00:01
We're looking at a redox reaction here where we have nad plus plus a proton plus two electrons, reducing nid to nadh.
00:11
And then we have fad with two, or fad plus with two, actually fad isn't an ion, is it? with two protons and two electrons forming fadh2, and those have potentials of negative .32 and negative .29 respectively.
00:34
And so what we are looking at is the oxidation of nadh to nad plus using fad.
00:41
And so what we have to do is flip the reaction around.
00:45
So we have nadh going to nad plus.
00:53
Plus two electrons, and we are assuming acidic conditions, so we can more or less neglect these.
01:04
And so what we end up with is positive 0 .32 volts on this half reaction because we are reversing it, our electrons are going to cancel out.
01:21
And so what we do is we just, add our potentials together for our two half reactions.
01:30
So that is going to be 0 .101, so 0 .101 for the overall oxidation by fad of nad to nad plus and reducing the fad to fadh2.
01:52
And so that is our potential, our e, e, not.
02:00
And then we have to determine what the delta g is.
02:05
So delta g is equal to n -f -e, where n is the number of electrons that is being transferred...