00:01
Okay, so for each of these examples, we're going to write the two half reactions and use them to help us find our e0, and then we'll use that to find our k.
00:13
So for this first example, i'm just writing my two half reactions.
00:21
So we've got an oxidation and a reduction, and from the table, i can get my voltages, 0 .408 and 0 .154 volts.
00:33
So that's going to give us an overall e0 of 0 .562 volts.
00:42
And i think we need to multiply this by 2 so we can see that n equals 2 for that.
00:49
So then we use our relationship between e0 and k, which is 0 .257 over n times the ln of k.
01:00
So 0 .562 .0257 over n, which was 2 times our ln of k.
01:10
So ln of k comes out to be 43 .7...