[Big-Daddy]

The oxidation of thiocyanate ions by permanganate in acid solution is applied in quantitative analysis. Express the reaction equation.

I worked out that the half-equations are

MnO₄^- + 8H⁺ + 5e^-  Mn²⁺ + 4 H₂O
SCN^- + 4H₂O  SO₄^2- + CN^- + 8H⁺ + 6e^-
2CN^- + 4H₂O  2CO₂ + N₂ + 8H⁺ + 10e^-

I don't know if I should actually need 5 equations because both S and C oxidation numbers change in the second, and both C and N oxidation numbers change in the third. Let's just go with the above three for now.

How do I balance them to find the overall reaction equation?

[Borek]

Are you sure it is not oxidized to (SCN)₂?

But assuming you are right, why not just

SCN^-  CO₂ + N₂ + SO₄^2-

[Big-Daddy]

Are you sure it is not oxidized to (SCN)₂?

I am not sure of that, but the question came under heading "balancing more than 2 half-equations" and the oxidation you propose is just a simple case of 2 half-equations so I think we should proceed with the 3 in the OP or maybe more if necessary.

But assuming you are right, why not just

SCN^-  CO₂ + N₂ + SO₄^2-

So you eliminated CN^-. How do you know CN^- is not a final product though?

If it could be discerned what all the products are, in a given case, we could balance by making sure any other species (including e^-) cancels out for the final equation. But while it is self-evident for 2 half-equations which species need to cancel - only e^- necessarily - it is not so obvious to me here that CN^- should be cancelled.

[Borek]

Quote entire question.

If CN^- is the final product, your third equation doesn't make sense as well.

3 posts, and it already starts to look like another ghost chase.

[Big-Daddy]

Quote entire question.

3 posts, and it already starts to look like another ghost chase.

These things look like a ghost chase because I am trying to understand, not just answer one given homework question.

The entire question is this:

"The oxidation of thiocyanate ions by permanganate in acid solution is applied in quantitative analysis. Express the reaction equation."

Three half-equations identical to mine in the first post are provided for the this reaction in the solutions manual.

If CN^- is the final product, your third equation doesn't make sense as well.

Just to be clear, the three half-equations should be assumed correct for this question.

If we know to eliminate CN^- then it is clear how to balance it. But how did you know, it is not the case that e.g. some CN^-, of that produced in the second half-equation, is then converted in the third half-equation whereas some ends up in the final product? Was it oxidation numbers?

[Borek]

Poorly worded question, and poorly defined chemistry IMHO. SCN^- can be oxidized not only to (SCN)₂, but also to OSCN^- (plus things you have listed earlier), so there are plenty of possibilities, selecting half equations they listed as a "correct" answer seems completely random to me.

[Big-Daddy]

Perhaps it was experimentally found that when permanganate is the oxidizing agent, the SCN- is oxidized via the two-step half-reactions shown in the solutions. I don't know about that.

But as I said I am trying to understand how to balance for three half-equations. What I want to know is, how you knew that we need to eliminate CN^- completely - that it would not be a final product at all?

Edit: Ahhh - perhaps it is oxidation numbers? If there are 3 half-equations, maybe it is accurate to say only 3 elements should change their oxidation numbers overall. If we accept these 3 half-equations, Mn, N and S can all be seen to change their oxidation numbers. But C has the same oxidation number in CO₂ as in SCN^- (4+) but has 2+ oxidation state in CN^-. Therefore, to keep C as having the same oxidation state overall, we must eliminate CN^- because that would be a change in oxidation state for C, whereas producing CO₂ has no change in oxidation state. Decent explanation for how to balance these 3?

[Borek]

You can't balance for three separate half reactions. What they do is they assume oxidation is a two step process - first, SCN^- is oxidized to CN^-, then - if there is still enough permanganate present - CN^- is oxidized.

[Big-Daddy]

I see. Thanks for clearing this one up.

What about this similar sum:

"Balance reaction XO₂⁺ + YO⁺  X₂O₄^3- + Y^- + Y₃O₇^2- occurring in basic solution." (Problem 5 on http://www.chemteam.info/Redox/Redox-ThreeEquations.html)

They've left out the OH^- and H₂O from their reaction equation of course, we'll have to put that back in.

