coconut64
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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!
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charco
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(Original post by coconut64)
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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
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coconut64
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(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.
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charco
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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-
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coconut64
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(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?
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Cian1998
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(Original post by coconut64)
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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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