00:03
For the following systems of the equations, let's calculate the cell potential after the addition of 15 moles of ammonia.
00:11
Let's restart by creating a redox reaction.
00:21
So we get the reduction of silver.
00:25
The e -not reduction here is 0 .80 volts.
00:30
We have the oxidation of cadmium.
00:40
Enod oxidation here is positive 0 .40 volts.
00:46
Multiply the reduction half reaction by two.
00:51
So we can cancel out the electrons.
00:55
And we'll get 2ag plus, cadmium solid, 2ag solid, cadmium 2 plus, e .0 cell is equal to 1 .20 volts.
01:16
Now let's create our equilibrium here.
01:19
Ag plus at nh3 an equilibrium with agnh3 k1 here is equal to 2 .1 times 10 to the 3 and we have agnh3 at nh3 equilibrium with ag nh32 k2 k2 here is equal to 8 .2 times 10 10 to the 3.
02:11
Let's combine these two together.
02:13
H .e.
02:14
And h3 would cancel.
02:16
We would get ag plus 2 in h3, equilibrium with ag and h3, aqueous.
02:31
Fulibrium constant here is equal to k1 times k2...