[jsmith613]

please could you tell me if this is these are the correct half-equations for the electrolysis of water

Electrolysis of water:
Cathode: 2H+ + 2e- --> H₂ (g)
Anode: 4OH- --? 2H₂O + O₂ + 4e-

[Schrödinger]

No

see http://en.wikipedia.org/wiki/Electrolysis_of_water

[Borek]

Yes

see http://en.wikipedia.org/wiki/Electrolysis_of_water



This is one of alternate ways of describing process of electrolysis.

[cliverlong]

No

see http://en.wikipedia.org/wiki/Electrolysis_of_water

The article contradicts itself

http://en.wikipedia.org/wiki/Electrolysis_of_water#Equations

(well it is wikipedia)

In one place it gives the equation quoted by jsmith (which seems reasonable to me)

Clive

[savy2020]

Electrolysis reactions depend on whether the medium is acidic or basic.
The eqns quoted by jsmith aren't of the same medium.
He mixed the cathodic reaction of acidic medium and anodic reaction of basic medium

[Borek]

Electrolysis reactions depend on whether the medium is acidic or basic.
The eqns quoted by jsmith aren't of the same medium.
He mixed the cathodic reaction of acidic medium and anodic reaction of basic medium

Good point.

What if the water is neutral?

To some extent selection of reaction equations is arbitrary, especially as long as we don't have precise knowledge about mechanism.

[savy2020]

Pure distilled water isn't a good conductor of electricity. So in order to electrolyze water, you need to add an electrolyte to increase conductance.

[Borek]

Pure distilled water isn't a good conductor of electricity. So in order to electrolyze water, you need to add an electrolyte to increase conductance.

I have added sodium sulfate, pH is still neutral.

[savy2020]

I have added sodium sulfate, pH is still neutral.

@Borek  -- Pls see this http://www.infoplease.com/chemistry/simlab/electrolpt1.html

Now,
Probable reactions at cathode and Anode
Cathode(1) : 2 H⁺_(aq) + 2e⁻ → H_2(g)                      -- Reacn C₁
Cathode(2) : 2 H₂O_(l) + 2e⁻ → H_2(g) + 2 OH^-_(aq)       -- Reacn C₂
Anode(1) : 2 H₂O_(l) → O_2(g) + 4 H⁺_(aq) + 4e⁻           -- Reacn A₁
Anode(2) : 4 OH^-_(aq) → O_2(g) + 2 H₂O_(l) + 4 e⁻         --Reacn A₂

Some conclusions with reasons
If the solution is acidic  -  Reacns C₁ and A₁ occur {Since there is significant H⁺ but not OH^- in the solution}
If the solution is neutral- Reacns C₂ and A₁ occur.{Since there is no significant H⁺ or OH^- ions in the neutral solution}
If the solution is basic  - Reacns C₂ and A₂ occur. {Since there is no significant H⁺ but there is significant OH^- in the solution}

[Borek]

I have added sodium sulfate, pH is still neutral.

@Borek  -- Pls see this http://www.infoplease.com/chemistry/simlab/electrolpt1.html

It doesn't add anything new - all it tells is that it is water that reacts on electrodes and that solution becomes LOCALLY acidic or basic, depending on which electrode we are close to. After stirring solution will be neutral again.

As for the reactions - trick is, you can write them, but you will be not able to tell which one occurs. Water autodissociation is a very fast process so

Cathode(2) : 2 H₂O_(l) + 2e⁻ → H_2(g) + 2 OH^-_(aq)       -- Reacn C₂

can be in fact

H₂O -> H⁺ + OH^-

followed by

2 H⁺_(aq) + 2e⁻ → H_2(g)

Same can be said about anode reaction. These cases can be distinguished by some fancy techniques, but as both electrode reaction and water autodissociation are very fast, that sounds like a difficult task. Very likely it has already been done, I have no access to literature so I can't check.

Interestingly, and perhaps importantly, approach that takes into account water autodissociation doesn't require different electrode mechanism for water decomposition in low/high pH.

Note: I am not stating I am right and you are wrong, you have just failed so far to show convincing evidence for your approach

[savy2020]

.As for the reactions - trick is, you can write them, but you will be not able to tell which one occurs Water autodissociation is a very fast process so

Cathode(2) : 2 H₂O_(l) + 2e⁻ → H_2(g) + 2 OH^-_(aq)       -- Reacn C₂

can be in fact

H₂O -> H⁺ + OH^-

followed by

2 H⁺_(aq) + 2e⁻ → H_2(g)

Same can be said about anode reaction. These cases can be distinguished by some fancy techniques, but as both electrode reaction and water autodissociation are very fast, that sounds like a difficult task. Very likely it has already been done, I have no access to literature so I can't check.

