[asaaal]

http://pdf.ocr.org.uk/download/pp_10..._gce_f325.pdf?

How do i do question 7a?

[thegodofgod]

(Original post by asaaal)
http://pdf.ocr.org.uk/download/pp_10..._gce_f325.pdf?

How do i do question 7a?

Okay, so you know the permanganate reduction half-equation (the first equation).

You also know the full equation.

Since two half-equations make up the full equation, you can deduce what the hydrogen peroxide oxidation half-equation is by subtracting the permanganate half-equation species from the overall equation, leaving you with the hydrogen peroxide half-equation

[TiTo20]

(Original post by asaaal)
http://pdf.ocr.org.uk/download/pp_10..._gce_f325.pdf?

How do i do question 7a?

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 as two 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?


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[asaaal]

(Original post by TiTo20)
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 as two 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?


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I dont understand how you got this part

OH WAIT you mean cancel eaach side of the equation by itself not opposite sides?!!

[asaaal]

(Original post by thegodofgod)
Okay, so you know the permanganate reduction half-equation (the first equation).

You also know the full equation.

Since two half-equations make up the full equation, you can deduce what the hydrogen peroxide oxidation half-equation is by subtracting the permanganate half-equation species from the overall equation, leaving you with the hydrogen peroxide half-equation

(Original post by TiTo20)
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 as two 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 very much for your help
For b i done:
4.69x10-4/2 =2.345x10-4 moles
1.1725x10-3/0.025 =0.0469 mol dm-3
but for gdm im assuming you: mass x volume?
so mass is nx mr = 1.1725x10-3x34 = 39.865 and volume is 5.499 x10-5

39.865 x 5.499x10-5

so i got 2.19x10-3 gdm-3


ive really messed it up but i dont know how or where such a fail.

[TiTo20]

(Original post by asaaal)
Thank you very much for your help
For b i done:
4.69x10-4/2 =2.345x10-4 moles
1.1725x10-3/0.025 =0.0469 mol dm-3
but for gdm im assuming you: mass x volume?
so mass is nx mr = 1.1725x10-3x34 = 39.865 and volume is 5.499 x10-5

39.865 x 5.499x10-5

so i got 2.19x10-3 gdm-3


ive really messed it up but i dont know how or where such a fail.

The first two bits are right, but you've forgotten the volume that has that about of moles in it. Remember, you USED 25cm^3 when there was 250cm^3 to START off with, so the number of moles of H2O2 is times by ten.

Think you can take it from here?


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[asaaal]

(Original post by TiTo20)
The first two bits are right, but you've forgotten the volume that has that about of moles in it. Remember, you USED 25cm^3 when there was 250cm^3 to START off with, so the number of moles of H2O2 is times by ten.

Think you can take it from here?



This was posted from The Student Room's iPhone/iPad App

Wait so how do i get gdm?

[TiTo20]

(Original post by asaaal)
Wait so how do i get gdm?

What did you get for your answer?


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