Question about E values in the half equations for electrolysis of water

[gobin]

Hi

I understand that in the electrolysis of water, there are these half equations

1. Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH−(aq) E=-0.83

2. Anode (oxidation) 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− E=1.23

3. Cathode(reduction) 2H+(aq) + 2e− → H2(g) E=0

4. Anode(oxidation) 4 OH- (aq) → O2(g) + 2H2O(l) + 4e- E=0.4

And from what I understand, all four occur. But two will be the main ones that occur.

In Neutral conditions, we'd get (1,2) as the main ones

In Acidic conditions we'd get (3,2) as the main ones

In Basic conditions we'd get (1,4) as the main ones.

And from what I understand, the E values  there are SEP - standard electrode potentials, particualrly, E(red), so, for reduction.  And they need to be converted into  "Actual Potentials" in order to be of any use, is that right?

'cos from what I understand,  in order to predict the products that will be the main products, so the half equations that are the main ones.  One would look at the half equations for the Cathode, and pick the one with the highest reduction potential.  And since  E(ox) = E(red)*-1    The half equation with the lowest E(red) value is the one with the highest E(ox), and thus the main one at the Anode.  So for the Anode I should pick the lowest E value.  But the problem I am having here, is that the E values don't reflect the fact that in neutral conditions I should get (1,2) .  In acidic (3,2).  And in Basic (1,4). 

 e.g. I get 

3. Cathode(reduction) 2H+(aq) + 2e− → H2(g) E=0    <--- Highest E for cathode  as 0> -0.83

4. Anode(oxidation) 4 OH- (aq) → O2(g) + 2H2O(l) + 4e- E=0.4 <---  Lowest E for Anode  as  0.4 < 1.23

But i'm not sure that any conditions are meant to produce main ones of (3,4).

So, i'm thinking the E values would be different for e.g. pH 3 (for acidic),   or  pH(10) for basic.

Let's assume room temperature and atmospheric pressure.

I'm wondering what the actual potentials are, and whether that rule about highest E for cathode, and lowest E for Anode, will then apply?

Thanks

 

 

[studiot]

9 hours ago, gobin said:

Hi

I understand that in the electrolysis of water, there are these half equations

1. Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH−(aq) E=-0.83

2. Anode (oxidation) 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− E=1.23

3. Cathode(reduction) 2H+(aq) + 2e− → H2(g) E=0

4. Anode(oxidation) 4 OH- (aq) → O2(g) + 2H2O(l) + 4e- E=0.4

And from what I understand, all four occur. But two will be the main ones that occur.

In Neutral conditions, we'd get (1,2) as the main ones

In Acidic conditions we'd get (3,2) as the main ones

In Basic conditions we'd get (1,4) as the main ones.

And from what I understand, the E values  there are SEP - standard electrode potentials, particualrly, E(red), so, for reduction.  And they need to be converted into  "Actual Potentials" in order to be of any use, is that right?

'cos from what I understand,  in order to predict the products that will be the main products, so the half equations that are the main ones.  One would look at the half equations for the Cathode, and pick the one with the highest reduction potential.  And since  E(ox) = E(red)*-1    The half equation with the lowest E(red) value is the one with the highest E(ox), and thus the main one at the Anode.  So for the Anode I should pick the lowest E value.  But the problem I am having here, is that the E values don't reflect the fact that in neutral conditions I should get (1,2) .  In acidic (3,2).  And in Basic (1,4). 

 e.g. I get 

3. Cathode(reduction) 2H+(aq) + 2e− → H2(g) E=0    <--- Highest E for cathode  as 0> -0.83

4. Anode(oxidation) 4 OH- (aq) → O2(g) + 2H2O(l) + 4e- E=0.4 <---  Lowest E for Anode  as  0.4 < 1.23

But i'm not sure that any conditions are meant to produce main ones of (3,4).

So, i'm thinking the E values would be different for e.g. pH 3 (for acidic),   or  pH(10) for basic.

Let's assume room temperature and atmospheric pressure.

I'm wondering what the actual potentials are, and whether that rule about highest E for cathode, and lowest E for Anode, will then apply?

Thanks

 

 

 

I see you have studied on this somewhat.

So I will give you a pointer to help you answer your own question, beforelaunching into any detail.

 

Pure water has a very low conductivity on account of the very low concentration of ions, both hydroxyl and hydrogen (or hydroxonium).

Pure water is neutral, and there are no other ions present.

In order to acidify or alkify the water (ie change the pH) you have to add other ions.

These afect the half cell reactions.

[gobin]

3 hours ago, studiot said:

 

I see you have studied on this somewhat.

So I will give you a pointer to help you answer your own question, beforelaunching into any detail.

