# Inorganic chemistry

So, I was just recapping the equations for the equations of the halogen ions with sulfuric acid and on chem revise it says the half reduction equation is: H2SO4 + 2 H+ + 2 e- --> SO2 + 2 H2O. But shouldn't it be H2SO4 + 2 H+ + 4 e- --> SO2 + 2 H20 for the charges on either side to be balanced because you are supposed to do that?
The charges are balanced on both sides, Oxidising number of sulphu in H2SO4 is +6, and in SO2 is +4, so it gains 2 electrons.
Original post by Taisha.p
The charges are balanced on both sides, Oxidising number of sulphu in H2SO4 is +6, and in SO2 is +4, so it gains 2 electrons.

Yes that is what I originally thought but when looking at the REDOX section and getting taught REDOX it says that you need to take the protons into consideration before adding the electrons.
Original post by Taisha.p
The charges are balanced on both sides, Oxidation number of sulphur in H2SO4 is +6, and in SO2 is +4, so it gains 2 electrons.

Correct.

Original post by skyeforster15
So, I was just recapping the equations for the equations of the halogen ions with sulfuric acid and on chem revise it says the half reduction equation is: H2SO4 + 2 H+ + 2 e- --> SO2 + 2 H2O. But shouldn't it be H2SO4 + 2 H+ + 4 e- --> SO2 + 2 H2O for the charges on either side to be balanced because you are supposed to do that?

If you have 4e^- on the LHS as well as 2H^+, then the overall charge on the LHS is -2 because 2 + 4(-1) = -2. The RHS would need a net charge of -2 as well, but you have already written out all the correct species on the RHS and they are all neutral (so the net charge on the RHS is 0), thus you only need 2e^- on the LHS.

Note: RHS and LHS are exam shorthand for right-hand side and left-hand side, respectively. Thought I’d mention it in case it seemed weird.
Original post by TypicalNerd
Correct.

If you have 4e^- on the LHS as well as 2H^+, then the overall charge on the LHS is -2 because 2 + 4(-1) = -2. The RHS would need a net charge of -2 as well, but you have already written out all the correct species on the RHS and they are all neutral (so the net charge on the RHS is 0), thus you only need 2e^- on the LHS.

Note: RHS and LHS are exam shorthand for right-hand side and left-hand side, respectively. Thought I’d mention it in case it seemed weird.

Yes. I totally understand where you are coming from but when doing the topic of Redox in aqa it says you need to take into consideration the protons for example. (It is in the attached link).
I am just confused about how you are just looking at the electrons for some and for others you need to consider the protons in order to balance out the charges on either side of the equations.
Original post by skyeforster15
Yes. I totally understand where you are coming from but when doing the topic of Redox in aqa it says you need to take into consideration the protons for example. (It is in the attached link).
I am just confused about how you are just looking at the electrons for some and for others you need to consider the protons in order to balance out the charges on either side of the equations.

Alright. I’ll follow the same four steps to derive the equation in question.

(1. H2SO4 has sulphur in the +6 oxidation state, whereas SO2 has sulphur in the +4 oxidation state. The number of electrons we need to add is (+6) - (+4) = +2.

So we get H2SO4 + 2e^- —> SO2

But obviously, this is incomplete.

(2. We need to add water to the right-hand side. As two of the four oxygens from the H2SO4 are on the SO2, we need to account for the two missing oxygens. These have been turned into water, so we get:

H2SO4 + 2e^- —> SO2 + 2H2O

However, this still isn’t complete. Looking at both sides of the equation, there are two hydrogens on the LHS and four on the RHS.

(3. Notice how we’ve got a charge on the LHS but not the RHS? We need enough positive charges on the LHS to cancel each electron. A single hydrogen ion has a charge of +1, so we need two of them. This also means that we balance out the hydrogens. Thus:

H2SO4 + 2H^+ + 2e^- —> SO2 + 2H2O

(4. Let’s check to see if what we have is right.
On the LHS, there is 1 sulphur, 4 hydrogens, 4 oxygens and no net charge.

The RHS has exactly the same, so the equation is balanced correctly.
Original post by TypicalNerd
Alright. I’ll follow the same four steps to derive the equation in question.

(1. H2SO4 has sulphur in the +6 oxidation state, whereas SO2 has sulphur in the +4 oxidation state. The number of electrons we need to add is (+6) - (+4) = +2.

So we get H2SO4 + 2e^- —> SO2

But obviously, this is incomplete.

(2. We need to add water to the right-hand side. As two of the four oxygens from the H2SO4 are on the SO2, we need to account for the two missing oxygens. These have been turned into water, so we get:

H2SO4 + 2e^- —> SO2 + 2H2O

However, this still isn’t complete. Looking at both sides of the equation, there are two hydrogens on the LHS and four on the RHS.

