Exemplar Problems
Problem 3 : Balance the following redox equation occurring in acidic solution:
$$[H_2S + KMnO_4 \rightarrow K_2SO_4 + MnSO_4 + H_2O]$$
Solution :
To balance this equation, follow these steps:
Step 1: Assign oxidation states to each element: $$[H_2S : S^{2-},]$$ $$[KMnO_4 : K^{+}, Mn^{7+}, O^{-2},]$$ $$[K_2SO_4 : K^{+}, S^{6+}, O^{-2},]$$ $$[MnS_2O_8 : Mn^{4+}, S^{6+}, O^{2-},]$$ $$[H_2O : 2H^{1+}, O^{-2}.]$$
Step 2: Write down the unbalanced equation: $$[H_2S + KMnO_4 \rightarrow K_2SO_4 + MnSO_4 + H_2O.]$$
Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: H_2S \rightarrow S^{2+},]$$ $$[Reduction: KMnO_4 \rightarrow Mn^{2+}.]$$
Step 4: Balance each half-reaction: Oxidation is a chemical reaction in which a substance loses electrons. $$[H_2S \rightarrow S^{2-}]$$ Add 8 electrons (eā») to the right side to balance the charge.
Reduction: $$[KMnO_4 \rightarrow MnO_4^{-}]$$ Add 5 electrons (eā») to the right side to balance the charge.
Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $$[5(H_2S \rightarrow S^{2-})]$$ $$[8(KMnO_4 \rightarrow Mn^{4+})]$$
Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[5H_2S + 8KMnO_4 \rightarrow K_2SO_4 + 8MnSO_4 + 5S + 10H_2O.]$$
BETA
