Which of the following is the correctly balanced chemical equation for the reaction: io3−(aq) + h2so3(aq) → i2(aq) + so42−(aq)?

• A. io3−(aq) + 3 h2so3(aq) → i2(aq) + 3 so42−(aq) + 3 h2o(l)
• B. io3−(aq) + h2so3(aq) → i2(aq) + so42−(aq) + h2o(l)
• C. 2 io3−(aq) + 5 h2so3(aq) → i2(aq) + 5 so42−(aq) + 5 h2o(l)
• D. 2 io3−(aq) + 5 h2so3(aq) → i2(aq) + 5 so42−(aq) + 6 h2o(l)

Answer: (D)

Verified step by step guidance

1. Identify the oxidation states of iodine in the reactant ion \( \mathrm{IO_3^-} \) and in the product \( \mathrm{I_2} \) to determine the change in oxidation number, which will help in balancing the redox reaction.
2. Write the half-reactions for oxidation and reduction separately. For this reaction, \( \mathrm{IO_3^-} \) is reduced to \( \mathrm{I_2} \), and \( \mathrm{H_2SO_3} \) (sulfurous acid) is oxidized to \( \mathrm{SO_4^{2-}} \) (sulfate ion).
3. Balance each half-reaction for all elements except hydrogen and oxygen first, then balance oxygen atoms by adding \( \mathrm{H_2O} \) molecules, and balance hydrogen atoms by adding \( \mathrm{H^+} \) ions (if in acidic solution).
4. Balance the charges in each half-reaction by adding electrons \( (e^-) \) to the appropriate side, ensuring that the total charge is the same on both sides of each half-reaction.
5. Multiply the half-reactions by appropriate coefficients so that the number of electrons lost in oxidation equals the number gained in reduction, then add the half-reactions together, canceling out electrons and any other species that appear on both sides, to get the balanced overall equation.