00:01
Hello students, let's begin with this question.
00:02
In this question one redox reaction is given that is two moles of silver ions.
00:07
It will react with the silver and this will result to the formation of the silver plus magnesium ion and we have to find out the e cell for this reaction and some data is provided in the question.
00:23
So, the concentration of the mg2 positive ion is already given in the question which is 7 .50 into 10 raised to power minus 4 molar and the concentration of the silver ion is given which is equals to 1 .27 molar.
00:38
Now, in this firstly we have to find out the reduced and the oxidized species.
00:42
Now, here this silver plus is changes to silver.
00:46
So, in this case there this silver the oxidation number of silver will here it will decreasing that is it is going from plus 1 to 0.
00:57
So, here the reduction occurs.
01:00
So, generally the reduction occurs at the cathode and the e naught for this reaction is equals to plus 0 .80 volt.
01:10
Now, we have to find out the oxidized species.
01:12
Now, the magnesium 0 which is having a zero oxidation state is converted into a magnesium 2 positive.
01:19
So, there is an increase in oxidation state.
01:21
So, here the oxidation occurs.
01:23
So, oxidation generally occurs at the anode and the e naught for this reaction is minus 2 .37 volt.
01:31
Now, in order to find out the e cell we have to find we have this can be find out by using the nernst equation.
01:41
So, nernst equation is equals to e cell is equals to e naught cell minus 0 .059 upon n log q.
01:52
So, here comes the concentration...