Reaction balance

I got this redox reaction on my exam that I didn't know how to balance.
$$Sodium~iodate+sulphur~dioxide+water\ce{->}sodium~sulphate+sulphuric~acid+iodine$$, thus it would be:
$$\ce{NaIO3 +SO2 +H2O->Na2SO4 +H2SO4 +I2}$$
How does one go about balancing this using the oxidation number method?
I understand that S goes from +4 to +6 losing 2 electrons and I from +5 to 0 gaining 5 electrons.


Do we need to consider that there's 2 S atoms on the products (thus each atom loses 1 electron)? So I guess we need to only multiply the reactants' side sulphur with 5/2, while in other cases we need to multiply sulphur both in the reactants and the products side.




@user14537 sent a comment asking if we can solve the exercise using only oxidation number method:
$$\ce{NaIO4 + SO2 +H2O ->Na2SO4 +H2SO4 +I2}$$


* The change in oxidation number between periodate ion and iodine molecule is $7\times 2=14$ (as we have two atoms in iodine molecule).
So, the first step is to multiply $\ce{NaIO4 }$ by $2$.
* The change in oxidation number between $\ce{SO2}$ and sulfate ion is $2$.
* As the changes must be equal, we have to multiply $\ce{SO2}$ by $7$ and $\ce{H2SO4}$ by
$6$. (As you have already a sulfate ion in $\ce{Na2SO4}$.
* Now, as you have $12$ hydrogen atoms in the right side, you add $6$ molecules of water to the left side:
$$\ce{2NaIO4 + 7SO2 +6H2O ->Na2SO4 +6H2SO4 +I2}$$


The equation is balanced.
I hope it's clear now.


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$\ce{2NaIO3 + 10e^- + 12H+ ->2Na+ +I2 + 6H2O}$


$\ce{5SO2 +10H2O ->5SO4^2- +10e^- + 20H+}$


$\ce{2NaIO3 + 5SO2 + 4H2O ->Na2SO4 + 4H2SO4 + I2}$


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Let's try to solve the question starting by iodate ion and then by periodate ion (as the post changes several times while I am answering the question).


* Iodate ion:
$$\ce{2IO3^- + 10e^- + 6H2O ->12OH^- +I2}$$


$$\ce{ SO2 +4OH^- ->SO4^2- +2e^- +2H2O}$$


If we multiply the second equation by $5$:


$$\ce{ 5SO2 +20OH^- ->5SO4^2- +10e^- +10H2O}$$
 Now, we add the the first and the last equation:


$$\ce{2IO3^- + 5SO2 +8OH^- ->5SO4^2- +4H2O +I2}$$
Now, we can rearrange this equation:
Let's begin by adding $8\ce{H+}$ to the two sides of the equation ( and combining $\ce{H+}$ and $\ce{OH-}$) to make water:
$$\ce{2IO3^- + 5SO2 +4H2O ->5SO4^2- +8H+ +I2}$$
We combine $\ce{4SO4^2- +8H+ }$ to make sulfuric acid:
$$\ce{2IO3^- + 5SO2 +4H2O ->SO4^2- +4H2SO4 +I2}$$
By adding $2\ce{Na+}$ to the two sides of the equation, we find:
$$\ce{2NaIO3 + 5SO2 +4H2O ->Na2SO4 +4H2SO4 +I2}$$


* Periodate ion:


Now, if we try to answer the question with periodate ion:
$$\ce{2IO4^- + 14e^- + 8H2O ->16OH^- +I2}$$
$$\ce{ SO2 +4OH^- ->SO4^2- +2e^- +2H2O}$$


If we multiply the second equation by $7$:
$$\ce{ 7SO2 +28OH^- ->7SO4^2- +14e^- +14H2O}$$
Now, we add the the first and the last equation:
$$\ce{2IO4^- + 7SO2 + 12OH^- -> 6H2O + I2 + 7SO4^2-}$$


Now, we can rearrange this equation:
Let's begin by adding $8\ce{H+}$ to the two sides of the equation ( and combining $\ce{H+}$ and $\ce{OH-}$) to make water:
$$\ce{2IO4^- + 7SO2 + 6H2O -> 6H2SO4 + I2 + SO4^2-}$$
By adding $2\ce{Na+}$ to the two sides of the equation, we find:
 $$\ce{2NaIO4 + 7SO2 +6H2O ->Na2SO4 +6H2SO4 +I2}$$
Conclusion: I think the question must be asked with periodate ion.


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