00:01
The net equation given here is n -a -d -h plus h -plus half -o -2 and the other side has h -2 plus plus n -a -d -plus.
00:21
Now here the first reaction we can say is the n -a -d -h -plus plus -h plus -h plus.
00:30
Here the nadh is oxidized and it becomes nad plus and there is two free hydrogen and two electron.
00:42
And here the reduction potential said is the minus 0 .320 volt.
00:51
And the second reaction has the half oxygen plus the hydrogen atoms plus.
01:00
The two electrons.
01:03
This makes the h2o.
01:06
And here the reduction potential given is equal to 0 .816 volt.
01:15
Now in this reaction, as we know that the electron, it transports from the negative charge to positive charge.
01:28
So we can say that the water here, oxygen water, this is the electron acceptor.
01:39
And the nad plus nadh, this is the electron donor here.
01:46
So first we have to calculate the free energy.
01:52
And it has a equation here free energy that is delta g zero.
01:57
This is equal to minus n f e zero here n is the number of electron that is transferred f is equal to ferradi's constant and e here this is the electromotive force under standard condition.
02:36
So if we calculate the free energy for the acceptor, electron acceptor, then this will be minus n f, e0.
02:51
Here, n is equal to 2 because two electrons are transferred.
02:55
So this makes it minus 2 into f that is farad is constant into e0 given.
03:02
Is 0 .816 is equal to minus 1 .632f.
03:10
And the delta g0 for the donor here will be minus 2 into f into minus 0 .320, the reduction potential given...