Hi all, I'm afraid I need help again. D:
Given, a simple cell for the production of electrical energy. One electrode is copper and the other is one of aluminium, copper, iron, magnesium or zinc. Suggest a suitable solution to use as electrolyte in this cell.
Answer: Copper (II) sulphate.
My concern here is, why can't I use an electrolyte like aluminium sulphate, iron (II) sulphate, magenisium sulphate or zinc sulphate? Isn't it that for voltage to be produced, as long as one metal is more reactive than the other and electrons are allowed to flow then there'll be a electrical energy, no? I know it sounds like a basic concept question but apparantly my textbook only has a small section on 'simple cells' and their concepts are completely wrong (as verified by our teachers as well).
Anyone care to enlighten? (: Thank you!
Hi again :D
Firstly, this isn't electrolysis. This is a simple cell question, although it is under the topic: Electrolysis.
Hey, you're right. It CAN be any salt solution. Hell, it can even be sodium chloride solution.
Or am I missing something?
EDIT: Ok wait, it depends on what products you want. I believe the product is copper metal.
And are you sure that you are using TWO copper electrodes in one simple cell experiment?
Hello :D Haha.
It's actually the first part of a full question and this is the only part I'm unsure of, that's why there's the copper there too, ignore that actually.
But then, it is a simple cell right, what does it has to do with the products? My concept on the electrical energy produced is right, right?
For the products, it is only a matter of concern if I'm purifying copper or electroplating something with copper, so I have to use copper (II) sulphate so that the Cu2+ can be discharged right? So in the case of a simple cell, it shouldn't matter, will it? Can anyone confirm? ^^ Thanks!
Your concept that a current flows between two metals with different positions on the reactivity series is correct. Electrons flow from the more reactive metal to the less reactive metal.
For example, the more reactive metal is zinc. Thus zinc ions are discharged into the electrolyte. Since zinc has been oxidised, there are electrons, right? Electrons go onto the zinc rod to make it negative. The electrons then go along the connecting wire to the copper electrode. Here, copper and hydrogen cations are attracted to the copper electrode. By preferential discharge, therefore, copper cations are reduced to copper metal, hence explaining why the need for a copper salt electrolyte in the first place.
In this case, I would assume that this is a ELECTROPLATING reaction using a simple cell (weird, never heard/seen such question before)
Yes you can use any electrolyte.
But you normally use an electrolyte of the same metal ion so that there wouldnt be displacement i guess. It may happen when your cathode is zinc and the electrolyte is any thing less than zinc.
Even for electroplating of copper, you can still use sodium sulphate electrolyte and nothing wrong will happen except that you will have very little copper ions to be discharged and thus very little copper to be electroplated on the object.
Please correct me if i am wrong XD
First, do realise that this is a simple cell question. It is not electrolysis.
Even for electroplating of copper, you can still use sodium sulphate electrolyte and nothing wrong will happen except that you will have very little copper ions to be discharged and thus very little copper to be electroplated on the object.
Correct. I'm not sure whether sodium will be discharged at the initial start of reaction (since it is the only cation, unless you mean solution, then it's hydrogen), but it most likely will. Once copper cations are enter the solution from the ionized anode, then copper will be plated.
In short, using an electrolyte different to the metal plated will screw up the process.
hmm i realised, which leads me to ask, does it matter what electrolyte you use in a simple cell? If the metal at the cathode is zinc and the electrolyte is copper sulphate.
Originally posted by davidche:hmm i realised, which leads me to ask, does it matter what electrolyte you use in a simple cell? If the metal at the cathode is zinc and the electrolyte is copper sulphate.
If the cathode is zinc, wouldn't the metal displace the copper from the salt solution, so instead of the electrons flowing to the anode, the metals would be taken in by the copper ions directly at the cathode?
True~
This is why I emphasised on it being a SIMPLE CELL question.
If the cathode is zinc, then the anode cannot be copper. I believe (based on what I have learnt) that in a SIMPLE CELL, cathode = +ve while anode = -ve
So electrons flow from anode to cathode.
Please, don't confuse this with electrolysis.
A good analysis of a displacement reaction taking place (I didn't realise), now this makes the question debatable. Using copper (II) sulphate causes a displacement reaction taking place, hence copper metal forms on the zinc electrode. Which really spoils the whole idea of using a simple cell.
Or maybe there aint a ''cathode'' in a simple cell?
it depends actually..
Alright, let's not use cathode and anode.
More reactive metal (MR) = -ve
Less reactive metal (LR) = +ve
Electrons flow from MR to LR, hence cations are discharged there to form elements.
But since this is a copper (II) sulphate solution electrolyte, by right copper metal should be formed, but then, displacement takes place between zinc and copper, hence after a while, reaction stops (you don't throw the whole zinc metal into the electrolyte). Or rather, copper doesn't form anyway. Hydrogen gas instead, is evolved.
