7.3.2. Balancing of Redox Reactions

Two methods are used to balance chemical equations for redox processes. One of these methods is based on the change in the oxidation number of reducing agent and the oxidising agent and the other method is based on splitting the redox reaction into two half reactions — one involving oxidation and the other involving reduction. Both these methods are in use and the choice of their use rests with the individual using them.

In order to balance redox reactions, there are primarily two approaches: the oxidation number method and the half-reaction method. The oxidation number method focuses on adjusting the oxidation states of the reactants to ensure the same total charge on both sides of the equation. In contrast, the half-reaction method involves breaking down the overall reaction into oxidation and reduction components, balancing each separately before combining them back into a complete equation. Students can choose either method based on personal preference or the specific reaction characteristics.

Think of balancing a seesaw. The oxidation number method is like adjusting weights on either side to ensure balance, while the half-reaction method is like first making sure the left side is stable before addressing the right side. Once each side is balanced individually, they can come together to form a balanced seesaw, similar to how individual half-reactions are combined for a complete redox equation.

Step 1: The skeletal ionic equation is: Cr2O72–(aq) + SO32–(aq) → Cr3+(aq) + SO42–(aq) Step 2: Assign oxidation numbers for Cr and S +6 –2 +4 –2 +3 +6 –2 Cr2O72–(aq) + SO32–(aq) → Cr(aq)+SO42–(aq) This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.

In this example, we start with a skeletal ionic equation. First, we identify which species are involved and what their oxidation states are. For chromium in Cr2O72–, the oxidation state is +6, while in Cr3+, it is +3, indicating a reduction. For sulphur in SO32–, the oxidation state is +4, which changes to +6 in SO42–, indicating an oxidation process. Balancing involves ensuring that the total increase of oxidation states equals the total decrease.

Imagine a simple game where players need to collect points. If one player's points decrease (they lose points), another player's points must increase (they gain points) to keep the total score constant. This balancing act of points represents the changes in oxidation states in redox reactions.

Step 3: Calculate the increase and decrease of oxidation number, and make them equal: from step-2 we can notice that there is change in oxidation state of chromium and sulphur. Oxidation state of chromium changes from +6 to +3. There is decrease of +3 in oxidation state of chromium on right hand side of the equation. Oxidation state of sulphur changes from +4 to +6. There is an increase of +2 in the oxidation state of sulphur on right hand side.

After assigning oxidation numbers, the next step is to calculate how much the oxidation state has changed for each element. Here, chromium's oxidation state decreases by 3, while sulphur's increases by 2. To balance the reaction, we must ensure that the total increase in oxidation state matches the total decrease. We balance the coefficients of the species involved accordingly.

This can be compared to a savings and spending account. If you spend $3 from your savings (decreasing), you should have someone else receiving it (increasing their funds) so that the total money remains constant. This reflects the balance needed in oxidation states during redox reactions.

Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both sides, add 8H+ on the left to make ionic charges equal

In acidic environments, we may need to add acidic components (like H+) to one side of the equation to ensure that both sides of the equation are balanced not only in terms of mass but also in terms of charge. This adjustment ensures that the overall reaction adheres to the conservation of charge principle in addition to the conservation of matter.

Consider mixing different colored paints to achieve an even color across a canvas. If one side is too dark (more charge), adding lighter paint (H+) can help achieve a uniform shade across the canvas (balance). This mirrors how we adjust ion concentrations in a redox reaction.

Step 5: Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4H2O) on the right to achieve balanced redox change.

The last step involves ensuring that all atoms, including hydrogen from any H+ added prior, are balanced by adding water molecules where needed. This process ensures that both the number of atoms and overall charge in the balanced equation match up, completing the full redox equation.

Think of adjusting ingredients in a recipe. If too much salt was added (like adding H+), you would also need to add more water (H2O) to achieve the recipe’s intended flavor and balance, similar to final adjustments in a chemical equation to achieve balance.

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