(3. Notice how we’ve got a charge on the LHS but not the RHS? We need enough positive charges on the LHS to cancel each electron. A single hydrogen ion has a charge of +1, so we need two of them. This also means that we balance out the hydrogens. Thus:

H2SO4 + 2H^+ + 2e^- —> SO2 + 2H2O

(4. Let’s check to see if what we have is right.
On the LHS, there is 1 sulphur, 4 hydrogens, 4 oxygens and no net charge.

The RHS has exactly the same, so the equation is balanced correctly.

Another note. When you are not dealing with an oxyanion (an anion with oxygens on it, like SO4^2- or MnO4^-) or an oxide, you ignore H^+ ions, as there won’t be any oxygens to convert to water.
Hello. I do not think that you understand where I am coming from. The equation that I quickly screenshotted on chemrevise clearly shows that they are taken the protons into account when balancing the equations because If they didn't then the resultant half equation would be: MnO4- --> Mn2+ + 4H2O +3e-. Which is clearly incorrect.

However, I used to always do equations like how you did previously and will continue to do so in exams for the reaction of concentrated sulfuric acid and hydrogen bromide as I have looked on mark schemes that this is theoretically correct. But I came across equations in the REDOX topic where you do in fact have to balance the hydrogens. If for example if I take the half equation that you derived of: 2e- + 2H+ + H2SO4 --> SO2 + 2H2O. So if you were to take into account the charges on the LHS it would be: 2+ (due to the 2 positive protons) add 2- (due to the two negative electrons) and the +6 charge of the sulfur in sulfuric acid. This would add to a total of +6. Compared to the RHS which would have a charge of 4+ because of the +4 charge of sulfur in sulfur dioxide and water, which does not have a charge. So when revising these topics I have realised that some equations, such as the one screenshotted above, they have taken into account the protons because if they didn't then there would be a charge of 1- on the LHS and a charge of 2+ on the RHS so to balance it you would have to put 3 electrons on the RHS. Now you can clearly see that they did not do this and took into account the protons as they have added 8 protons and 5 electrons to the LHS so overall it will have a charge of 2+ due to the -ve MnO4- ion as well as a 2+ charge in the RHS.
Original post by skyeforster15
Hello. I do not think that you understand where I am coming from. The equation that I quickly screenshotted on chemrevise clearly shows that they are taken the protons into account when balancing the equations because If they didn't then the resultant half equation would be: MnO4- --> Mn2+ + 4H2O +3e-. Which is clearly incorrect.

However, I used to always do equations like how you did previously and will continue to do so in exams for the reaction of concentrated sulfuric acid and hydrogen bromide as I have looked on mark schemes that this is theoretically correct. But I came across equations in the REDOX topic where you do in fact have to balance the hydrogens. If for example if I take the half equation that you derived of: 2e- + 2H+ + H2SO4 --> SO2 + 2H2O. So if you were to take into account the charges on the LHS it would be: 2+ (due to the 2 positive protons) add 2- (due to the two negative electrons) and the +6 charge of the sulfur in sulfuric acid. This would add to a total of +6. Compared to the RHS which would have a charge of 4+ because of the +4 charge of sulfur in sulfur dioxide and water, which does not have a charge. So when revising these topics I have realised that some equations, such as the one screenshotted above, they have taken into account the protons because if they didn't then there would be a charge of 1- on the LHS and a charge of 2+ on the RHS so to balance it you would have to put 3 electrons on the RHS. Now you can clearly see that they did not do this and took into account the protons as they have added 8 protons and 5 electrons to the LHS so overall it will have a charge of 2+ due to the -ve MnO4- ion as well as a 2+ charge in the RHS.

Ok so does the question boil down to ‘why put hydrogens on the LHS and not electrons on the RHS’?

In this case, it’s because there are two oxygens on the sulphate ion that need to go. In redox chemistry, the only real way to do this is to turn them into water, but you can’t have two water molecules without four atoms of hydrogen. That’s why you have to use hydrogen ions on one side.

Let’s even construct another equation to show the difference between what happens when you balance with hydrogen ions, rather than electrons. How about the reduction of manganese (IV) oxide in acidic solution to form manganese (II) ions?

(1. The Mn goes from (IV) —> (II), so 2e^- are needed, hence: MnO2 + 2e^- —> Mn^2+.

(2. The two oxygens on the LHS are converted to water, and so far we have: MnO2 + 2e^- —> Mn^2+ + 2H2O

If we try to balance the charges by adding electrons only:

(3. The total charge on the LHS is -2 and on the RHS it’s +2, so adding 4e^- to the RHS will balance the charges:

MnO2 + 2e^- —> Mn^2+ + 2H2O + 4e^-

(4. Cancelling:

MnO2 —> Mn^2+ + 2H2O + 2e^-

Now this has to be incorrect. It suggests that four hydrogen atoms have popped out of nowhere and also would suggest that MnO2 is actually losing electrons, rather than gaining them.

Let’s now use the correct steps:

(3. Let’s add hydrogen ions to the LHS, as they simultaneously balance out the charge and fix the oxygen problem. We’ll need 4 of them, as there are 4 hydrogens on the RHS.