I wrote the three half-equations

2 XO₂⁺ + 5e^-  X₂O₄^3-
YO⁺ + 2 H⁺ + 4e^-  Y^- + H₂O
3 YO⁺ + 4 H₂O  Y₃O₇^2- + 8 H⁺ + 3e^-

So we've got some form of disproportionation being enabled by XO₂⁺ acting as an oxidizing agent.

According to the given answers, we just need to add up the "two reductions" and then balance them by multiplying the oxidation reaction by 3 (to cancel e^-). But why is this odd procedure being taken?

[AWK]

But why is this odd procedure being taken?

This equation has many solutions. Autor, in cited link, gives us only one solution. Try other possible combinations of these half equations, eg 1, 4, 7 (1x5e^-+4x4e^-=7x3e^-) or 2, 5, 10 (2x5+5x4=10x3) and so on.

[Borek]

IOW: sad to say, but apparently whoever did that page didn't understand what they were doing. This equation can't be balanced in a unique way. They are doing some unjustified tricks so they get a random answer.

[Big-Daddy]

IOW: sad to say, but apparently whoever did that page didn't understand what they were doing. This equation can't be balanced in a unique way. They are doing some unjustified tricks so they get a random answer.

Are you sure it's not that we're all missing a constraint here?

I tried solving an analogous problem, this time in IChO: "O₂ oxidizes S₂O₃^2- to S₃O₆^2- and SO₄^2-. Write the reaction equation." By using the same method - I added together the two oxidations, with coefficients of 1 each, and then multiplied the reduction of O₂ to match it - I finally ended up with the right answer for this, 2 O₂ + 2 S₂O₃^2-  SO₄^2- + S₃O₆^2-. If this is a unique solution (and indeed no others are listed, I checked the answers) then doesn't that mean there is some logic behind the method followed? Either that, or there must be another, logical way to do this IChO sum which only so happens to give the same result as the (wrong) method from the website.

[Borek]

Are you sure it's not that we're all missing a constraint here?

Yes I am. AWK already explained why.

O₂ oxidizes S₂O₃^2- to S₃O₆^2- and SO₄^2-

This one has the unique answer, the other one has no unique answer:

2NO₂⁺ + 40YO⁺ + 74OH^- N₂O₄^3- + 7Y^- + 11Y₃O₇^2- + 37H₂O

is as correct as the one they listed. I could list many more "correctly" balanced equations, but that would be just a cheap trick:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-failure

I believe you wrote in the past you know how to use the algebraic method, so it should be obvious for you: when using algebraic method, thiosulfate equation gives three separate balances (S, O and charge), and requires finding 4 coefficients - in general it is not possible to find a solution, but we have a constraint as we know we look for the set of lowest positive integers, that's equivalent of having 4th equation and in general system can be solved. The former equation needs 7 coefficients, there are 5 balances (X, Y, H, O and charge), so even after adding the lowest integers constraint we are still one equation short. That means infinitely many solutions, period.

In both cases it may happen there is no solution at all, but if there is no solution, there is no way to balance the equation correctly anyway.

[Big-Daddy]

I believe you wrote in the past you know how to use the algebraic method, so it should be obvious for you: when using algebraic method, thiosulfate equation gives three separate balances (S, O and charge), and requires finding 4 coefficients - in general it is not possible to find a solution, but we have a constraint as we know we look for the set of lowest positive integers, that's equivalent of having 4th equation and in general system can be solved.

I see. So (without knowing of any pro or against experimental evidence) I will assume you guys are right about the reactions on the three half-reactions page.

I am familiar with the algebraic method. And it does give the unique solution here. But you have also assumed in setting up the equations that there is no H⁺ or H₂O. If we had to involve these, then suddenly we have 4 equations (plus the Diophantine restraint) but 6 coefficients. So then, will there be infinite possible solutions? Or is there a method to use redox balancing in some way or other to find the coefficients anyway? (and in such a method, for this particular question, H⁺ and H₂O will cancel in the end - but at least we did not assume so from the beginning.)

[Borek]

But you have also assumed in setting up the equations that there is no H⁺ or H₂O

I didn't assume anything, I just checked it works.

If we had to involve these

What for? Reaction is already balanced.