Exemplar Problems
Problem 4 :Balance the following redox equation occurring in basic solution:
$$[MnO_4^- + H_2S \rightarrow MnO_2 + S^{2-} + OH^-.]$$
Solution :
To balance this equation in basic solution, follow these steps:
Step 1: Assign oxidation states to each element: $$[MnO_4^- : Mn^{7+}, O^{-2},]$$ $$[H_2S : H^{1+}, S^{-2},]$$ $$[MnO_2 : Mn^{4+}, O^{-2},]$$ $$[S^{2-} : S^{-2},]$$ $$[OH^- : O^{-2}, H^{1+}.]$$
Step 2: Write down the unbalanced equation: $$[MnO_4^- + H_2S \rightarrow MnO_2 + S^{2-} + OH^-.]$$
Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: H_2S \rightarrow S^{2-},]$$ $$[Reduction: MnO_4^- \rightarrow MnO_2.]$$
Step 4: Balance each half-reaction: Oxidation: $$[H_2S \rightarrow S^{2-}]$$ Add 2 electrons (eā») to the left side to balance the charge.
Reduction: $$[MnO_4^- \rightarrow MnO_2]$$ 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: $$[5(H_2S \rightarrow S^{2-})]$$ $$[2(MnO_4^- \rightarrow MnO_2)]$$
Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[5H_2S + 2MnO_4^- \rightarrow 5S^{2-} + 2MnO_2 + 10OH^-.]$$
BETA
