00:01
Hello, in this question here we are given that is the half reaction.
00:07
First, here we cannot add the reduction potential directly.
00:10
So here we need to convert in the form of the gifts free energy so that we can add it.
00:15
So therefore here in this case that is the half reaction.
00:21
Here we have that is v plus 2 on gaining this 2 electron it get converted into v and here this electron potential value in this case it is 1 .13 voltrity scale it is there so therefore in this case gives free energy in this case as we know that that is this is nothing but minus n f e not 1 for this so now putting the value here that is here n is the number of moles number of electrons which is used here so as we can see that is 2 electron here so minus 2 f is the let us suppose f let it be f is the faraday constant let it write it in this way only and e0 value it is given that is minus 1 .13 so therefore on solving this here we will have that is plus 2 .26f now the next half cell reaction here we have that is v plus 3 on adding one electron here we have v plus 2 and here this e not value for this we have minus 0 .2.
01:28
26 volt it is given.
01:30
So therefore here in this case data let us suppose this is g1 and for the gift free energy for this second one here this using the same formula and here it is one electron so minus 1 multiplied by f multiplied by minus 0 .26 and here we will have the value that is plus 0 .26 f now adding this both reactions here we will have, that is, v plus 3 plus 3 electron, and here it will get converted to v, which is in the solid form, and here it is in the form of equest.
02:15
So now, in this case, so delta g -not here, it will become minus n is 3, multiply it by f, and here this e -0 value that we need to find out, and therefore here we can write it as minus 3f e0 so now also we will add we will have this value data g not also when we get the value if we add this value so therefore here data g0 is equal to the data g1 value and this second value and here we have that is 2 .26f plus 0 .26f and here we will have the value in this case that is 2 .52f.
03:06
Now using this formula, here we have this delta g -not...