# Help needed on half-equation!!

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#1

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!
0
3 years ago
#2
(Original post by coconut64)

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
0
#3
(Original post by charco)
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.
0
3 years ago
#4
(Original post by coconut64)
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-
0
#5
(Original post by charco)
Alkaline conditions are not usually used. but:

H2O2 + 2e --> 2OH-
How do you know that water is not formed in this reaction and OH- is the only product?
0
3 years ago
#6
(Original post by coconut64)

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!
Can I ask what paper this is from? Trying to do the question wanted to check if I am right. Thanks
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