electrolysis of aqeuous solutions

sgstudent

When I use inert electrodes for a solution of Copper (ii) sulfate. So the oxidation at the anode is OH- while reduction at the cathode is the copper ions. However, what will happen if I continue the experiment? There is a definite amount of sulfate so does the hydroxide oxidise until the amount of H+ is two lines of the sulfate cos the formula is H2SO4. Then once this criteria is met the creation ends.

This seems very complicate to determine what will happen if I leave it on. So in exams or tests will they ask you such questions? Thanks for all the help!

Borek

What will be present in the solution once you got all copper electrodeposited on the electrode?

sgstudent

H+, OH- and SO4 2- ions. So in order to maintain the balanced H2SO4 as OH- ions will be discharged. So all the H+ and OH- will discharge until there is so little water that only H2SO4 gas remains? Meaning the other OH- gets discharged leaving behind Cu O2 and H2SO4 gas?

Borek

Your explanation is a little bit convoluted, you should start with the overall equation of the initial process (in the presence of copper).

But you are right about water being electrolyzed - yes, after all copper was used, and there is nothing but sulfuric acid present, it is water that will start to decompose. That is, assuming potential is high enough.

sgstudent

Oh sorry about that
Oh ok. Here it goes: when all the copper ions get reduced at the cathode, the remaining ions present in the solution are H+, SO4 2- and OH-. As sulfate ions do not get discharged, hence only the OH- are discharged along with the H+. As they get discharged, there must always be 2 units of the H+ ions remaining with the SO4 2- ion as SO4 2- cannot exist alone as an ion. Once all the water is being dissociated, only gaseous H2SO4 exists as there is no more water present. So in the entire electrolysis process the produced substances are: O2, Cu, H2, and the gaseous H2SO4.

Equation: 2H+ +2e-->H2
This is so with 2 units of H+ as determined by the SO4 2- not being discharged.

Is this correct? Thanks for the below Borek. Really appreciate the help! :)

Borek

Sulfuric acid is not gaseous, it is a liquid at room temperature. Also your use of terms "discharge" (when you mean either reduction or oxidation - electrode reaction with a charge transfer) and "dissociation" (when you mean electrolytic decomposition) is incorrect. But in general seems like you understand what is going on.

sgstudent

But there must always be 2:1 ratio of H+ and SO4 2- ions right? So if I continue the electrolysis experiment then eventually I'll get a pure H2SO4 in gaseous state? Is this concept right?

Thanks for the help!

Borek

What makes you think about gaseous state?

sgstudent

I thought its a covelant substance so it has a low boiling point? And there's no more water so it should exist as a gas? Or is pure H2SO4 a liquid in rtp?

Borek

I already told you it is a liquid at STP. Pure sulfuric acid melts at 10°C and boils at 337°C (give or take a deg or two, different sources give slightly different values).

sgstudent

Oh so the hydroxides and hydrogen ions will really get oxidised and reduced until the point that only pure sulfuric acid is left? Also it does not have any H+ ions in it anymore as there's no more water?

Thanks for clarifying this with me.:)

sgstudent

Hi Borek, I was thinking, can water ever be the limiting reagents in this case? Cos, if it is possible, then eventually we could also get a mixture of solid copper (ii) sulfate and liquid H2SO4?

Borek

Why solid copper sulfate, if we already reduced copper(II) to copper(0)?

And water is not a limiting reagent, or at least it is not a correct use of the term "limiting reagent". Water doesn't react with something else, it is electrolysed till it is all used up.

sgstudent

I mean if amount of water is less than the amount of copper (ii) sulfate. Eg 1 mole if Cu2+, 1 mole of SO4 2-, 1 mole of H+ and 1 mole of OH-. In this case, 1/2 of Cu2+,0 OH-, 1 H+ and 1SO4 2-?

So eventually we are left with some CuSO4 and H2SO4? Thanks!

Borek

No, copper will be reduced BEFORE water electrolysis starts. These are two separate processes.

sgstudent

I'm sorry, I meant that could it be possible that copper (ii) ions do not run out first? Could the OH- run out before through Cu2+ completely runs out? For example I have 2 moles of Cu2+, 2 moles of SO4 2- 1 moles of H+ and 1 mole of OH-. In this case, only half a mole of Cu2+ will get reduced while 1 mole of OH- gets oxidised. So in the end you are left with some unreduced Cu2+ and since there is no more water, so the copper (ii) sulfate will be formed at a solid with liquid hydrogen sulfate?

So in essence I'm asking whether the water can run out before the added substance or will there always be enough of it? Thanks for the help Borek!

Borek

This is not trivial.

First of all, copper sulfate is usually a pentahydrate, so it contains enough water and then some:

2CuSO₄^.5H₂O -> 2Cu + 2H₂SO₄ + O₂ + 8H₂O

The only problematic situation is when you add anhydrous copper sulfate to some solvent other than water - then the problem is not whether there is enough water, but whether the sulfate is soluble at all (and it won't electrolyze as long as it is not dissolved).