O' level Chemistry Electrolysis Concept

Garrick_3658 SG

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?

UltimaOnline SG

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).

davidche SG

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.

UltimaOnline SG

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.

davidche SG

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

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.

UltimaOnline SG

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.

>>> So at the aluminium anode, both reduction and oxidation is taking place. <<<

At the anode, ONLY oxidation occurs. Nothing is reduced there (since nothing chemically accepts electrons there). Reduction occurs only at the cathode.

(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 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. <<<

Take the electrolysis of dilute sulfuric acid using platinum (inert) electrodes.

At the anode, oxidation of OH- ions (alternatively, water) occurs.

At the cathode, reduction of H+ ions (alternatively, water) occurs.

davidche SG

Oh yeah i thought wrongly.

So reduction and oxidation cannot take place at the same electrode eh?

I was asking wad if aluminuim anode was placed in copper sulphate solution. Your implied answer should be no displacement since the electrons given out by oxidation of aluminium is immediately sucked into the wire right? Thus no electron goes into the electrolyte to reduce Cu2+ in the electrolyte.

UltimaOnline SG

Oh yeah i thought wrongly.

So reduction and oxidation cannot take place at the same electrode eh?

(EDITED : FOR THE ELECTRIC CELL TO WORK, COMPARMENTALIZE ELECTRODES & USE 2 ELECTROLYTES & USE SALT BRIDGE TO AVOID ANY DIRECT REDOX BETWEEN ELECTROLYTE AND ELECTRODE)

Always remember, "Red Cat riding An Ox". So ANY reduction MUST take place ONLY at the cathode. ANY oxidation MUST take place ONLY at the anode.

Any further questions?

davidche SG

not at all XD

UltimaOnline SG

davidche, let's revisit this for a bit.

TinTin wrote :

>>> 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? <<<

This is true (meaning redox reaction directly between electrode and electrolyte, current does not pass through wire, no usable electricity is generated; and this problem is usually avoided by means of a salt bridge).

But at the same time :

>>> Afterall, "Red Cat riding An Ox". So ANY reduction MUST take place ONLY at the cathode. ANY oxidation MUST take place ONLY at the anode. <<<

This is also always true, IF the electric cell functions as an electric cell.

So let's consider what happens in such a setup.

Assuming standard molarities for all solutions (although molarities would change immediately and rapidly as the reactions progress) and at 25 deg C.

Say we have aluminium electrode, zinc electrode, copper(II) sulfate solution.

And no salt bridge.

Aluminium electrode would react in a redox reaction with the copper(II) sulfate solution.

Al ---> Al3+ + 3e- (+1.66V)

Cu2+ + 2e- ---> Cu (+0.34V)

Zinc electrode would react in a redox reaction with the copper(II) sulfate solution.

Zn ---> Zn2+ + 2e- (+0.76V)

Cu2+ + 2e- ---> Cu (+0.34V)

Both redox reactions are feasible and do occur. And because there is no wire separating the reactants (for these 2 redox reactions), no electrons go into the external wire, and hence no usable electricity is generated.

Because of this, neither electrode qualifies as either "cathode" or "anode", because the setup has failed as an electric/Galvanic/Voltaic cell.

Say we have another setup. Copper electrode, zinc electrode, copper(II) sulfate solution. And again no salt bridge.

Zinc electrode would react in a redox reaction with the copper(II) sulfate solution.

Zn ---> Zn2+ + 2e- (+0.76V)

Cu2+ + 2e- ---> Cu (+0.34V)

Let's see if the copper electrode would react with the copper(II) sulfate solution.

Oxidation of copper electrode :

Cu ---> Cu+ + e- (-0.52V)

Cu ---> Cu2+ + 2e- (-0.34V)

Reduction of copper(II) electrolyte :

Cu2+ + e- ---> Cu+ (+0.15V)

Cu2+ + 2e- ---> Cu (+0.34V)

Case 1 :

Cu ---> Cu+ + e- (-0.52V)

Cu2+ + e- ---> Cu+ (+0.15V)

Not feasible.

Case 2 :

Cu ---> Cu+ + e- (-0.52V)

Cu2+ + 2e- ---> Cu (+0.34V)

Not feasible.

Case 3 :

Cu ---> Cu2+ + 2e- (-0.34V)

Cu2+ + e- ---> Cu+ (+0.15V)

Not feasible.

Case 4 :

Cu ---> Cu2+ + 2e- (-0.34V)

Cu2+ + 2e- ---> Cu (+0.34V)

Not feasible.

So copper electrode does not directly react with copper(II) electrolyte.

Now, if you had prevented (eg. by means of a salt bridge) the direct redox reaction between zinc electrode and copper(II) sulfate electrolyte, then at the anode, the electrons would indeed flow via the external wire to the cathode, where reduction of Cu2+ ions from the electrolyte to Cu atoms/metal would occur at the cathode.

So to solve this problem, either USE A SALT BRIDGE, or use an electrolyte that will not react with the electrodes.

davidche, in retrospect, it was my bad that I neglected that you guys were discussing this particular aspect of electrochem (of direct redox reaction between electrode and electrolyte). I may have been distracted by your (apparent) confusion regarding the terms "cathode" and "anode", when you made statements like, "so reduction and oxidation occur simultaneously at the anode? then why did you say reduction only occurs at cathode and oxidation only occurs at the anode?"; and now I realize why you had this confusion - because you didn't realize (and unfortunately no one else was able to point this out, till my post now) that when such direct redox between electrode and electrolyte occurs, and no current flows in the external wire, IN SUCH A CASE THERE IS NO "CATHODE" AND NO "ANODE".

So davidche (and anyone else reading this thread who might have been confused by its contents), my bad that I neglected to explain this earlier. So in light of this latest development, again I ask you (and anyone else reading this thread),

Any Further Questions?

UltimaOnline SG

Ahhh ok, I see this question as the Nov 97 P2 Section A Q7.

This question is actually trickier than it seems :

Because there is no compartmentalization and salt bridge, you cannot use copper(II) sulfate electrolyte; because as explained in a previous post (and first brought up by "Tin Tin" - the Charted Financial Analyst who tutors "after work nia"; due credit to her), there will be a redox reaction between the metal (eg. Al) electrode and the Cu2+ ions in solution.

Ideally, compartmentalize the two electrodes and connect with a salt bridge. This will give much more flexibility in regard to the electrolytes to be used (copper(II) sulfate may be used as electrolyte for the copper electrode; the electrolyte for the anode could be sodium nitrate; since Na+ cannot be reduced by Al). The ions in the salt bridge could be (choose ions that are very 'safe') sodium nitrate or ammonium nitrate, etc. (Google "salt bridge" to understand more about it.)

HOWEVER, since the diagram setup in this past 'O' level exam question does not include compartmentalization and a salt bridge, I suggest using a electrolyte such as aqueous sodium chloride solution (tis the best SOLUTION... pun intended!!! Woooot!!!)

The anode (reactive metal such as aluminium) will undergo oxidation :

2Al ---> 2Al3+ + 6e-

The cathode (copper) will transfer electrons (from anode via external wire) to reduce the protons (H+ ions) in the sodium chloride solution electrolyte, to H2 gas :

6H+ + 6e- --> 3H2

Any Futher Question?

eagle SG

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

UltimaOnline is the power chemist here :D

davidche SG

Hes the man!

No questions for now XD

« Previous Page