Electrolysis of Water: How Does the Electric Field Affect Covalent Bonds?

[erocored]

What does break bounds in CuSO4? Is it just charge of the electrons in the left electrode? And why does the right electrode getting a positive charge?

 

[Frigus]

What does break bounds in CuSO4?

I am not able to understand this statement.
I am interpreting this as "why does CuSO₄ hydrolyse?"
It is a salt and they dissociate in H₂O as they get more stable after dissociating.

And why does the right electrode getting a positive charge?

Because initially atoms of right electrode have equal protons and electrons and when electrons leave the atom it became positively charged(atom).

 

[erocored]

I am not able to understand this statement.

I meant why CuSO4 does split into Cu and SO4?

 

[Frigus]

I meant why CuSO4 does split into Cu and SO4?

I have tried to answer this question in post #2.

 

[Merlin3189]

What does break bounds in CuSO4?

It is a salt and they dissociate in H₂O as they get more stable after dissociating.

Yes. I was going just to say, "water"!
$Cu^{++} and \ SO_4^{--}$ obviously are attracted and need some energy to break them apart. But once they are surrounded by water they return more than that energy by binding to the water molecules. $Cu^{++}$ attracts 6 water molecules. $SO_4^{--}$ attacts up to 12 water molecules.

So your solution already contains the (hydrated) $Cu^{++} and \ SO_4^{--}$ ions, free to move in solution.

The $Cu^{++}$ ions are attracted to the -ve electrode, increasing their concentration there, and they are repelled from the +ve electrode, reducing the concentration there.

And why does the right electrode getting a positive charge?

Simply because the battery is there.
When $Cu$ atoms dissolve to become $Cu^{++}$ ions, it leaves behind electrons, which would actually cause the electrode to become negatively charged. But the battery removes them - or makes them move through the wires and go to the left electrode.

The negative charge on the left elctrode (caused by the battery) attracts the $Cu^{++}$ ions and supplies the electrons needed to convert them to $Cu$ atoms on its surface.

I meant why CuSO4 does split into Cu and SO4?

It doesn't ! CuSO4 is actually crystals of $Cu^{++} \ ions, \ and \ SO_4^{--} \ ions \$ which are electrostatically attracted together. The water enables the ions to separate, as explained at the start.

 

[etotheipi]

Because initially atoms of right electrode have equal protons and electrons and when electrons leave the atom it became positively charged(atom).

I don't think this is quite correct; the electrodes are metals whose electronic structure is like an 'electron sea', i.e. there is an overlap of the valence and conduction bands, and these bands are filled up to the Fermi level.

In vague terms, when you apply a voltage across an electrolytic cell, on one side of the cell the Fermi level of the electrode exceeds the energy of the lowest unoccupied molecular orbital (LUMO) of the species in the solution which means that the free energy can be reduced by electron transfer into the solution [and vice versa for the other electrode and the highest occupied molecular orbital (HOMO) of the species in the solution].

In reality, electrolysis is quite a complicated kinetic process. Maybe it would be better to first study and understand the simpler case with ideal polarised electrodes, where you have no charge transfer across the metal-solution interface and as such this interface can be treated as a capacitor.

 

[DaveE]

As others have said, you don't have to break the bonds of this salt, the electric field is just moving the ions apart.

However, for other molecules, you can break bonds with a strong enough electric field (i.e. the voltage between the electrodes). For example, if you only put water in the beaker you can break the bond between the H₂ and O. H₂O is a polar molecule (meaning one side has more electric charge than the other). So, if the field is strong enough, it will try to pull the side with more positive charge towards the cathode and also pull the more negative side towards the anode. If the e-field is strong enough this force can overcome the force of the covalent bond and separate it: 2H₂O → 2H₂ + O₂.

edit: This doesn't only apply to polarized molecules. With a strong enough field you can rip apart anything that isn't an elementary particle.

 

[Frigus]

I don't think this is quite correct; the electrodes are metals whose electronic structure is like an 'electron sea', i.e. there is an overlap of the valence and conduction bands, and these bands are filled up to the Fermi level.

In vague terms, when you apply a voltage across an electrolytic cell, on one side of the cell the Fermi level of the electrode exceeds the energy of the lowest unoccupied molecular orbital (LUMO) of the species in the solution which means that the free energy can be reduced by electron transfer into the solution [and vice versa for the other electrode and the highest occupied molecular orbital (HOMO) of the species in the solution].

In reality, electrolysis is quite a complicated kinetic process. Maybe it would be better to first study and understand the simpler case with ideal polarised electrodes, where you have no charge transfer across the metal-solution interface and as such this interface can be treated as a capacitor.

Thanks for pointing this out.
My teacher always says me to stop making my own theories but I always do it unknowingly.

 

[etotheipi]

However, for other molecules, you can break bonds with a strong enough electric field (i.e. the voltage between the electrodes). For example, if you only put water in the beaker you can break the bond between the H2 and O. H2O is a polar molecule (meaning one side has more electric charge than the other). So, if the field is strong enough, it will try to pull the side with more positive charge towards the cathode and also pull the more negative side towards the anode. If the e-field is strong enough this force can overcome the force of the covalent bond and separate it: 2H2O → 2H2 + O2.

I don't believe this is really what happens, I think a few things are being mixed up here.

On the one hand you have self-ionisation of water, which is an equilibrium process $\mathrm{{2H_2O}_{(l)} \rightleftharpoons {H_3O^+}_{(aq)} + {OH^-}_{(aq)}}$ with $\mathrm{K_w = [{H_3O^+}_{(aq)}][{OH^-}_{(aq)}] / (c^o)^2} \approx 10^{-14}$ at R.T.P. But this is a very small effect, and these low concentrations are basically why you cannot electrolyse pure water easily [people have achieved electrolyte-free, pure water electrolysis using nanogap electrochemical cells - although that's quite different to what we're talking about here!]

You can use water containing an electrolyte to increase the efficiency of water-electrolysis. The half equations are sometimes quoted with $\mathrm{H^+}$, but these are simplifications of the mechanism since isolated protons do not actually exist in solution - they are always hydronium $\mathrm{H_3O^+}$ ions. The proper half equations are,$$\mathrm{6H_2O_{(l)} \rightarrow {O_2}_{(g)} + {4H_3O^+}_{(aq)} + 4e^-}$$at the anode and$$\mathrm{{2H_2O}_{(l)} + 2e^- \rightarrow {H_2}_{(g)} + {2OH^-}_{(aq)}}$$at the cathode. At no point during this process does the electric field between the electrodes rip apart any covalent bonds. What happens really is that the hydronium ions migrate via the Grotthuss mechanism and one of the protons adsorbs to the cathode, is reduced to a $\mathrm{H^{\bullet}}$ radical and reacts with another $\mathrm{H^{\bullet}}$ radical to form $\mathrm{H_2}$. The anode half equation is a bit more complicated, but that essentially involves another form of migration followed by the oxidation of $\mathrm{OH^-}$ to molecular oxygen.

Overall you do have a cell reaction $\mathrm{{2H_2O}_{(l)} \rightarrow {O_2}_{(g)} + {2H_2}_{(g)}}$, but I don't think it's really got anything to do with the electric field splitting covalent bonds.