Question:
Observe the following unbalanced equation \(aS_{8}+b~OH^{-}(aq)\rightarrow cS^{2-}(aq)+dS_{2}O_{3}^{2-}(aq)+eH_{2}O(l)\). In the balanced equation, the ratio of c and d is:
Observe the following unbalanced equation \(aS_{8}+b~OH^{-}(aq)\rightarrow cS^{2-}(aq)+dS_{2}O_{3}^{2-}(aq)+eH_{2}O(l)\). In the balanced equation, the ratio of c and d is:
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In disproportionation reactions, ensure the atom being disproportionated is balanced in both the reduced and oxidized products before finalizing the electron count.
Updated On: Jun 8, 2026
- \(1:2 \)
- \(2:1 \)
- \(1:3 \)
- \(3:1 \)
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The Correct Option is D
Solution and Explanation
Concept:
This is a disproportionation reaction where the element sulfur (\(S_{8}\)) is simultaneously oxidized and reduced. To balance it, we use the ion-electron method by splitting the reaction into two half-reactions.
Step 1: Split into half-reactions.
Reduction: \(S_{8} + 16e^{-} \rightarrow 8S^{2-}\) Oxidation: \(S_{8} + 24OH^{-} \rightarrow 4S_{2}O_{3}^{2-} + 12H_{2}O + 24e^{-}\)
Step 2: Equalize the number of electrons transferred.
To balance the electrons, we multiply the reduction half-reaction by \(3\) and the oxidation half-reaction by \(2\): * Reduction: \(3 \times (S_{8} + 16e^{-} \rightarrow 8S^{2-}) \Rightarrow 3S_{8} + 48e^{-} \rightarrow 24S^{2-}\) * Oxidation: \(2 \times (S_{8} + 24OH^{-} \rightarrow 4S_{2}O_{3}^{2-} + 12H_{2}O + 24e^{-}) \Rightarrow 2S_{8} + 48OH^{-} \rightarrow 8S_{2}O_{3}^{2-} + 24H_{2}O + 48e^{-}\)
Step 3: Combine and simplify.
Adding the two balanced half-reactions gives: \(5S_{8} + 48OH^{-} \rightarrow 24S^{2-} + 8S_{2}O_{3}^{2-} + 24H_{2}O\) In the balanced equation, the coefficient for \(S^{2-}\) (which is \(c\)) is \(24\), and the coefficient for \(S_{2}O_{3}^{2-}\) (which is \(d\)) is \(8\).
Step 4: Calculate the ratio.
The ratio of \(c\) to \(d\) is: \[ \frac{c}{d} = \frac{24}{8} = \frac{3}{1} \]
Step 1: Split into half-reactions.
Reduction: \(S_{8} + 16e^{-} \rightarrow 8S^{2-}\) Oxidation: \(S_{8} + 24OH^{-} \rightarrow 4S_{2}O_{3}^{2-} + 12H_{2}O + 24e^{-}\)
Step 2: Equalize the number of electrons transferred.
To balance the electrons, we multiply the reduction half-reaction by \(3\) and the oxidation half-reaction by \(2\): * Reduction: \(3 \times (S_{8} + 16e^{-} \rightarrow 8S^{2-}) \Rightarrow 3S_{8} + 48e^{-} \rightarrow 24S^{2-}\) * Oxidation: \(2 \times (S_{8} + 24OH^{-} \rightarrow 4S_{2}O_{3}^{2-} + 12H_{2}O + 24e^{-}) \Rightarrow 2S_{8} + 48OH^{-} \rightarrow 8S_{2}O_{3}^{2-} + 24H_{2}O + 48e^{-}\)
Step 3: Combine and simplify.
Adding the two balanced half-reactions gives: \(5S_{8} + 48OH^{-} \rightarrow 24S^{2-} + 8S_{2}O_{3}^{2-} + 24H_{2}O\) In the balanced equation, the coefficient for \(S^{2-}\) (which is \(c\)) is \(24\), and the coefficient for \(S_{2}O_{3}^{2-}\) (which is \(d\)) is \(8\).
Step 4: Calculate the ratio.
The ratio of \(c\) to \(d\) is: \[ \frac{c}{d} = \frac{24}{8} = \frac{3}{1} \]
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