Balancing a Redox Reaction
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Balance the following reaction using oxidation number method. $S + OH^{-} \rightarrow S^{2-} + S_{2}O_{3}^{2-}$ (A) $6OH^{-} + 4S \rightarrow 2S^{2-} + S_{2}O_{3}^{2-} + 3H_{2}O$ (B) $6OH^{-} + 5S \rightarrow 2S^{2-} + 3S_{2}O_{3}^{2-} + 2H_{2}O$ (C) $6OH^{-} + S \rightarrow 2S^{2-} + 2S_{2}O_{3}^{2-} + H_{2}O$ (D) None of these
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Hi Mohammad, let's solve this together. We need to balance the reaction of elemental sulfur with hydroxide ions in a basic medium.
Balancing a Redox Reaction
First, let's identify the oxidation states of sulfur in each species.
Sulfur starts at zero. In sulfide, it is negative two, a reduction. In thiosulfate, with oxygen at negative two, sulfur has an oxidation state of plus two, which is an oxidation.
Reduction: S^0 \rightarrow S^{-2} \text{ (gain of 2e^-)}
Oxidation: 2S^0 \rightarrow S_2O_3^{-2} \text{ (loss of 4e^-)}
To balance the electrons, we multiply the reduction by two so the electron gain equals the electron loss.
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