# 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.

• Welcome to chemistry.SE! If you had any questions about the policies of our community, you can ‎visit the help center or take a ‎‎tour of the website.‎ – M.A.R. Feb 20 '15 at 16:12
• Firstly, sodium periodate is NaIO4, not NaIO3. – Shafter Feb 20 '15 at 16:34
• @MARamezani I think you're editing to aggressively. If his/her exam said periodate, and he/she doesn't understand what periodate is, that could be addressed in the answer. If you change "periodate" to "iodate" it totally changes the question. It's easier to give a good answer if someone can see what the person is actually asking with out trying to dig through the edits and find the Dead Sea scrolls. – DavePhD Feb 20 '15 at 17:43
• @DavePhD Sorry, I thought that was too minor and was a typo rather than a misbelief. "Inexperiencedness" causes trouble. – M.A.R. Feb 20 '15 at 17:46

@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.

• I'm trying to find an algorithm for this kind of problems, so the first step is equating the iodine atoms, shouldn't we be doing the same for sulphur (multiplying SO2 by 2)? And then why do we raise the coefficient only of H2SO4 to get the 7 ions? I suppose that is a combination of charge and inspection type balancing? – user14537 Feb 20 '15 at 19:16
• The first step is to calculate the "change of oxidation number" for the element iodine. It's from 7 to zero, i.e. 7 . As we have two atoms of iodines in the left side, the change is 14. – Yomen Atassi Feb 20 '15 at 19:25
• You mean right side, yes I understand that. – user14537 Feb 20 '15 at 19:30
• The first step is to calculate the "change of oxidation number" for the element iodine. It's from 7 to zero, i.e. 7 . As we have two atoms of iodines in the left side, the change is 14. So, we multiply the periodate by two, to balance the element iodine. On the other hand, the change of oxidation number" for the element sulfur is 2, and as the changes must be equal for the two elements (sulfur and iodine), we multiply SO2 by 7 (so the change will be 14). To balance the element sulfur, we multiply sulfuric acid by 6 as we have already the element sulfur in Na2SO4. Then we add water to balance – Yomen Atassi Feb 20 '15 at 19:31
• Yes I understand that, but why don't we e.g. multiply Na2SO4 by 3 and H2SO4 by 4? – user14537 Feb 20 '15 at 19:37

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

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

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

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.

• Can't it be solved directly by the oxidation number method then? I am not really familiar with this method with the hydroxide ions. By the way I meant to write Sodium iodate, I just got confused and wrote periodate the first time. – user14537 Feb 20 '15 at 17:51
• @user14537 please see my second answer. I hope it's clear now. – Yomen Atassi Feb 20 '15 at 18:48