Productive discussion going on here. Not much to add, other than :
1) Always visualize a "Red Cat riding An Ox". Reduction occurs at the Cathode; at the Anode, Oxidation occurs.
This will always hold true regardless of whether you're talking about an electrolytic cell (ie. in which the setup includes a battery or electrical source, and the direction you choose to place the battery, which side of the battery negative and which side positive, will determine which electrode is the cathode and which electrode is the anode), or a galvanic/voltaic/simple cell which generates electricity.
2) As to the electrolyte, for the purpose of a simple cell, as you guys have already pointed out correctly, you can use any solution that conducts electricity, as long as the ions in the electrolyte solution does not react (IN A REDOX REACTION) with, or precipitate out with, the new ions generated by at the anode (if this happens, it would RESULT IN NO CURRENT FLOWING IN THE EXTERNAL WIRE, OR reduce the availability of mobile ions, and the current would decrease to zero in a short amount of time, defeating the purpose of the electric cell setup).
(EDITED : FOR THE ELECTRIC CELL TO WORK, COMPARMENTALIZE ELECTRODES & USE 2 ELECTROLYTES & USE SALT BRIDGE TO AVOID ANY DIRECT REDOX BETWEEN ELECTROLYTE AND ELECTRODE)
If you're not sure whether the electrolyte you think of can be used, simply write the expected reduction half equation (at the cathode) and the expected oxidation half equation (at the anode), and check that the redox reaction of the cell is feasible. At 'O' levels, you need not calculate using reduction and oxidation potentials (ie. numerical values), you only need to apply your recall of the reactivity series to conclude if the setup is feasible or not.
Back to the original post's question then. How feasible is that?
Originally posted by Garrick_3658:Back to the original post's question then. How feasible is that?
The original post gave too many possible setups (permutations and combinations). Link a specific suggested electrolyte to a specific setup (ie. specify both electrodes), and write out the reduction and oxidation half equations, and then I'll tell you if your equations are correct and feasible or not.
Hey ultimaonline, that was a very clear explanation, thank you! =) Let's say I let one of the electrode to be copper and the other one to be aluminium, and I choose aluminium sulphate as the electrolyte.
Aluminium being more reactive will be the anode and copper will be the cathode.
At the anode(+):
Al(s) -> Al3+(aq) + 3e
At the cathode(-):
Cu(s) + 2e -> Cu2+(aq)
Overall reaction:
2Al(s) + 3Cu(s) -> 2Al3+(aq) + 3Cu2+(aq)
Correct? >< Then how do I know if it's feasible or not.
(Hmmm... my post kena truncated and data lost... wait... )
>>> Let's say I let one of the electrode to be copper and the other one to be aluminium, and I choose aluminium sulphate as the electrolyte.
Aluminium being more reactive will be the anode and copper will be the cathode.
At the anode(+):
Al(s) -> Al3+(aq) + 3e
At the cathode(-):
Cu(s) + 2e -> Cu2+(aq)
Overall reaction:
2Al(s) + 3Cu(s) -> 2Al3+(aq) + 3Cu2+(aq)
Correct? >< Then how do I know if it's feasible or not. <<<
Your anode equation is correct, but your cathode equation is wrong. You can only reduce metal cations to metal atoms, you cannot reduce metal atoms any further.
The problem with using aluminium sulfate electrolyte, is that you'e not providing a feasible reduction process at the cathode. Why would the universe bother to painstakingly extract electrons from aluminium metal at the anode, simply so that these electrons can be used to reduce aluminium cations at the cathode back to aluminium metal? Where's the gain (in stability) in such a silly act?
If human beings behaved senselessly like this, then the world financial markets would collapse... oh... ummm... ok.
So basically, iamapebble, you'd want to use copper(II) sulfate as the electrolyte because being an unreactive metal, Cu2+ ions would be eager to be reduced to Cu metal at the cathode, meaning they happily accept the electrons that came from the oxidation process (Al ---> Al3+ + 3e-) at the anode, and so if you've a happy (electron) donor and a happy (electron) recipient, the current can flow and we're all good and happy.
Note : if you had used aqueous aluminium sulfate electrolyte, protons from water, and/or water molecules themselves, would be reduced at the cathode and a current would still flow, but the current would be weaker. Besides, at 'O' levels, your teachers/examiners want you to be able to think sharply to produce an optimal current and therefore to give the most appropriate answer of "copper(II) sulfate electrolyte" in which Cu2+ ions are reduced to copper metal and joins/coats the copper cathode; rather than rely on brute force (just because your aluminium anode very reactive, so you think you can bully unhappy-to-be-reduced water into being reduced at the cathode! Buay sui lah!).
