00:01
Nernst equation computes for the non -standard cell potential and it is resolved at 25 degrees celsius.
00:09
It is written as e equals the standard cell potential minus 0 .0592 divided by n.
00:21
So e, because once again e is equal to e naught which is the standard cell potential minus 0 .0592 divided by n which is the mass of electron transferred times log of the reaction quotient q.
00:33
So for the given cell notation cu, cu2 plus, let's write it first as the given and then we have au3 plus au.
00:49
It is understood that here we have the oxidation half reaction to be the one that is the first part of the cell notation.
01:00
This is the one that is in the anode, the other is in the cathode.
01:03
So oxidation is for the anode.
01:06
So the oxidation half reaction then is going to be cu converting into cu2 plus with a loss of two electrons.
01:17
The reduction half reaction is the gain of three electrons by ion by whole three ions to form gold metal.
01:30
Now we have to make sure that this equation is balanced especially in terms of the number of electrons transferred.
01:37
So we multiply the first equation by three, second equation by two in order to make number of electrons the same.
01:45
So doing that the result will be 3cu to form 3cu2 plus plus six electrons and we have 2au3 plus plus six electrons forming 2au.
02:02
We add these two equations, cancel the six electrons and we get 3cu plus 2au3 plus forming 3cu2 plus plus 2au.
02:20
Now the number of electrons here is indeed equal to six.
02:25
We have seen that we just compute this standard cell potential for this reaction and we do that by subtracting the reduction potential for the reduction half reaction which is the one given here.
02:40
So the reduction potential for this is given in the problem to be equal to 1 .50.
02:49
Whereas for the reduction potential involving cu2 plus cu is 0 .34.
02:57
So once again reduction minus oxidation.
03:00
So ered of the reduction minus ered of the oxidation.
03:08
So that's going to be 1 .50 minus 0 .34.
03:18
So the standard cell potential is equal to 1 .16.
03:24
So we have this and then let's finally solve for the q.
03:32
The q is the ratio of the equilibrium concentration based on, it's just how it's just the same as how we write the equilibrium constant k that we will write it as the constitution of the product over the reactant and only those machine aq's and gaseous state appear in the equation...