00:01
Okay, this question is similar to 111 .11.
00:03
We need to consider whether or not the cation is going to be reduced or water is going to be reduced.
00:11
Or whether the anion is going to be, or we need to consider whether the anion is going to be oxidized or oxygen is going to be oxidized.
00:20
So for the first salt, we have potassium to cation with a reduction potential of negative 2 .92 volts.
00:27
The water reduction potential is negative .83 volts because water has a greater reduction potential than potassium than water will be reduced at the cathode.
00:39
The oxidation potential for fluoride going to fluorine is the opposite of the reduction potential of fluorine going to fluoride.
00:51
So it's negative 2 .87 instead of positive 2 .87 as it is shown in the table.
00:56
And then water has an oxidation potential of negative 1 .23 volts.
01:02
Because negative 1 .23 volts is greater than the negative 2 .87 volts, than water also is now going to be oxidized at the anode.
01:14
Then going on to b, the potential reduction, half reaction, is copper 2 plus going to copper solid from the salt.
01:24
But there's also a potential reduction reaction for water going to hydrogen gas and hydroxide.
01:32
We look up the reduction potential for copper at positive 0 .34 volts, and then we look at the reduction potential for water at negative 0 .83 volts, because copper has a higher reduction potential, then it will be the one that is reduced.
01:46
It wants to be reduced more than water...