(EDITED : FOR THE ELECTRIC CELL TO WORK, COMPARMENTALIZE ELECTRODES & USE 2 ELECTROLYTES & USE SALT BRIDGE TO AVOID ANY DIRECT REDOX BETWEEN ELECTROLYTE AND ELECTRODE)
>>> Hey ultimaonline, that was a very clear explanation, thank you! =) <<<
You're welcome, iamapebbel! If you've any further questions on this, feel free to write out more scenarios (ie. specify the electrolyte, the cathode, the anode, and the reduction half equation, the oxidation half equation, the overall equation), and I will comment and advise you on them.
Hey thanks :D Your explanation made me laugh, which helped me absorb it better, which is a good thing! Chemistry paper 2 in exactly a week's time >< I'll practice more electrolysis questions and clear my doubts here :D
Isn't anode supposed to be negative while the cathode is positive in an ELECTRIC CELL (or simple cell if you like it to be simple). It is opposite to an electrolysis process, although the Red Cat and Angry Ox theory still applies.
Another thing is, an electric cell experiment is meant to carry out what it is supposed to do in the first place: get current to be used on other appliances. It does not REALLY have anything to do with electroplating, and that is another matter altogether... Or is it?
Originally posted by Garrick_3658:Isn't anode supposed to be negative while the cathode is positive in an ELECTRIC CELL (or simple cell if you like it to be simple). It is opposite to an electrolysis process, although the Red Cat and Angry Ox theory still applies.
Another thing is, an electric cell experiment is meant to carry out what it is supposed to do in the first place: get current to be used on other appliances. It does not REALLY have anything to do with electroplating, and that is another matter altogether... Or is it?
Yes, in a Galvanic/Voltaic/electric cell, because oxidation at the anode generates electrons, so the anode is negative.
While you can attempt to carry out electroplating using a Galvanic/Voltaic/electric cell, it's not as efficient as using an electrolytic cell (because in the latter you can control the current and hence better control the electroplating process).
yee.
so in a simple cell, if the negative electrode is aluminium while the positive one is copper, and the electrolyte is copper sulphate, would copper metal be formed at the aluminium electrode?
Because in the chem textbook i think they used this set up to explain how a simple cell works.
Originally posted by davidche:yee.
so in a simple cell, if the negative electrode is aluminium while the positive one is copper, and the electrolyte is copper sulphate, would copper metal be formed at the aluminium electrode?
Because in the chem textbook i think they used this set up to explain how a simple cell works.
(EDITED : FOR THE ELECTRIC CELL TO WORK, COMPARMENTALIZE ELECTRODES & USE 2 ELECTROLYTES & USE SALT BRIDGE TO AVOID ANY DIRECT REDOX BETWEEN ELECTROLYTE AND ELECTRODE)
In a Galvanic/Voltaic/simple cell, "negative electrode" means "electrons are generated there" means "oxidation occurs there" means "this is the anode".
(it is important you are able to understand the above, and mentally work out the above process yourself, so as to correctly deduce where oxidation occurs and where reduction occurs. Try it with an electrolytic cell, using the same logic, you'll notice the anode for an electrolytic cell is called the "positive electrode", because electrons from the anode are 'sucked' into the positive pole of the battery; and the cathode for an electrolytic cell is called the "negative electrode", because electrons 'rush out' from the negative pole of the battery towards the cathode).
Al is oxidized to Al3+, releasing 3 electrons, at the aluminium anode.
Al ---> Al3+ + 3e-
2Al ---> 2Al3+ + 6e-
Cu2+ ions from the electrolyte are accept these 6 electrons, and are hence reduced at the cathode (the "positive electrode).
3Cu2+ + 6e- --> 3Cu
Hence, the copper cathode increases in mass over time, and the blue colour of the copper(II) sulfate electrolyte solution becomes pale over time, as Cu2+ ions (responsible for the blue colour) are removed from solution.
Originally posted by UltimaOnline:
In a Galvanic/Voltaic/simple cell, "negative electrode" means "electrons are generated there" means "oxidation occurs there" means "this is the anode".
Al is oxidized to Al3+, releasing 3 electrons, at the aluminium anode.
Al ---> Al3+ + 3e-
2Al ---> 2Al3+ + 6e-
Cu2+ ions from the electrolyte are accept these 6 electrons, and are hence reduced at the cathode (the "positive electrode).
3Cu2+ + 6e- --> 3Cu
Hence, the copper cathode increases in mass over time, and the blue colour of the copper(II) sulfate electrolyte solution becomes pale over time, as Cu2+ ions (responsible for the blue colour) are removed from solution.
So at the aluminium anode, both reduction and oxidation is taking place. We still call it the anode because the electrode itself undergoes oxidation thru Al -> Al3+ + 3e- (angry Ox)
In view of this, how do you justify an inert electrode in a electrolysis set-up to be a anode/cathode? Since the electrode itself doesnt undergo anything.