00:01
Hello students in this question the potential of a cell which is capped in the compartment was the buffer of ph 4 then the potential of that cell is 0 .2094 word so when the following potentials they were obtained as given in the question that the buffer solution was replaced with the unknown so we have to find the value of ph for these unknowns so for a which is a which is a 3 .0 .3011.
00:36
So the reaction involved in this is two hydrogen in the eco solution with two electrons so it gives hydrogen which is in a gas form.
00:55
So now the nernest equation this will be e equals to 0 .0591 divided by n lobe 10 multiply h, hydrogen ion square divided by ph2.
01:22
So here the value of ph 2, this is 1.
01:26
So the value of e -cell, this will be equal to cell potential, wherein this is number of equivalence electron.
01:55
So the value of e -cell or cell potential, this will be equal to 0 .0591 divided by n, multiplied 2, into log 10 into hydrogen ion so this is equation one now the ph this is minus logg 10 into hydrogen ion which is equation two so we have to multiply the equation one by negative both the sides then the value of minus e cell this will be equal to 0 .0 591 divided by n multiplied 2 minus log 10 into hydrogen ion so therefore the minus log hydrogen ion this will be equal to ph as we have told here so the value of minus e -cell this will be equal to minus will come in this side so we get e -cell equal to minus 0 .0591 multiplied 2 divided by n into ph.
03:29
This is equation 3.
03:31
So put all the values in the equation 3 where we calculated.
03:37
So here for a, which is 0 .3011v...