Question

In the reaction, $\mathrm{As}_{2} \mathrm{S}_{3} + \mathrm{xNO}_{3}^{-} + \mathrm{yH}^{+} \rightarrow 2 \mathrm{AsSO}_{4} + \mathrm{zNO}_{2} + \mathrm{aS} + 2 \mathrm{H}_{2} \mathrm{O}$.

Find the coefficients $x$, $y$, $z$, and $a$ respectively:

A $10, 4, 10, 3$ B $5, 4, 5, 3$ C $4, 10, 3, 10$ D None of these

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Answer

The correct option is $\mathbf{A}$: $10, 4, 10, 3$

This reaction is a special case of a redox reaction. Specifically, it involves two oxidation reactions and one reduction reaction.

Oxidation Half-Reactions:

  1. For Arsenic:$$ \text{As}^{3+} \rightarrow \text{AsO}_{4}^{3-} $$

  2. For Sulfur:$$ \text{S}^{2-} \rightarrow \text{S} $$

Let's balance these two oxidation half-reactions first: 1. $$ 2 \text{As}^{3+} + 8 \text{H}_2\text{O} \rightarrow 2 \text{AsO}_{4}^{3-} + 16 \text{H}^{+} + 4 \text{e}^{-} $$ 2. $$ 3 \text{S}^{2-} \rightarrow 3 \text{S} + 6 \text{e}^{-} $$

Adding these two half-reactions: $$ 2 \text{As}^{3+} + 3 \text{S}^{2-} + 8 \text{H}_2\text{O} \rightarrow 2 \text{AsO}_{4}^{3-} + 3 \text{S} + 16 \text{H}^{+} + 10 \text{e}^{-} $$

Reduction Half-Reaction:

For Nitrate:$$ \text{NO}_{3}^{-} \rightarrow \text{NO}_{2} $$

Balancing the reduction half-reaction: $$ \text{NO}_{3}^{-} + 2 \text{H}^{+} + 1 \text{e}^{-} \rightarrow \text{NO}_{2} + \text{H}_2\text{O} $$

Combining the Reactions:

To combine the reactions, multiply the reduction half-reaction by 10 to equal the number of electrons in the oxidation half-reactions. Then, add: $$ \text{As}_2\text{S}3 + 10 \text{NO}{3}^{-} + 4 \text{H}^{+} \rightarrow 2 \text{AsO}_{4}^{3-} + 10 \text{NO}_{2} + 3 \text{S} + 2 \text{H}_2\text{O} $$

Thus, the coefficients are $x = 10$, $y = 4$, $z = 10$, and $a = 3$, matching option A.


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