Writing overall equations from half equations
Chemistry discussion, revision, exam and homework help.
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Writing overall equations from half equations
Does anyone know how to write half equations from the overall equation?
For example:
2MnO4^- + 6H^+ + 5 H2O2 --> Mn^2+ + 8h2o + 5o2
Write the simplest whole number half equation for the oxidation of h2o2
You are also given MnO4^- + 8h+ 5e- --> Mn^2+ + 4h20
F25 June 2010 paper question 7 a -
There's a forum post about your question already...(Original post by Kkkkkkkk)
Does anyone know how to write half equations from the overall equation?
For example:
2MnO4^- + 6H^+ + 5 H2O2 --> Mn^2+ + 8h2o + 5o2
Write the simplest whole number half equation for the oxidation of h2o2
You are also given MnO4^- + 8h+ 5e- --> Mn^2+ + 4h20
F25 June 2010 paper question 7 a
But never mind, I'll just copy what I wrote on it.
Here we go:
The way I did this question was to look at the first equation on the page.
You can see that it is MnO4- + 8H+ +5 e- --> Mn2+ + 4H2O.
In the second equation, it is 2MnO4-... Etc etc.
You have to times the first equation by two to get 2MnO4- for the second equation, so do that now.
2MnO4- + 16H+ + 10e- --> 2Mn2+ +8H2O.
If you think of it like two math equations, you can take away the new equation above from the second equation, so you get :
2MnO4- + 6H+ + 5H2O2 - 2MnO4- - 16H+ - 10e- --> 2Mn2+ + 8H2O + 5O2 - 2Mn2+ - 8H2O
Cancel anything out if you can and you get left with :
5H2O2 -10H+ - 10e- --> 5O2
Move all the 'negative' signs to the other side (in those case, it's the H+ and the e-) and change them to 'positives'
5H2O2 ---> 5O2 + 10H+ + 10e-
Simplify (everything goes into 5 so divide by 5)
H2O2 --> O2 + 2H+ + 2e-
That's your answer! =]
Did you understand the process I went through?
This was posted from The Student Room's iPhone/iPad App -
Re: How to write overall equations from half equations?Thank you so much! One day before the exam and its all just fallen into place. Better late than never(Original post by TiTo20)
There's a forum post about your question already...
But never mind, I'll just copy what I wrote on it.
Here we go:
The way I did this question was to look at the first equation on the page.
You can see that it is MnO4- + 8H+ +5 e- --> Mn2+ + 4H2O.
In the second equation, it is 2MnO4-... Etc etc.
You have to times the first equation by two to get 2MnO4- for the second equation, so do that now.
2MnO4- + 16H+ + 10e- --> 2Mn2+ +8H2O.
If you think of it like two math equations, you can take away the new equation above from the second equation, so you get :
2MnO4- + 6H+ + 5H2O2 - 2MnO4- - 16H+ - 10e- --> 2Mn2+ + 8H2O + 5O2 - 2Mn2+ - 8H2O
Cancel anything out if you can and you get left with :
5H2O2 -10H+ - 10e- --> 5O2
Move all the 'negative' signs to the other side (in those case, it's the H+ and the e-) and change them to 'positives'
5H2O2 ---> 5O2 + 10H+ + 10e-
Simplify (everything goes into 5 so divide by 5)
H2O2 --> O2 + 2H+ + 2e-
That's your answer! =]
Did you understand the process I went through?
This was posted from The Student Room's iPhone/iPad App
thank you so much!
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No probs.(Original post by Kkkkkkkk)
Thank you so much! One day before the exam and its all just fallen into place. Better late than never
thank you so much!
F325? =P
This was posted from The Student Room's iPhone/iPad App
thank you so much!