00:01
Hi, so for this problem the first thing that you need to remember is that the e0 for the cell is going to be equal to e0 of the cathode minus e0 for the anode.
00:17
And both of these, the potentials are taken from their reduction reactions.
00:21
The origin of this minus sign comes from the fact that we switch up the equation of the anode because we are going to have an oxidation.
00:37
But since in the tables we are going to have always the reduction potentials, this is the way that we can find directly the e0 for the cell by just looking at the tables.
00:49
So, out of the possible options that we are going to have, in the case for the cathode, remember that in the cathode occurs the reduction.
00:57
So, the reaction that we are going to employ for the cathode, it is going to be silver plus, plus one electron, producing silver solid.
01:09
And the potential for this case, e0, is going to be 0 .8 volts.
01:19
Now, for the anode, the equation that we are going to use, it is going to be that of zinc.
01:28
But we need to switch it up.
01:32
So, we are going to have zinc solid that is going to turn into zinc two plus, plus two electrons.
01:42
And the e0 for that cell corresponds to minus 0 .76.
01:48
E0 for reduction.
01:49
If we did the e0 for the oxidation, you would need to have the opposite value.
01:57
So, to find an e0 of that cell, the value is going to be 0 .8 volts minus, minus 0 .76 volts, which is going to give us the desired 1 .56 volts.
02:14
Here you may realize that for this to be balanced, you would need to multiply the equation of the cathode by two to get the overall reaction.
02:25
But remember that the e0 potentials are not going to depend on the number that you need to multiply that equation.
02:35
They are independent of the stoichiometry.
02:40
So, now the diagram that we are going to use.
02:43
Let's draw two different flasks.
02:51
Each of them are going to have a solution.
02:55
First of all, on the left -hand side, we are going to write down the anode.
02:59
And i'm going to draw in purple a bar of zinc.
03:05
So, on the left side, we are going to have here a zinc bar.
03:12
On the right side, we are going to have a gold one.
03:23
Oh, sorry, a silver one.
03:29
Silver one, solid.
03:30
And in between, connecting them, we need to draw a salt bridge.
03:42
And finally, in order to be able to measure what is happening here, we need to get a voltmeter.
03:53
And the flow of electrons goes in this direction.
04:07
So, this is going to be our diagram.
04:09
Remember that here the oxidation reaction occurs.
04:12
Over here, the reduction...