Exemplar Problems

Problem 4 : Balance the following redox equation occurring in basic solution:

$$[Br_2 + OH^- \rightarrow BrO_3^- + Br^- + H_2O]$$

Solution :

To balance this equation in basic solution, follow these steps:

Step 1: Assign oxidation states to each element: $$[Br_2 : Br^0,]$$ $$[OH^- : O^{-2}, H^{-1},]$$ $$[BrO_3^- : Br^{5+}, O^{-2},]$$ $$[Br^- : Br^{-1},]$$ $$[H_2O : H^{1+}, O^{-2}.]$$

Step 2: Write down the unbalanced equation: $$[Br_2 + OH^- \rightarrow BrO_3^- + Br^- + H_2O.]$$

Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: Br_2 \rightarrow Br^-,]$$ $$[Reduction: OH^- \rightarrow BrO_3^-.]$$

Step 4: Balance each half-reaction: Oxidation: $$[Br_2 \rightarrow 2Br^-]$$ Add 2 electrons (e⁻) to the left side to balance the charge.

Reduction: $$[OH^- \rightarrow BrO_3^-]$$ Add 6 electrons (e⁻) to the left side to balance the charge.

Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $$[3(Br_2 \rightarrow 2Br^-)]$$ $$[1(OH^- \rightarrow BrO_3^-)]$$

Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[3Br_2 + 6OH^- \rightarrow 2BrO_3^- + 6Br^- + 3H_2O.]$$