balance elements other than O and H.

Step 3: Add H₂O to balance oxygen.

Step 4: Balance hydrogen with protons.

Step 5: Balance the charge with e^–.

Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.

Step 7: Add the reactions and cancel the electrons.

Step 8: Add OH^– to balance H⁺. There are 2 net protons in this equation, so add 2 OH^– ions to each side.

Step 9: Combine OH^– ions and H⁺ ions that are present on the same side to form water.

Step 10: After cancelling common terms, we have.

Neutral conditions example

Balance the following reaction

Solution

Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:

The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:

Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire Cu⁺_(aq)+e− → Cu_(s)Cu+(aq)+e−→Cu(s) half-reaction by 3 and leaving the other half reaction as it is. This gives:

Step 3: Adding the equations give:

The electrons cancel out and the balanced equation is left.