question about a half equation

Original post by gazbo1

Looking at

O₂(g) + 4H⁺(aq) + 4e^- -> 2H₂O(l)1.23
from http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html

Is that only one half of a redox reaction?

And in that one half reaction there.

Are both Oxygen and Hydrogen being reduced?

Or is it only Oxygen that is being reduced?

what's happening to Hydrogen.. Is it being reduced or staying the same or being oxidised?

Thanks

It is indeed just half a redox reaction- after all, where do the electrons on the LHS come from?

Only the oxygen is reduced- look at the oxidation states. Oxygen goes from 0 in O2 to -2 in H2O, implying reduction. Hydrogen’s oxidation state is unchanged as it is +1 in both H^+ and H2O.

(edited 10 months ago)

Reply 2

OP

Original post by TypicalNerd

It is indeed just half a redox reaction- after all, where do the electrons on the LHS come from?

Only the oxygen is reduced- look at the oxidation states. Oxygen goes from 0 in O2 to -2 in H2O, implying reduction. Hydrogen’s oxidation state is unchanged as it is +1 in both H^+ and H2O.

Thanks..

So how about

2H2O(l) + 2e- -> H2(g) + 2OH-(aq)

I understand then that it's a reduction reaction since the +e- is on the right.. Presumably no oxidation occurring anywhere?

The H on the left is 2- That's oxidised in the H2. That seems strange 'cos isn't it meant to be a reduction reaction?

In the case of OH- , I understand that's an overall negative (OH)- and the - is actually on the O, so more like HO-.. But to get an overall negative, given the - on the O. The H must be oxidation number of 0.

So isn't the H getting oxidised there too in the HO-. From oxidation number of -2 on the left,in the H2O, to being oxidation number 0 on the right in the HO-?

Reply 3

Original post by gazbo1

Thanks..

So how about

2H2O(l) + 2e- -> H2(g) + 2OH-(aq)

I understand then that it's a reduction reaction since the +e- is on the right.. Presumably no oxidation occurring anywhere?

The H on the left is 2- That's oxidised in the H2. That seems strange 'cos isn't it meant to be a reduction reaction?

In the case of OH- , I understand that's an overall negative (OH)- and the - is actually on the O, so more like HO-.. But to get an overall negative, given the - on the O. The H must be oxidation number of 0.

So isn't the H getting oxidised there too in the HO-. From oxidation number of -2 on the left,in the H2O, to being oxidation number 0 on the right in the HO-?

Indeed there is no oxidation taking place- it is just reduction.

Hydrogen is reduced from +1 in H2O to 0 in H2. You may note that in OH^-, the hydrogen also has an oxidation state of +1, so not all of it is reduced.

I think it may help to give a quick reminder of some rules with oxidation states:

-When you have an element, i.e Fe, H2, O2 etc, the oxidation state is 0.

-In compounds, oxygen is almost always -2.

-In compounds, hydrogen and any group 1 element is almost always +1,

-The sum of all the oxidation states is the overall charge on the species.

So, for example, let us consider H2O and deduce the oxidation state of the hydrogen within it.

Given O in compounds like H2O is practically always -2 and H2O has 0 charge, let the oxidation state of the hydrogens be x.

If we use the rule that the sum of the oxidation states is the charge in the species, then:

x + x + (-2) = 0 (there are 2 hydrogens, hence x has been added twice)

==> 2x - 2 = 0
==> 2x = +2
==> x = +1, so hydrogen is in the +1 oxidation state

Reply 4

OP

ah I see, thanks..

I don't know why I made the error of thinking H was -2 on the left..

Yeah so H is +1 on the left. That explains the H2.. , H getting reduced by one there.

OH- i'm not sure of though.. 'cos it's not a compound 'cos a compound is neutral by definition.. It's a polyatomic ion.

But if we say that the rules are the same for compounds as for polyatomic ions.. (though with polyatomic ions,, sum of charges add to overall charge of non-zero)..

So H has +1, and O has 2- So that makes sense. We get an overall charge of -1

I had in mind at one point HO- and the negative charge being on the oxygen but that was definitely wrong then. because a)O there has an oxidation state of 2- not 1- and b)the actual charge on O would be a "partial charge", it'd be less than 1, like with covalent compounds, 'cos polyatomic ions are covalently bound internally

Ia that all right?

Thanks

Reply 5

Original post by gazbo1

ah I see, thanks..

I don't know why I made the error of thinking H was -2 on the left..

Yeah so H is +1 on the left. That explains the H2.. , H getting reduced by one there.

OH- i'm not sure of though.. 'cos it's not a compound 'cos a compound is neutral by definition.. It's a polyatomic ion.

But if we say that the rules are the same for compounds as for polyatomic ions.. (though with polyatomic ions,, sum of charges add to overall charge of non-zero)..

So H has +1, and O has 2- So that makes sense. We get an overall charge of -1

I had in mind at one point HO- and the negative charge being on the oxygen but that was definitely wrong then. because a)O there has an oxidation state of 2- not 1- and b)the actual charge on O would be a "partial charge", it'd be less than 1, like with covalent compounds, 'cos polyatomic ions are covalently bound internally

Ia that all right?

Thanks

Ok, I should really have clarified that the general rules apply for compounds and ions.

So oxygen in the OH^- ion is -2 and the hydrogen is +1. Adding the two oxidation states together gives an overall charge of -1, as expected.

Indeed polyatomic ions are internally covalently bound. The oxygen both has a negative formal charge from the one electron it has picked up from another atom and a partial charge from drawing the electrons in the covalent bond closer to itself. So it’s perhaps a little more complex than at first glance.

Reply 6

OP

Original post by TypicalNerd

Indeed polyatomic ions are internally covalently bound. The oxygen both has a negative formal charge from the one electron it has picked up from another atom and a partial charge from drawing the electrons in the covalent bond closer to itself. So it’s perhaps a little more complex than at first glance.

ah I see thanks.. So O in OH- is a great example of where formal charge differs from oxidation state. So it has formal charge, oxidation state, and partial charge, all different.. Great.. Thanks

Reply 7

OP

Thanks..

So for OH-

Looking at oxidation states. The O has oxidation state of -2, The H has oxidation state of +1 . So overall oxidation state of -1

Looking at formal charges. The O has a formal charge of -1. The H has a formal charge of 0. The overall formal charge is -1

Is the minus after the OH, indicating that the overall formal charge is -1, or is it indicating that the overall oxidation state is -1?

Reply 8

Original post by gazbo1

Thanks..

So for OH-

Looking at oxidation states. The O has oxidation state of -2, The H has oxidation state of +1 . So overall oxidation state of -1

Looking at formal charges. The O has a formal charge of -1. The H has a formal charge of 0. The overall formal charge is -1

Is the minus after the OH, indicating that the overall formal charge is -1, or is it indicating that the overall oxidation state is -1?

The minus after the OH indicates the overall formal charge on the ion is -1.

Reply 9

OP

thanks.. do we ever look at total/overall oxidation state, is there even such a thing?

and do we ever determine the overall formal charge by summing the oxidation states on each atom?

(edited 8 months ago)

Reply 10