[Arpayon]

Hi everyone,
I'm studying for my Chemistry exam at the university.
I've got to balance the following redox
ICl₃ + HNO₃  HIO₃ + Cl₂ + NOCl
Usually i find which element oxidates and which reducts, and then i balance the 2 semi reactions. A sum term by term leads me to the result. Problem here is that I have 2 elements that oxidates (Iodine and Chlorine), should I make 3 semi reactions?
If i split in 3 semireactions, the ones involving Chlorine and Nitrogen are not that easy...

[Borek]

No solution.

Do you know algebraic method?

aICl₃ + bHNO₃ -> cHIO₃ + dCl₂ + eNOCl

From hydrogen balance b = c.

From oxygen balance 3b = 3c + e.

That means 3c = 3c + e, or e = 0.

[Hunter2]

First the given equation is not balanced.
To make it more easy seperat the chlorine  ICl₃ => Cl₂ + ICl

Now you have to deal only with Iodine and the Nitrogen in nitric acid. Only two half reactions are necessary.

[AWK]

First the given equation is not balanced.
To make it more easy seperat the chlorine  ICl₃ => Cl₂ + ICl

Now you have to deal only with Iodine and the Nitrogen in nitric acid. Only two half reactions are necessary.

Wrong assumption - see Borek prove.

[Hunter2]

What do you mean

It is easier to solve the equation to use ICl + HNO₃ => HIO₃ + NOCl

Of course the equation is not balanced, on the right side one O to much.

Ready solutions here not allowed I learned.

Half reaction is ICl => HIO₃

and HNO₃ => NOCl

Maybe more easier I⁺ => HIO₃

HNO₃ => NO⁺ even wie dont have ions I⁺ + NO⁺

[Borek]

Ready solutions here not allowed I learned.

Feel free to post solution if you find it, you have my permission to do so.

Did I tell you there is no solution, so you will be not able to post it?

[Hunter2]

ICl₃ + 3 H₂O => HIO₃ + 5 H⁺ + Cl^- + 4 e^- + Cl₂ Oxidation

HNO₃ + 3 H⁺ + 2 e^- => 2 H₂O + NO⁺
Reduction

The rest is mathematics. Second equation times 2 and elimination of electrons water and H⁺

NO⁺ + Cl^- => NOCl

[Borek]

No hand waving please, give the final answer.

[Hunter2]

ICl₃ + 3 H₂O => HIO₃ + 5 H⁺ + Cl^- + 4 e^- + Cl2 Oxidation

HNO₃ + 3 H⁺ + 2 e^- => 2 H₂O + NO⁺

Second equation times 2

2 HNO₃ + 6 H⁺ + 4 e^- => 4 H₂O + 2 NO⁺

Addition

ICl₃ +  2 HNO₃ + H⁺ => HIO₃ +  NO⁺  + NOCl +  Cl₂ + H₂O

H⁺ comes from HNO₃ and the Nitrate will react with NO⁺ and form NO₂

ICl₃ +  3 HNO₃  => HIO₃ +  2 NO₂  + NOCl +  Cl₂ + H₂O

The given equation wasn't complete.

[AWK]

You changed equation by addition of  NO2 on the right side (note, eventual addition of water is allowed)

[Hunter2]

That is corrext.

I change it in this way, eliminate water what gives HNO₂

ICl₃ +  2 HNO₃  => HIO₃ +  HNO₂  + NOCl +  Cl₂ 

But anyway nobody knows what species in this kind of mixture are existing. NO, NO₂ HNO₂ NOCl , HCl etc.

The given equation isn't complete.

[Arpayon]

Solved. My teacher doesn't give full equations, we have to know that in water H+ and Cl- form HCl to balance the equation.
I wrote the 2 oxidations and the reduction, summed the first 2 to form ICl₃ and then summed the reduction. Final equations results in
2ICl₃ + 5HNO₃ + 5HCl 2HIO₃ + 3Cl₂ + 2NOCl + 4H₂O

I think the result of Hunter2 is correct too

[Borek]

The given equation isn't complete.

That's what I was aiming at from the very beginning - equation as given can't be balanced.

we have to know that in water H+ and Cl- form HCl to balance the equation.

In water you can use H⁺, OH^- and H₂O, as they are there. But assuming presence of Cl^- is completely unsupported. Why Cl^- and not Au³⁺?

[Hunter2]

Why gold, there is no gold in the equation. Chlorid by decomposition of the ICl₃. Even NOCl will decompose in water.

But the result from Arpayon also shows that HCl was missing in the first equation.

[Arpayon]

I could leave H⁺ and Cl^- ions, I just put them together because I had the same number of them (5) after the balancing.