 

Pure water has a very low conductivity on account of the very low concentration of ions, both hydroxyl and hydrogen (or hydroxonium).

Pure water is neutral, and there are no other ions present.

In order to acidify or alkify the water (ie change the pH) you have to add other ions.

These afect the half cell reactions.

 

I know that when pH is more acidic ,  you have lots of H+ ions relative to HO- ions.  And when water is more basic  you have lots of HO- ions over H+ ions.

Also just because water is a poor conductor of electricity wouldn't mean that nothing happens in the electrolysis of water.. There'd be some reactions taking place.

I think it is valid to talk of the electrolysis of water with pH 7 - neutral water. Even though it's maybe a slow reaction. In theory I suppose the reaction could be sped up even with neutral water, if the H+ ions and HO- ions were increased while maintaining the 50/50 proportion. So maintaining the neutral pH7. But that probably adds another complexity to the question.  Also I understand that in chemistry there's no bottle of H+ ions or bottle of OH- ions.  People add a salt into it to get more conductivity(which isn't what i'm asking about). And from what I understand , it is valid to speak of the electrolysis of pure water.  It's mentioned here https://en.wikipedia.org/wiki/Electrolysis_of_water "Electrolysis of water is using electricity to split water into oxygen (O2) and hydrogen (H2) gas by electrolysis.".  Though it's perhaps not often encountered by people.

 

 

 

[studiot]

1 hour ago, gobin said:

 

I know that when pH is more acidic ,  you have lots of H+ ions relative to HO- ions.  And when water is more basic  you have lots of HO- ions over H+ ions.

Also just because water is a poor conductor of electricity wouldn't mean that nothing happens in the electrolysis of water.. There'd be some reactions taking place.

I think it is valid to talk of the electrolysis of water with pH 7 - neutral water. Even though it's maybe a slow reaction. In theory I suppose the reaction could be sped up even with neutral water, if the H+ ions and HO- ions were increased while maintaining the 50/50 proportion. So maintaining the neutral pH7. But that probably adds another complexity to the question.  Also I understand that in chemistry there's no bottle of H+ ions or bottle of OH- ions.  People add a salt into it to get more conductivity(which isn't what i'm asking about). And from what I understand , it is valid to speak of the electrolysis of pure water.  It's mentioned here https://en.wikipedia.org/wiki/Electrolysis_of_water "Electrolysis of water is using electricity to split water into oxygen (O2) and hydrogen (H2) gas by electrolysis.".  Though it's perhaps not often encountered by people.

 

 

 

Sorry I didn't make myself very clear.

Firstly let me repeat

You cannot get acid ar basic consitions in pure water, you can only get neutral ones.

Quote

In Neutral conditions, we'd get (1,2) as the main ones

In Acidic conditions we'd get (3,2) as the main ones

In Basic conditions we'd get (1,4) as the main ones.

You can get acid or basic conditions only by adding something.

 

If you add something that is ether acid (eg sulphuric or hydrochloric acid) or basic  (potassium or sodium hydroxide) you will get and acidic or basic mixture that now contains other ions than OH- or H+

If, as you suggest, you add a neutral salt then you will also change the pH according, more or less, to the solubility and the Henderson-Hasselbalch equation

Either way you will have additional ions in what is now  a solution.

How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO? | Socratic

To then get to the effect on the cell potentials you then need to apply the Nernst Equation

What is the equation that connects pH and its effect on electric potential of an electrochemical cell? | Socratic

[exchemist]

1 hour ago, gobin said:

 

I know that when pH is more acidic ,  you have lots of H+ ions relative to HO- ions.  And when water is more basic  you have lots of HO- ions over H+ ions.

Also just because water is a poor conductor of electricity wouldn't mean that nothing happens in the electrolysis of water.. There'd be some reactions taking place.

I think it is valid to talk of the electrolysis of water with pH 7 - neutral water. Even though it's maybe a slow reaction. In theory I suppose the reaction could be sped up even with neutral water, if the H+ ions and HO- ions were increased while maintaining the 50/50 proportion. So maintaining the neutral pH7. But that probably adds another complexity to the question.  Also I understand that in chemistry there's no bottle of H+ ions or bottle of OH- ions.  People add a salt into it to get more conductivity(which isn't what i'm asking about). And from what I understand , it is valid to speak of the electrolysis of pure water.  It's mentioned here https://en.wikipedia.org/wiki/Electrolysis_of_water "Electrolysis of water is using electricity to split water into oxygen (O2) and hydrogen (H2) gas by electrolysis.".  Though it's perhaps not often encountered by people.

 

 

 

Yes, that article mentions that an overpotential is needed, which increases the degree of ionisation from its equilibrium value, thus speeding up the electrolysis.