Electrolysis Question for Chemistry

anpanman SG

I have some queries regarding a question here.

What happens when aqueous copper(II)sulphate is electrolysed using reactive electrodes?

If you are talking about the copper anode, the SO4 2- and OH- migrate to anode. Since copper is not an inert electrode, the following reaction occers at node

Cu(s) ->Cu2+(aq) + 2e-

The anode dissolves to form Cu2+ in aqueous solution

1 May I know what happens to the sulphate and hydroxide ions that have migrated? Why aren't they involved in any reaction?

Aqueous copper(II)sulphate is electrolysed using carbon electrodes. What would you expect to see duing electrolysis

At the cathode (-) At the anode(+)

a. pink solde forms anode dissolves

b.pink solid forms colourless gas forms

The answer is b. But I do not know why it can't be A given that the explanation for the earlier question seems to give the idea that (a) is the correct answer. I remember my teacher said something about temperature being high... and won't burn kind of stuff. Would someone kindly explain? Thanks.

UltimaOnline SG

Several points.

Carbon (graphite) electrodes will burn (ie. combustion / oxidation) when the temperature is very high (ie. when molten compounds electrolyzed, not aqueous compounds), and when oxygen gas is produced at the anode (eg. electrolyzing metal oxides).

Since carbon cathode and carbon anode are specified, hence carbon anode (being inert) won't be oxidized, therefore it won't dissolve. Instead, hydroxide ions (and/or water) are being oxidized at the anode to produce oxygen gas.

>>> May I know what happens to the sulphate and hydroxide ions that have migrated? Why aren't they involved in any reaction? <<<

Why should they be? Since the anode is made of copper, so copper is oxidized (into Cu2+(aq) which dissolves into solution) INSTEAD of hydroxide ions (note that sulfate ions are never oxidized under aqueous conditions in the 'O' level syllabus on electrolysis). So nothing happens to the sulfate and hydroxide ions - they remain in solution.

dkcx SG

Since there are H+ (left after evolving of O2) and SO42- ions in the solution, would the solution turn acidic since the remaining ions can form sulphuric acid?

I know the solution will loss its copper and the blue would get lighter and lighter but i never got a chance to test whether the acidity of the solution would be increase which would account for what happens to the SO42- ions.

UltimaOnline SG

Since there are H+ (left after evolving of O2) and SO42- ions in the solution, would the solution turn acidic since the remaining ions can form sulphuric acid?

Whenever protons (H+ cations) are reduced at the cathode, the ratio of hydroxide ions to protons is increased, and the electrolyte becomes more alkaline (ie. increase in pH).

Whenever hydroxide ions (OH- ions) are oxidized at the anode, the ratio of protons to hydroxide ions is increased, and the electrolyte becomes more acidic (ie. decrease in pH).

If protons are reduced at the cathode and hydroxide ions are oxidized at the anode, which occurs in solutions such as dilute sulfuric(VI) acid, then of course the ratio of protons to hydroxide ions remain constant, and the pH remains unchanged.

Also note that pH = 7 is neutral only at 25 deg C. (More about this at 'A' levels).

UltimaOnline SG

Since there are H+ (left after evolving of O2) and SO42- ions in the solution, would the solution turn acidic since the remaining ions can form sulphuric acid?

One more point.

It's actually technically inaccurate to say that this "would account for what happens to the SO42- ions". But it's good that you can think this far (which is further than what the average 'O' level student would.)

Yes, you can say sulfuric acid is formed, because hydroxide ions are removed (from water, leaving protons; autoionization of water into protons H+ and hydroxide OH- ions).

And yes, the -ve charges of the sulfuric(VI) ions are counterbalanced by the +ve protons that are left in solution.

But technically speaking, the sulfate(VI) ions are spectator ions, so NOTHING HAPPENS TO THEM.

You see, (and this is the part that the majority of 'O' level students do not understand), when sulfuric(VI) acid is formed, the sulfate(VI) ions do NOT "combine" or "react" with the protons (that are left in solution from the oxidation of hydroxide ions from water molecules) to form sulfuric(VI) acid.

The sulfate(VI) ions are SIMPLY LEFT IN AQUEOUS STATE THROUGHOUT THE PROCESS, untouched, unreacting, spectators.

To be even more technically accurate, the sulfate(VI) ions are slowly protonated as pH decreases, so what you get is a gradual stepwise process, as molarity of protons increase in the electrolyte :

SO4 2- + H+ --> HSO4 -

HSO4 - + H+ --> H2SO4

The second equation above will not occur until molarity of protons is very high, when pH is very low, and when you use state symbol (l) for liquid concentrated sulfuric acid, and not aqueous.

The point that most 'O' level students don't properly understand, is the concept of spectator ions in solution, and that it is incorrect to say "the sulfate ions REACTED with the H+ ions to form sulfuric acid". The sulfate(VI) ions were always there (originally from the copper(II) sulfate solution), and simply remained in solution throughout.

Since protons are now present in greater amounts, we call the resulting solution "sulfuric(VI) acid". But remember the sulfate(VI) ions are spectator ions throughout.

An example to help 'O' level students understand this point :

If you have an electric cell (not electrolytic cell) setup that consists of a zinc electrode, a copper electrode, sodium chloride solution; describe what happens, and name the compounds (both solid and aqueous) present in the electrolyte as the reaction proceeds.

Solution :

Zn(s) is oxidized to Zn2+(aq); oxidation at zinc anode.

H+(aq) (from water) is reduced to H2(g); reduction at copper cathode.

Ions present as the reaction proceeds :

Zn2+, Na+, OH-, Cl-.

