# What are the half reactions for the redox reaction between SO2 and iodine?

What are the half reactions for the redox reaction between $$\ce{SO2}$$ and iodine? The relevant chemical equation: $$\ce{ SO2 + H2O + I2 -> H2SO4 + 2HI}$$

This is what I think it is, not sure if it's correct.

Oxidation: $$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$$

Reduction: $$\ce{I2 + 2e- -> 2I-}$$

Your initial comment exchange made me realize that you have some difficulties in balancing redox equations. Thus, I decided to include some clues for your benefit.

An important part of writing half-reactions is making sure they're balanced by mass and charge. Since most redox reactions are done in aqueous medium, you can always balance $$\ce{O}$$ by $$\ce{H2O}$$. By doing so, you contribute extra $$\ce{H}$$ to the equation so you can balance them by $$\ce{H+}$$ since it is mostly an acid medium reactions (if they are in base or neutral conditions, you may cancel $$\ce{H+}$$ by adding the same amount of $$\ce{OH-}$$ to both sides of the reaction). Finally, cancel the net plus charge by adding $$\ce{e-}$$s to the appropriate side.

Let's see the easy one first, in your case reduction half reaction where $$\ce{I2}$$ reduce to $$\ce{I-}$$: $$\ce{I2 -> I-}$$. Balance its mass, which gives you: $$\ce{I2 -> 2I-}$$. Now, balance the negative charges by $$\ce{e-}$$s. So, you got balanced reduction half-reaction ($$\ce{e-}$$s are in LHS): $$\ce{I2 + 2e- -> 2I-} \qquad \mathrm{E^\circ = \pu{0.536 V}} \qquad \text{(1)}$$

Now, see the more difficult second equation, the oxidation half reaction where $$\ce{SO2}$$ oxidizes to $$\ce{SO4^2-}$$: $$\ce{SO2 -> SO4^2-}$$. Its $$\ce{S}$$ is already balanced, but $$\ce{O}$$ is not. So, balance it with $$\ce{H2O}$$, which gives you: $$\ce{SO2 + 2H2O -> SO4^2-}$$. Now, balance additional $$\ce{H}$$ by $$\ce{H+}$$, which gives you a mass-balanced equation: $$\ce{SO2 + 2H2O -> SO4^2- + 4H+}$$. Now, balance the negative charges by $$\ce{e-}$$s. So, you got balanced oxidation half-reaction ($$\ce{e-}$$s are in RHS): $$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-} \qquad \mathrm{E^\circ = \pu{0.157 V}} \qquad \text{(2)}$$ If you add (1) and (2) together in order to cancel $$\ce{e-}$$s, you get the redox reaction you are looking for: $$\ce{SO2 + 2H2O + I2 -> SO4^2- + 4H+ + 2I-} \qquad \mathrm{E^\circ_{cell} = \pu{0.693 V}} \qquad \text{(3)}$$

The value of $$\mathrm{E^\circ_{cell}}$$ is positive means the reaction is spontaneous.

(Note: The value of $$\mathrm{E^\circ_{\ce{SO2/SO4^2-}}}$$ is from Ref.1)

Reference:

1. J. A. O’Brien, J. T. Hinkley, S. W. Donne, “Electrochemical Oxidation of Aqueous Sulfur Dioxide II. Comparative Studies on Platinum and Gold Electrodes,” J. Electrochem. Soc. 2012, 159(9), F585–F593 (DOI: 10.1149/2.060209jes).

One path I accept is as proceeding via the created presence of HOI from the interaction of iodine and SO2 with water in a series of reactions:

$$\ce{I2 + H2O = H+ + I- + HOI}$$

$$\ce{SO2 + H2O = H+ + HSO3-}$$

$$\ce{HOI + HSO3- -> H+ + I- + HSO4-}$$

However, the reaction could also proceed to some extent via the following reactions following a disproportionation of hypoiodous acid:

$$\ce{HOI -> 2/3 HI + 1/3 HIO3}$$

$$\ce{1/3 HIO3 + H2SO3 -> 1/3 HI + H2SO4}$$

Bottom line, the system is in reality, I suspect, a little more complex as compared to the reported action of HOCl on aqueous SO2 as only concentrated hypochlorous is likely to form HClO3 (although UV light may provide a radical path with dilute hypochlorous acid and warming), which is not the case with HOI. For my proposed pathway, a supporting related reaction source:

"A further mechanism for SO2 oxidation in solution with sea water proposed via halogen compounds HOCl and HOBr (Vogt et al., 1996.... Simulations by Keene et al. (1998) found reactions with HOCl more rapid than with H2O2 at pH 3–5.5 and HOBr more rapid..."

The above reference also cites radical pathways.

Here is another HOCl/HOBr interaction with SO2 source 'Heterogeneous reactions of SO2 with HOCl and HOBr on ice surfaces'. Note, a reference to possible surface chemistry afoot as well at low temperatures.

The question "What are the half reactions for the redox reaction between SO2 and iodine?" should be more correctly stated as "Postulating a simple reaction pathway, speculate on a suggested half cell reactions for the REDOX of SO2 with aqueous I2 at room temperature."

On the electrolysis based response, I would caution students that electrolysis is based on net reactions, and not a proper source for understanding a reaction system mechanics, but can give measurement-based guidance.