[Borek]

As I wrote - that just means you should use H⁺ and H₂O to balance half reactions.

Looking at oxidation numbers it should be obvious to you iodine gets oxidized - that means H₂O₂ gets reduced. If it gets reduced, half reaction must consume electrons:

H₂O₂ + e^- -> .

This needs some guessing now. Reaction posted by Nobby consumes electrons, but it produces OH^-, so it doesn't fit. However, it is rather easy to guess how to use water and H⁺ to balance.

Note that any reaction that produces OH^- can be easily converted to one that doesn't contain OH^- and consumes H⁺ - it is enough to add H⁺ on both sides of the equation.

[Nobby]

@Bat3996 Compare post 4 and 6.

H₂O₂ + 2 e^- => 2 OH^-

can be converted by adding H⁺ to

H₂O₂ + 2 H⁺ + 2 e^- => 2 H₂O

But nevertheless the reaction takes place under basic conditions (Your first post) and that means the equations have to be developed by using OH^-.

@Borek >>Reaction posted by Nobby consumes electrons, but it produces OH-, so it doesn't fit<<

What does this means, it doesn't fit.

[Borek]

@Borek >>Reaction posted by Nobby consumes electrons, but it produces OH-, so it doesn't fit<<

What does this means, it doesn't fit.

It doesn't fit conditions of acidic solution - it is better to balance it using H⁺. Which you later did, I was just trying to explain Bat3996 why these things are equivalent.

[Nobby]

Ok, understand.

I agree in the most cases to balance with  H⁺ is more easier. But in my opinion the people should also learn how it works with OH^-.

A reaction in alkaline conditions maybe will not take place in acidic conditions, what I beleave is also the case by oxiding Iodide with peroxide or the other example of synproportion of ammonia and nitrate.

[Bat3996]

Thank you for all the help guys, I am still working on complete understand, but it helps a lot.