Hence, compounds present in electrolyte as the reaction proceeds are :

Sodium chloride solution, sodium hydroxide solution, zinc chloride solution, zinc hydroxide precipitate.

dkcx SG

I do have to think further than an O's student considering i'm a chemistry undergrad but somehow i'm asking myself why i'm here now since i lost most interest in chemistry the more i study...

I understand the spectator part since all ionic compounds dissociate in water and they behave as relatively individual ions then rather than as the original compound but for them to see it base on equation, it might be easier to just have it form sulphuric acid as that in water is still H+ and SO42- ions.

Can sulphuric acid ever exist in (l) even at very high concentrations? CuSO4 by itself contains water and its likely to normally be in excess to the extend that water should still be present in large amounts after all the Cu is gone and the resultant solution left should be sulphuric acid (aq).

Sulphuric acid from what i know doesn't have a solid or liquid form. It would loss it's acidic characteristic if there is no water since acids only become acidic when it dissociates to give H+. Never ever try to evaporate a sample of sulphuric acid to dryness to see if it can occur as a solid but i believe it would turn to gas in the end and unless very high pressure is used, i don't really think it will ever exist as a liquid without water.

UltimaOnline SG

I do have to think further than an O's student considering i'm a chemistry undergrad but somehow i'm asking myself why i'm here now since i lost most interest in chemistry the more i study...

My sympathies that you've "lost interest in chemistry the more you study". It's always tragic when one is forced to study a discipline that one does not enjoy, or perhaps in a way (eg. your particular undergraduate course in your particular university) that one does not enjoy. May this change for you soon, in one way or the other (eg. either you somehow begin to enjoy Chemistry more, or you switch course, or you take Chemistry in another University, etc). Studying at undergraduate and graduate levels must be enjoyable if one is to excel in it.

Yes, you're right that liquid H2SO4 (dihydrogen sulfate(VI), not sulfuric(VI) acid) would not have dissociated aqueous protons, and hence does not have acidic properties. But it does indeed exist in liquid state (if there are no water molecules present) at r.t.p due to intermolecular hydrogen bonding, and gaseous state when heated.

In aqueous state then, the 1st acidic proton is strongly acidic, while the 2nd acidic proton is weakly acidic.

It's fine to write H2SO4(aq) in a chemical reaction, because what it is usually taken to mean is 2H+(aq) + SO4 2-(aq), but since only the 1st acidic proton is strongly acidic, so the species that exist in aqueous state is really usually H+(aq) + HSO4 -(aq).

And as you say, carrying out electrolysis on aqueous copper(II) sulfate implies a large excess of water. Which is why you would not under experimental conditions get liquid H2SO4.

My point in my previous post wasn't that one should write H2SO4(l) for the electrolysis experiment. If you thought that was what I meant, you misunderstood me.

My point was that the only scenario in which the SO4 2- sulfate(VI) ions actually "react" or "combine" with the protons (as is usually erroneously conceptualized and written by 'O' level students), is when there is no water solvent molecules present, and all protons present are taken up by the SO4 2- and HSO4 - basic ions; in other words, complete protonation of all SO4 2- and HSO4 - ions present, to give H2SO4(l), which is more properly called dihydrogen sulfate(VI) rather than sulfuric(VI) acid.

In other words, I was pointing out a common conceptual error in 'O' level students, which of course, isn't an issue for you. Same goes for the "spectator ions" bit, which you of course understand. That part was also meant for 'O' level students.

May you find the joy in whatever you choose to study or do.

(edited typo error : "lost interest" not "most interest")

anpanman SG

Several points.

>>> May I know what happens to the sulphate and hydroxide ions that have migrated? Why aren't they involved in any reaction? <<<

That's a lot of help. Thank you!

dkcx SG

When u go further up in chemistry and u get to study alot more theory behind everything it gets boring since all that is learn gets more research usage than application usage. Besides the lab modules, most of the theoretical content would be rather useless in terms of actual usage in work since an uni grad works mostly as a chemist and not a researcher. Lets just say that physical chemistry especially is causing alot of people in my cohort to really wonder why we chose chemistry since its killing most of us and i wonder whether i can clear it this year...

Liquid form of H2SO4 is something new to me and yeah it won't be called sulphuric acid anymore just like HCl is called hydrogen chloride in liquid, gaseous form and not hydrochloric acid.

Yeah i misunderstood that part since i don't think we'll ever get sulphuric acid in liquid form through electrolysis.

anpanman SG

Several points.

>>> May I know what happens to the sulphate and hydroxide ions that have migrated? Why aren't they involved in any reaction? <<<

Hi. I want to ask what happens over at the Copper Cathode.

Cu2+ + 2e- -> Cu

May I know whether the Cu2+ comes from that in the solution and also from those disharged by anode into the solution? After all, the cathode is copper and not Cu2+.

thanks.

UltimaOnline SG

Hi. I want to ask what happens over at the Copper Cathode.

Cu2+ + 2e- -> Cu

May I know whether the Cu2+ comes from that in the solution and also from those disharged by anode into the solution? After all, the cathode is copper and not Cu2+.

thanks.

At the cathode, reduction occurs.

If there are Cu2+(aq) in solution, they will be reduced at the cathode.

The solid copper cathode itself is already in the reduced state, and cannot be reduced further.

If the anode is made of copper, then it will be oxidized and dissolved from Cu(s) to Cu2+(aq).

In such a setup, the Cu2+(aq) that is reduced at the cathode, will come from both the original copper(II) sulfate solution, as well as the copper anode.

But note that such a setup (ie. both electrodes made of copper) will not see any change in colour or molarity of the copper(II) sulfate electrolyte solution, because for every 1 Cu2+ that is removed at the cathode, 1 new Cu2+(aq) is formed at the anode.