Related Problems with Solution

Problem 2 : Balance the following redox equation occurring in basic solution:#

$$[MnO_4^- + H_2O_2 \rightarrow MnO_2 + O_2 + OH^-.]$$

Solution :#

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

Step 1: Assign oxidation states to each element: $$[MnO_4^- : Mn^{7+}, O^{-2}]$$ $$[H_2O_2 : H^{1+}, O^{-2}]$$ $$[MnO_2 : Mn^{4+}, O^{-2}]$$ $$[O_2 : O^{0}]$$ $$[OH^- : O^{-2}, H^{1+}]$$

Step 2: Write down the unbalanced equation: $$[MnO_4^- + H_2O_2 \rightarrow MnO_2 + O_2 + OH^-.]$$

Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: H_2O_2 \rightarrow O_2]$$ $$[Reduction: MnO_4^- \rightarrow MnO_2]$$

Step 4: Balance each half-reaction: Oxidation: $$[H_2O_2 \rightarrow O_2]$$ Add 4 electrons (e⁻) to the left side to balance the charge.

Reduction: $$[MnO_4^- \rightarrow MnO_2]$$ Add 3 electrons (e⁻) to the right side to balance the charge.

Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $$[3(H_2O_2 \rightarrow O_2)]$$ $$[4(MnO_4^- \rightarrow MnO_2)]$$

Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[3H_2O_2 + 4MnO_4^- \rightarrow 3O_2 + 4MnO_2 + 6OH^-.]$$