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Half Reactions

  • Page ID
    31721
  • You may think that the two sulfur atoms in the formula are identical, but they are different. You have to understand the chemistry of these ions and then start to investigate their chemical reaction. The structure of \(\ce{HS2O3-}\) may be compared to that of \(\ce{HSO4-}\):

           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 (=).

    1. 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, \(\ce{H2O}\).
    2. Only the key elements are given on the left in the half-reactions:

      \(\ce{HS(=S)O_3^- \rightarrow S +} \textrm{...}\ce{(HSO_4^- )}\)
      \(\ce{H_2O \rightarrow H_{2\large{(g)}}} + \textrm{...}\ce{(HSO4- )}\)

    3. Add electrons to compensate for the oxidation changes:

      \(\ce{HS(=S)O3- \rightarrow S + 2 e- + HSO4-}\)
      \(\ce{H2O + 2 e- \rightarrow H2(g) + HSO4-}\)

    4. Combining the two half-reactions gives the following balanced chemical equation:

      \(\ce{HS(=S)O3- + H2O \rightarrow S + H_{2\large{(g)}} + HSO4-}\)

    The following questions require one step at a time, but you may take any question and follow the four steps as illustrated in the above examples.

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