How is $thiosulfate$ $"thiosulfate"$ , $S_{2} O_{3}^{2 -}$ $S_2O_3^(2-)$ , oxidized to $dithionate$ $"dithionate"$ , $S_{4} O_{6}^{2 -}$ $S_4O_6^(2-)$ ?

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How is $thiosulfate$ $"thiosulfate"$ , $S_{2} O_{3}^{2 -}$ $S_2O_3^(2-)$ , oxidized to $dithionate$ $"dithionate"$ , $S_{4} O_{6}^{2 -}$ $S_4O_6^(2-)$ ?

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Answer 1 · 2016-06-12T10:09:26

$2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 e^{-}$ $2S_2O_3^(2-) rarr S_4O_6^(2-) + 2e^-$

Explanation:

Is the reaction balanced with respect to mass and charge. If not, then we know that it is wrong.

In thiosulfate the average sulfur oxidation state is $+ I I$ $+II$ . In dithionate, the average oxidation state is $+ I I \frac{1}{2}$ $+II1/2$ (the central sulfur linkage has oxidation state $0$ $0$ , and the termini are $S (V)$ $S(V)$ .

We could assign individual sulfur oxidation numbers in dithionate as... $^{-} O (O =) + V S - 0 S - 0 S - + V S (= O) O^{-}$ $""^(-)O(O=)stackrel(+V)S-stackrel(0)S-stackrel(0)S-stackrel(+V)S(=O)O^(-)$ . The internal sulfurs are conceived to share the electrons in the $S - S$ $S-S$ bonds equally...and are thus formally zerovalent.

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How is $thiosulfate$ $"thiosulfate"$ , $S_{2} O_{3}^{2 -}$ $S_2O_3^(2-)$ , oxidized to $dithionate$ $"dithionate"$ , $S_{4} O_{6}^{2 -}$ $S_4O_6^(2-)$ ?

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How is thiosulfate"thiosulfate", S2O2−3S_2O_3^(2-), oxidized to dithionate"dithionate", S4O2−6S_4O_6^(2-)?

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Explanation:

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How is $thiosulfate$ $"thiosulfate"$ , $S_{2} O_{3}^{2 -}$ $S_2O_3^(2-)$ , oxidized to $dithionate$ $"dithionate"$ , $S_{4} O_{6}^{2 -}$ $S_4O_6^(2-)$ ?