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

## 1 Answer

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).