00:01
So let's go ahead and start by writing our half reactions, so that we can find our electric potentials, and then use that to find k.
00:11
Okay, so here's our oxidation, and then our reduction, silver plus an electron, gives us that.
00:21
So our potentials here are 0 .127 volts and 0 .799.
00:28
So our overall voltage is going to be 0 .926 volts.
00:35
This first one has two electrons, so i'm going to need to multiply the second equation by 2.
00:41
And that shows us that n is 2.
00:43
Two electrons are being lost and gained.
00:47
And then e0 is simply 0257 over n times the ln of k.
00:56
So 0 .926 is our 0 .257 over 2 times the ln of k.
01:07
So the ln of k is going to come out to be 72 .1.
01:13
And k is 1 .98 times 10 of the 31.
01:21
So we'll do the other two examples in the same way.
01:25
So this first one i've got oxygen plus 4h plus.
01:32
It's being reduced with these four electrons to make two waters.
01:38
And then our oxidation is our fe2 plus, goes to fe3 plus, plus one electron.
01:48
Our voltages are 1 .229 volts and negative 0 .769, which gives us an overall voltage for our overall equation of 0 .460 volts...