00:01
Occasion and i'm going to go ahead and split this into my half reactions and with my mn02 let's go ahead and put one two three four one two i'm going to put two waters over here this is an acidic solution and four h pluses on this side and i got one two three four one two three four.
00:36
Then this will be here.
00:39
I've got four, so i'll leave three electrons on this side.
00:43
Then i have my no2, no3 minus plus h2o plus 2h plus.
00:56
And then i will need, i believe, one electron on this side.
01:10
Okay.
01:10
Let me see if that looks right.
01:12
I think that looks looks right to me.
01:14
Okay.
01:16
Now, let's take this equation times 3.
01:25
And this will give me 3, 3, 3, 6, and 3.
01:33
So now let me rewrite this.
01:35
My electrons will cross out.
01:36
So i want to end up with, here's 4h plus and 6h plus.
01:42
So i have 6h plus, 2h plus.
01:59
That'll be my h plus.
02:00
And here i'll have mno4 plus 3h2o.
02:10
Oops, no, i'll have 1h2o and 3no2.
02:26
That looks good.
02:27
And then here i'll have 3no3 minus an mno2.
02:40
Ok, so there's my balanced chemical equation.
02:43
That was part a.
02:46
Part b says, given the standard free entities of formation below, low.
02:57
Given our delta g, find the standard cell potential.
03:12
Okay.
03:13
So i'm just going to write down my given numbers here.
03:16
Write them down.
03:17
Mno4 is negative 425 .1.
03:26
N03 minus is negative 110 .5.
03:38
2 is no2 is 33 .2.
03:45
Mno2 is negative 466 .1.
03:53
H2ol is negative 237.
03:58
And this is zero.
04:00
Those are all in kilojoules per mole.
04:09
Okay, so then we're going to take our delta g for the reaction will be equal to the sum of our delta g's of formation for our products minus the sum of our delta g's of formation of our reactants.
04:28
So let's see if i can write this down.
04:30
Delta g, our reaction, will be equal to 2 times 0...