Exemplar Problems

Problem 1: Balance the following redox equation occurring in acidic solution:

$$[MnO_4^- + H_2C_2O_4 \rightarrow Mn^{2+} + CO_2]$$

Solution :

To balance this equation, follow these steps:

Step 1: Assign oxidation states to each element: $$[MnO_4^- : Mn^{+7},]$$ $$[H_2C_2O_4 : C^{+4},]$$ $$[Mn^{2+} : Mn^{2+},]$$ $$[CO_2 : C^{4+}.]$$

Step 2: Write down the unbalanced equation: $$[MnO_4^- + H_2C_2O_4 + H^+ \rightarrow Mn^{2+} + CO_2 + H_2O.]$$

Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: MnO_4^- \rightarrow Mn^{2+},]$$ $$[Reduction: H_2C_2O_4 \rightarrow CO_2 + H_2O.]$$

Step 4: Balance each half-reaction: Oxidation: $$[MnO_4^- \rightarrow Mn^{2+}]$$ Add 5 electrons (e⁻) to the left side to balance the charge.

Reduction: $$[H_2C_2O_4 \rightarrow 2CO_2 + H_2O]$$ Add 2 electrons (e⁻) to the left side to balance the charge.

Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $$[5(MnO_4^- \rightarrow Mn^{2+})]$$ $$[2(H_2C_2O_4 \rightarrow 2CO_2 + O_2)]$$

Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[5MnO_4^- + 16H^+ + 2H_2C_2O_4 \rightarrow 5Mn^{2+} + 8H_2O + 2CO_2.]$$