00:01
In this problem, we're asked to calculate the standard free energy change for this reaction, given standard reduction potentials for each of the half reactions.
00:08
So, of course, in any electrochemical cell, we should first divide the reaction into its half reactions.
00:16
So in this, we have oxygen gas going to water, and we have fad to fadh2, and we're given the reduction potentials for each of these.
00:30
This one being 0 .82 volts, and this one being negative 0 .22 volts.
00:39
Now, let's first give these a quick balance just to figure out how many moles of electrons we're going to transfer.
00:46
So we're going to balance oxygen first.
00:48
So this would be two, then hydrogen, so 4h plus.
00:53
Then we'll balance the charges with electrons.
00:56
And similarly here, we'll balance 2h plus, and two electrons.
01:03
And we see that this is a two electron transfer because of the relation here, one -half -02 instead of one -02.
01:12
So we have a two -electron transfer.
01:15
The next we have to recognize which direction each half -reaction is going to go.
01:23
We realize, of course, oxygen on the reactants, h -2 -0 on the products, that this reaction is going forward.
01:31
But we have fadh2 on the reactant side and fad on the product side so this one goes in reverse...