Interestingly, and perhaps importantly, approach that takes into account water autodissociation doesn't require different electrode mechanism for water decomposition in low/high pH.

Note: I am not stating I am right and you are wrong, you have just failed so far to show convincing evidence for your approach

That's enlightening
Water auto-dissociation is a fast process? how do you justify that?

I accept that I've failed to provide convincing evidence but I'll try to find some soon.

[cliverlong]

Now,
Probable reactions at cathode and Anode
Cathode(2) : 2 H₂O_(l) + 2e⁻ → H_2(g) + 2 OH^-_(aq)       -- Reacn C₂

Anode(1) : 2 H₂O_(l) → O_2(g) + 4 H⁺_(aq) + 4e⁻           -- Reacn A₁

Hi I'm intrigued by these two reactions.

I always thought that electrolysis in an electrolyte occurs because of

1. the presence of free ions that can move to the electrodes in the presence of a potential difference between the electrodes

2. At the electrodes the ions are discharged due to the gain or loss of electrons

3. if there are competing species at the electrodes such as Cu²⁺ and H⁺ at the cathode then the one with the more positive (?) electrode potential is preferentially discharged as this requires less energy

4. In the examples you have given above you are showing H₂O is discharged at the electrodes. I would have thought that undissociated covalent molecules will have no nett movement to the electrodes and only will be discharged if they undergo dissociation to H⁺ and OH^- (and H₃O⁺ and whatever) first.

What do you think?

Clive

[Borek]

In the examples you have given above you are showing H₂O is discharged at the electrodes. I would have thought that undissociated covalent molecules will have no nett movement to the electrodes and only will be discharged if they undergo dissociation to H⁺ and OH^- (and H₃O⁺ and whatever) first.

Uncharged molecules diffuse through solution just like any other molecules. Charged ones will additionally migrate influenced by the electric field, but in normal circumstances (electrolysis with just a few volts applied) this process is slow and can be neglected (that is, diffusion and mixing are much more important). So, if the current is limited by diffusion, it doesn't matter if molecules are charged or not.

Besides, if you think about it, electrode put in water will be in contact with water always, no matter what the diffusion speed is.

Finally, there is no need for the moleule to be charged before it can react on the electrode. See for example Chronoamperometry and pulse voltammetry of uncharged species at microelectrodes in the presence of a very low amount of supporting electrolyte (I happen to know both Ziggie Stojek and Maggie Ciszkowska in person, hence I know of this paper, but I am sure there are others).

[jsmith613]

Hi I'm intrigued by these two reactions.

I always thought that electrolysis in an electrolyte occurs because of

1. the presence of free ions that can move to the electrodes in the presence of a potential difference between the electrodes

2. At the electrodes the ions are discharged due to the gain or loss of electrons

3. if there are competing species at the electrodes such as Cu²⁺ and H⁺ at the cathode then the one with the more positive (?) electrode potential is preferentially discharged as this requires less energy

4. In the examples you have given above you are showing H₂O is discharged at the electrodes. I would have thought that undissociated covalent molecules will have no nett movement to the electrodes and only will be discharged if they undergo dissociation to H⁺ and OH^- (and H₃O⁺ and whatever)

the definition of an electrolyte is something that can split up into ions by the use of electricity in solution

I personally would have assumed that the electrolysis happens based on the polarity of each end of the molecule.
(e.g: NaCl = Na+ + Cl-) - Molten NaCl
The charged ends of the molecule, +, - are attracted to the respective electrode. BUT as there are loose ions in solution, there is a circuit and the compound can be separated into Na + Cl2
---------------------------------------------------------------------------------
Additionally, maybe the original post may have been slightly flawed - assuming the solution is neutral however, then my half equations seem to be correct. Assuming an acid / base is added to increase the conductivity of the solution, then clearly my half equation is wrong.

The main reason I ask, is that I am revising for my exams, and if I am asked to write half equations for certain compounds / molecules, I will obviously need to have practice and experience.

I understand how most half equations work BUT my only issue was the equation with water.

Cathode(1) : 2 H+(aq) + 2e− → H2(g)                      -- Reacn C1
Cathode(2) : 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)       -- Reacn C2
Anode(1) : 2 H2O(l) → O2(g) + 4 H+(aq) + 4e−           -- Reacn A1
Anode(2) : 4 OH-(aq) → O2(g) + 2 H2O(l) + 4 e−         --Reacn A2

If I showed ANY of the following eq, however then my answer should be correct

C1 & A1
C2 & A2
C2 & A1
C1 & A2

Agreed?