00:01
Okay, so for this problem, we need to write the half reactions for these.
00:06
So for a, we have the line notation is ag, ag +, salt bridge, h +, h2, and it had a platinum electrode.
00:24
We don't need to draw the cells, so that doesn't really make a difference.
00:28
But this always goes from anode to cathode.
00:34
So that means that the thing that is being oxidized is our silver.
00:39
So we'll start out with our ag.
00:40
On the other side will be the ag+.
00:43
We'll write out our half reaction here.
00:45
So this would be plus one electron.
00:48
So this is the oxidation.
00:51
The reduction is the h +, going to the h2.
00:56
There would be two of these, and this would be plus two electrons.
01:02
And let's see, the whole cell reaction then would be, we have to multiply this by two.
01:10
So it's going to be 2ag plus 2h +, goes to 2ag +, plus h2.
01:19
That's the first one.
01:21
For b, we have our dichromate.
01:26
So cr2o7, two minus.
01:30
I'm just going to go ahead and write out the half reactions.
01:34
And then here we have our cr3+.
01:37
We need to put a two in front of this so we have the same number of chromiums.
01:43
And then it's going from a plus seven to a plus three.
01:47
So that's, nope, it's not a plus seven.
01:53
So minus 14, 12, it's plus six.
01:57
Okay, so from a plus six to a plus three.
02:09
Okay, so that would be plus six electrons then.
02:13
And then the other half, we have our bromide going from minus one to zero.
02:19
This one is backwards.
02:21
This line notation is not correct.
02:24
So from bromide to bromine...