A level chemistry help

Hi guys,
I know this sounds a bit daft as we've been balancing equations since GCSE, but how do we balance really complex equations I just can't get my head around it.
Thanks x

Reply 1

Methene

Hey there,
What exam board are you doing? And feel free to post any examples that have you stumped... :smile:

Reply 2

UtterlyUseless69

Original post by LoneGiraffex

Hi guys,
I know this sounds a bit daft as we've been balancing equations since GCSE, but how do we balance really complex equations I just can't get my head around it.
Thanks x

As above, it depends on what question you have been given.

I would say the hardest equations to balance at A level (if you haven’t seen the trick I’m about to discuss) usually are the redox ones where you have say MnO4^- being reduced by something like (COOH)2 in acidic solution to form Mn^2+ and CO2.

The way to do that is to break it up into half-equations and scale them up appropriately.

The reduction half equation would be

MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O

Since we know that Mn^2+ is a product. We also know that to balance out the oxygens, we should use H^+ ions (as the solution is acidic) and water. Working out how many electrons can be done by either working out the change in oxidation state (+7 —> +2 implying a gain of 5 electrons) or by trying to balance the charges (-1 + 8 + x(-1) = 2, so 7 - x = 2 and thus x = 5).

The oxidation half-equation is as follows:

(COOH)2 —> 2CO2 + 2H^+ + 2e^-

In this case, we know CO2 is a product and getting rid of hydrogens is most easily done by turning them into H^+ and e^-.

This gives us the equations

MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O
(COOH)2 —> 2CO2 + 2H^+ + 2e^-

Scaling these up so the numbers of electrons match (i.e x the first equation by the number of electrons there are in the second and x the second equation by the number of electrons in the first):

2MnO4^- + 16H^+ + 10e^- —> 2Mn^2+ + 8H2O
5(COOH)2 —> 10CO2 + 10H^+ + 10e^-

Adding these together and cancelling down things that occur on both sides:

2MnO4^- + 16H^+ + 10e^- + 5(COOH)2 —> 2Mn^2+ + 8H2O + 10CO2 + 10H^+ + 10e^-

2MnO4^- + 6H^+ + 5(COOH)2 —> 2Mn^2+ + 8H2O + 10CO2

And this is the correct final answer.

Reply 3

LoneGiraffex

Original post by Methene

Hey there,
What exam board are you doing? And feel free to post any examples that have you stumped... :smile:

Hi I'm doing OCR.
The question I'm stumped on should be quite simple but the thing is I don't really have a method when balancing.
Here's the question:
__Al +__CH3COOH --> __(CH3COO)3Al + __H2

Reply 4

LoneGiraffex

Original post by UtterlyUseless69

As above, it depends on what question you have been given.
I would say the hardest equations to balance at A level (if you haven’t seen the trick I’m about to discuss) usually are the redox ones where you have say MnO4^- being reduced by something like (COOH)2 in acidic solution to form Mn^2+ and CO2.
The way to do that is to break it up into half-equations and scale them up appropriately.
The reduction half equation would be
MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O
Since we know that Mn^2+ is a product. We also know that to balance out the oxygens, we should use H^+ ions (as the solution is acidic) and water. Working out how many electrons can be done by either working out the change in oxidation state (+7 —> +2 implying a gain of 5 electrons) or by trying to balance the charges (-1 + 8 + x(-1) = 2, so 7 - x = 2 and thus x = 5).
The oxidation half-equation is as follows:
(COOH)2 —> 2CO2 + 2H^+ + 2e^-
In this case, we know CO2 is a product and getting rid of hydrogens is most easily done by turning them into H^+ and e^-.
This gives us the equations
MnO4^- + 8H^+ + 5e^- —> Mn^2+ + 4H2O
(COOH)2 —> 2CO2 + 2H^+ + 2e^-
Scaling these up so the numbers of electrons match (i.e x the first equation by the number of electrons there are in the second and x the second equation by the number of electrons in the first):
2MnO4^- + 16H^+ + 10e^- —> 2Mn^2+ + 8H2O
5(COOH)2 —> 10CO2 + 10H^+ + 10e^-
Adding these together and cancelling down things that occur on both sides:
2MnO4^- + 16H^+ + 10e^- + 5(COOH)2 —> 2Mn^2+ + 8H2O + 10CO2 + 10H^+ + 10e^-
2MnO4^- + 6H^+ + 5(COOH)2 —> 2Mn^2+ + 8H2O + 10CO2
And this is the correct final answer.

Hi, thanks for this

Reply 5

aamina.hs

Original post by LoneGiraffex

Hi I'm doing OCR.
The question I'm stumped on should be quite simple but the thing is I don't really have a method when balancing.
Here's the question:
__Al +__CH3COOH --> __(CH3COO)3Al + __H2

The way I would do this is:

Count where there's an imbalance of atoms. Right away I can see there's three CH3COO- ions on the right (the little 3 after the bracket) but only one on the left so multiply the CH3COOH on the left by 3.
__Al + 3CH3COOH --> __(CH3COO)3Al + __H2

That would form 3 H+ ions on the left but there's 2 hydrogen atoms on the right. So multiply the H2 on the right by 3/2. It's now balanced.
__Al + 3CH3COOH --> __(CH3COO)3Al + 3/2H2

This obvs results in non integer numbers so you could multiply everything through by two to get:
2Al +6CH3COOH --> 2(CH3COO)3Al + 3H2
But generally, you should get marks for multiples (assuming here though, might depend on ur exam board).

Hope this helps 🙂

Reply 6

LoneGiraffex

Original post by aamina.hs

The way I would do this is:

That would form 3 H+ ions on the left but there's 2 hydrogen atoms on the right. So multiply the H2 on the right by 3/2. It's now balanced.
__Al + 3CH3COOH --> __(CH3COO)3Al + 3/2H2

Hope this helps 🙂

Thank you so much!
That actually makes so much sense.
I'll take this approach when I encounter future questions like this. x

Reply 7

UtterlyUseless69

Original post by lonegiraffex

Hi I'm doing OCR.
The question I'm stumped on should be quite simple but the thing is I don't really have a method when balancing.
Here's the question:
__Al +__CH3COOH --> __(CH3COO)3Al + __H2