Half Reactions
- Page ID
- 31721
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S O- O O- \\ / \\ / S S // \ // \ O OH O OH
Thus, one of the two \(\ce{S}\) atoms has an oxidation state of -2, and we represent this \(\ce{S}\) atom by (\(\ce{=S}\)) to indicate that it is attached to another \(\ce{S}\) atom by a double bond (=).
- In this reaction, one \(\ce{S}\) atom goes from -2 to 0, whereas the oxidation state of the other \(\ce{S}\) atom does not change. You have to assume that the \(\ce{S}\) atom is oxidized by a reducing agent, \(\ce{H2O}\).
- Only the key elements are given on the left in the half-reactions:
\(\ce{HS(=S)O_3^- \rightarrow S +} \textrm{...}\ce{(HSO_4^- )}\)
\(\ce{H_2O \rightarrow H_{2\large{(g)}}} + \textrm{...}\ce{(HSO4- )}\) - Add electrons to compensate for the oxidation changes:
\(\ce{HS(=S)O3- \rightarrow S + 2 e- + HSO4-}\)
\(\ce{H2O + 2 e- \rightarrow H2(g) + HSO4-}\) - Combining the two half-reactions gives the following balanced chemical equation:
\(\ce{HS(=S)O3- + H2O \rightarrow S + H_{2\large{(g)}} + HSO4-}\)
Contributors and Attributions
Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)