Exemplar Problems
Problem 4 : Balance the following redox equation occurring in acidic solution:
$$[HNO_3 + H_2S \rightarrow NO + S + H_2O.]$$
Solution :
To balance this equation, follow these steps:
Step 1: Assign oxidation states to each element: $$[HNO_3 : N^{5+}, O^{-2}, H^{1+},]$$ $$[H_2S : H^{1+}, S^{-2},]$$ $$[NO : N^{2+}, O^{-2},]$$ $$[S : S^{0},]$$ $$[H_2O : H^{1+}, O^{-2}.]$$
Step 2: Write down the unbalanced equation: $$[HNO_3 + H_2S \rightarrow NO + S + H_2O.]$$
Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: H_2S \rightarrow S^{0},]$$ $$[Reduction: HNO_3 \rightarrow NO.]$$
Step 4: Balance each half-reaction: Oxidation: $$[H_2S \rightarrow S^{0}]$$ Add 2 electrons (eā») to the left side to balance the charge.
Reduction: $$[HNO_3 \rightarrow NO]$$ Add 5 electrons (eā») to the left side to balance the charge.
Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $$[5(H_2S \rightarrow S^{0})]$$ $$[2(HNO_3 \rightarrow NO)]$$
Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[5H_2S + 2HNO_3 \rightarrow 5S + 2NO + 4H_2O.]$$
BETA
