00:01
In this problem we are given that the half shell reaction will be.
00:05
So firstly we are writing the reaction at anode and it will be 2 times ag that is silver and it is getting converted into ag plus and then it will be plus 2 electrons.
00:19
So let us write the reaction at cathode and it can be here written as cathode and the reaction will be hg then it is 2.
00:29
Now here it will be 2 plus and then it is plus 2 electrons and it is giving us 2 and then it will be hg so overall reaction can be written over here as 2 ag and then it is plus hg 2 now it is 2 plus so this is giving us 2 ag plus and then it will be plus 2 and it will be hg so given data includes that the value of e cell will be here equals to it is 0 .0455 volt and then the value of partial differentiation of e cell with respect to the temperature and that is here equals to 3 .38 times it will be 10 to the power negative 4 volt per kelvin and the value of n will be equals to 2.
01:30
So in the first part we will calculate the free energy change denoted by delta g and the expression for delta g will be negative n multiplied with f and here it is e cell so it will be here equals to let us write the value of delta g and that is negative 2 multiplied with 9 6 4 8 5 and then it will be be 0 .0455 and this is equals to 8780.
02:08
Now it is 0 .1 then it will be joules.
02:11
So we have obtained the value of free energy change and we can put this particular value inside the box and it will look like this.
02:21
Actually we forget to mention the negative sign.
02:25
Very sorry for that.
02:26
And then in the second part we will write the value of entropy change...