[Atome]

Hello everyone:

I was able to balance the following redox reaction using the oxidation-number method, but I am having some trouble with the half-reaction method.

Thank you very much!

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1. HIO₃ + FeI₂ + HCl -> FeCl₃ + ICl + H₂O

5HIO₃ + 4FeI₂ + 25HCl -> 4FeCl₃ + 8ICl + 5ICl + 15H₂O

● The compounds involved in oxidation and reduction are highlighted:

5HIO₃ + 4FeI₂ + 25HCl -> 4FeCl₃ + 8ICl + 5ICl + 15H₂O

● The redox equation in half-reaction form:

Reduction:

5HIO₃ -> 5ICl

5HIO₃ + 5HCl -> 5ICl + 15H₂O

5HIO₃ + 5HCl + 20H⁺ + 20e^- -> 5ICl + 15H₂O

● I know that this is correct since I determined that the oxidising agent, HIO₃, would lose 20 electrons to ICl.

Oxidation:

4FeI₂ -> 4FeCl₃

4FeI₂ + 12HCl -> 4FeCl₃ + 8HIO₃

4FeI₂ + 12HCl + 24 H₂O-> 4FeCl₃ + 8HIO₃

4FeI₂ + 12HCl + 24 H₂O-> 4FeCl₃ + 8HIO₃ + 52H⁺ + 52e^-

This is incorrect since I should have only 20 electrons in the oxidation, equal to the amount of electrons reduced.

Could someone please tell me where I may have erred?

[Borek]

5HIO₃ -> 5ICl

Why 5 and not 1?

4FeI₂ -> 4FeCl₃

Why 4 and not 1?

And first of all - why not net ionic? Are you specifically asked to balance the equation using uncharged molecules?

There is something fishy going on here, as I^- are not necessary to balance the reaction equation, they are forced into the reaction by the use of FeI₂. In the real solution ions that are spectators can't be forced into the reaction just because they are added.

[Atome]

Thanks for your reply, Borek.

I just tried it using a 1:1 ration and get 13 hydrogen ions and 13 elecrons at the end, instead of 20. I am still making a particular mistake.

Yes, I must balance it with the ion-electron method as this is a redox equation (and not like a normal chemical equation even though there are no charges).

[LQ43]

● The compounds involved in oxidation and reduction are highlighted:

5HIO₃ + 4FeI₂ + 25HCl -> 4FeCl₃ + 8ICl + 5ICl + 15H₂O

I can see the colors now! You highlight 4FeI2 --> 4FeCl3  +  8ICl

yet....

Oxidation:

4FeI₂ -> 4FeCl₃

4FeI₂ + 12HCl -> 4FeCl₃ + 8HIO₃

As Borek pointed out, ideally, the solution to the problem should start with an unbalanced equation as in

HIO3 + FeI2 + HCl -> FeCl3 + ICl + H2O
then  split up into the 2 half reactions. The aim is to arrive at the balanced equation, not to start with it.

[Atome]

I arrived at my final answer.

Thank you for your replies!

I would like to confirm that to obtain the same elements on both sides of the half-reactions, I can only add pure substances to the deficient side only if the pure substance appears in the deficient side in the original equation.

For example, if I have A + B + C -> D + E + F, and the half-reaction A -> D, I cannot add E to A even though E may help to balance.