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

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

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

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.

Original post by Methene

Hey there,

What exam board are you doing? And feel free to post any examples that have you stumped...

What exam board are you doing? And feel free to post any examples that have you stumped...

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

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.

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

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 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:

1.

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

2.

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

3.

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).

Original post by aamina.hs

The way I would do this is:

Hope this helps 🙂

1.

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

2.

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

3.

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).

Thank you so much!

That actually makes so much sense.

I'll take this approach when I encounter future questions like this. x

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 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

spoiler

(edited 3 weeks ago)

Original post by UtterlyUseless69

You can follow the same procedure as I described, though the approach described by aamina.hs is likely much easier.

spoiler

This makes perfect sense, thank you for your help x

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