Help needed on half-equation!!

Hi in this later part for this question, it asks for the half equation of the oxidation of the copper II ions and the half equation of hydrogen peroxide.

I really don't understand how to construct the equation for hydrogen peroxide. I know that H2O2 accepts electrons in reduction, but I find it hard to work out what are formed.

So the unbalanced equation will be H2O2 + e- -> products
Why are the products just OH- ? I thought it would be both OH- and H2O...

Thanks!

Hydrogen peroxide can be both an oxidising and a reducing agent.

As an oxidising reagent:

H₂O₂ + 2H⁺ + 2e --> 2H₂O

What about the alkaline condition in which only OH- is formed? Also how do you know what products will form? Is it just something you have to memorise as I am confused about why water is not formed in the alkaline condition thanks.

Alkaline conditions are not usually used. but:

H₂O₂ + 2e --> 2OH^-

Hi in this later part for this question, it asks for the half equation of the oxidation of the copper II ions and the half equation of hydrogen peroxide.

I really don't understand how to construct the equation for hydrogen peroxide. I know that H2O2 accepts electrons in reduction, but I find it hard to work out what are formed.

So the unbalanced equation will be H2O2 + e- -> products
Why are the products just OH- ? I thought it would be both OH- and H2O...

Thanks!