# Having difficulty with Ox-Redox

1. Nov 27, 2007

### superdude

Hello,

Can someone please explain to me something here please? In this oxidation problem

Na2Cr2O7 + HNO3 + Na2SO3 yields NaNO3 + Cr(NO3)3 + Na2SO4 + H2O

I know S is Oxidized and Cr is Reduced. However, when I look at the solution set, I am get completely confused.

They are putting

SO3^2- yields SO4^2- for oxidation and for reduction they have Cr2O7^2- yields 2 Cr ^ 3+.

I understand how to get what is Oxidized and what is reduced.

However, I am having trouble knowing what to eliminate. How can they just get read of the Na2 in the first one and (NO3)3 in the second one? Is there some sort of rule that tells you what you can get rid of? Also, once they do get rid of half of the original ion, how do they assign the charge? Is the charge just the polyatomic charge of sulfite and sulfate?

2. Nov 27, 2007

### superdude

open second review, it seems that they seperated polyatomic ions from the main elements. is this possible? and would you follow this step for other ox redox reactions?

3. Nov 27, 2007

### Gokul43201

Staff Emeritus
Write down each compound as a sum of its "ionic" species (example: $Na_2Cr_2O_7 = 2Na^+ + Cr_2O_7^{2-}$). Then you just throw away all the ions that do not change their oxidation states (example: $Na^+, NO_3^-, H^+)$.

4. Nov 27, 2007

### superdude

So would this turn out to be:

2 Na + Cr2O7^2- + H + NO3 2- + 2 Na + SO3 ^ 2- yields Na + NO3 + Cr(NO3)3 + 2 NA + SO4 ^ 3 + H2O

and when you refer to "ionic" species, what are you referring to exactly?

Thanks for the help

Last edited: Nov 27, 2007
5. Nov 28, 2007

### Gokul43201

Staff Emeritus
I mean that you have to include the charges on all species.

Na and H should both have a charge of 1+ (or simply +). Also, the charge on the nitrate ion (NO3) is 1-, not 2-.

And finally, you have yet to split up Cr(NO3)3 into its parts.

6. Nov 28, 2007

### superdude

I think I am beginning to understand this now.

1. Assign ox states, see what is oxidized and what is reduced.

2. "Split the elements", but if it is a polyatomic ion, you keep that together.

In this problem, you would split

Na2Cr2O7 into 2Na^+1 + 2Cr^6 + 7O^-2.

Then you would split HNO3 into H^+1 + NO3^-1.

Disregarding this problem, should you also keep all Oxygen's next to an element?

For Instance:

I2 + HNO3 -------> NaIO3 + NO3

I is oxidized and N is reduced.

would the equation for I be I2 ---------> I03

and for N be NO3^1 ---------> Na^+1.