MnO2 + 2e^- + 4H^+ —> Mn^2+ + 2H2O

(4. Let’s check everything. On both sides, we have the same total charge and the same numbers of each element. Clearly, this is right.
Original post by TypicalNerd
Ok so does the question boil down to ‘why put hydrogens on the LHS and not electrons on the RHS’?

In this case, it’s because there are two oxygens on the sulphate ion that need to go. In redox chemistry, the only real way to do this is to turn them into water, but you can’t have two water molecules without four atoms of hydrogen. That’s why you have to use hydrogen ions on one side.

Let’s even construct another equation to show the difference between what happens when you balance with hydrogen ions, rather than electrons. How about the reduction of manganese (IV) oxide in acidic solution to form manganese (II) ions?

(1. The Mn goes from (IV) —> (II), so 2e^- are needed, hence: MnO2 + 2e^- —> Mn^2+.

(2. The two oxygens on the LHS are converted to water, and so far we have: MnO2 + 2e^- —> Mn^2+ + 2H2O

If we try to balance the charges by adding electrons only:

(3. The total charge on the LHS is -2 and on the RHS it’s +2, so adding 4e^- to the RHS will balance the charges:

MnO2 + 2e^- —> Mn^2+ + 2H2O + 4e^-

(4. Cancelling:

MnO2 —> Mn^2+ + 2H2O + 2e^-

Now this has to be incorrect. It suggests that four hydrogen atoms have popped out of nowhere and also would suggest that MnO2 is actually losing electrons, rather than gaining them.

Let’s now use the correct steps:

(3. Let’s add hydrogen ions to the LHS, as they simultaneously balance out the charge and fix the oxygen problem. We’ll need 4 of them, as there are 4 hydrogens on the RHS.

MnO2 + 2e^- + 4H^+ —> Mn^2+ + 2H2O

(4. Let’s check everything. On both sides, we have the same total charge and the same numbers of each element. Clearly, this is right.

Hello, I am unsure if you are using a different example but if this is off the previous equation that I attached isn't it MnO4-. I really appreciate you trying to help I think my previous email sounded a bit aggressive so sorry about that. Obviously, I know that you are right because I mean me against all the chemists and scientists in the world but the issue isn't about the hydrogens and electrons, I have balanced many equations before, and it's only now that I have actually come across my issue I honestly just think that I am overthinking it and I need to take a step back.
Original post by TypicalNerd
Ok so does the question boil down to ‘why put hydrogens on the LHS and not electrons on the RHS’?

In this case, it’s because there are two oxygens on the sulphate ion that need to go. In redox chemistry, the only real way to do this is to turn them into water, but you can’t have two water molecules without four atoms of hydrogen. That’s why you have to use hydrogen ions on one side.

Let’s even construct another equation to show the difference between what happens when you balance with hydrogen ions, rather than electrons. How about the reduction of manganese (IV) oxide in acidic solution to form manganese (II) ions?

(1. The Mn goes from (IV) —> (II), so 2e^- are needed, hence: MnO2 + 2e^- —> Mn^2+.

(2. The two oxygens on the LHS are converted to water, and so far we have: MnO2 + 2e^- —> Mn^2+ + 2H2O

If we try to balance the charges by adding electrons only:

(3. The total charge on the LHS is -2 and on the RHS it’s +2, so adding 4e^- to the RHS will balance the charges:

MnO2 + 2e^- —> Mn^2+ + 2H2O + 4e^-

(4. Cancelling:

MnO2 —> Mn^2+ + 2H2O + 2e^-

Now this has to be incorrect. It suggests that four hydrogen atoms have popped out of nowhere and also would suggest that MnO2 is actually losing electrons, rather than gaining them.

Let’s now use the correct steps:

(3. Let’s add hydrogen ions to the LHS, as they simultaneously balance out the charge and fix the oxygen problem. We’ll need 4 of them, as there are 4 hydrogens on the RHS.

MnO2 + 2e^- + 4H^+ —> Mn^2+ + 2H2O

(4. Let’s check everything. On both sides, we have the same total charge and the same numbers of each element. Clearly, this

OH my goodness, I am such an idiot I know exactly what I have done. I was taking the charge of the whole thing so for example for the MnO4- as -1. Instead of just the charge for the Mn. I think this is enough chemistry for one day.
Original post by skyeforster15
Hello, I am unsure if you are using a different example but if this is off the previous equation that I attached isn't it MnO4-. I really appreciate you trying to help I think my previous email sounded a bit aggressive so sorry about that. Obviously, I know that you are right because I mean me against all the chemists and scientists in the world but the issue isn't about the hydrogens and electrons, I have balanced many equations before, and it's only now that I have actually come across my issue I honestly just think that I am overthinking it and I need to take a step back.

Yes it was a different example, and no, you didn’t come across as aggressive. Perhaps a little frustrated at most, but that’s only natural.

A level chemistry is hard, but you have at least made an effort to do something about weak points you have identified.

I reckon there is an element of overthinking involved. Perhaps give yourself a few hours break from studying full stop?