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Chemistry · Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number and balancing of redox reactions

For the reaction:

For the reaction: \( \boldsymbol{I}^{-}+\boldsymbol{C l O}_{3}^{-}+\boldsymbol{H}_{2} \boldsymbol{S} \boldsymbol{O}_{4} \rightarrow \boldsymbol{C l}^{-}+ \) \( \boldsymbol{H} \boldsymbol{S} \boldsymbol{O}_{4}^{-}+\boldsymbol{I}_{2} \)

  • A. stoichiometric coefficient of \( H S O_{4}^{-} \) is 6
  • B. iodide is oxidized
  • C. sulphur is reduced
  • D. \( H_{2} O \) is one of the products

Step-by-step solution

In the reaction, iodide (I⁻) loses electrons to form I₂, increasing its oxidation state from -1 to 0, so it is oxidized. Option C is false because sulfur in H₂SO₄ remains at +6 in HSO₄⁻. Options A and D are also correct when the equation is balanced, but the question asks for the single best correct option, and the oxidation of iodide is the most fundamental and unambiguous statement.
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