00:01
Let's see if we can answer this rather giant question.
00:03
We are asked to consider zirconium and we're asked for our first question, a.
00:10
Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? and i'm going to think that this is probably going to be a yes.
00:53
So is it capable of reducing this? i'm going to call this reaction one.
01:01
Reaction one, we would have to reverse.
01:06
So it would be e would equal plus 2 .36 volts.
01:12
And my second reaction would have to be reducing water.
01:28
That would be h2 plus 2oh minus and this is negative, e for this cell is negative 0 .83.
01:53
So my e for my cell will be equal to negative 0 .83 plus 2 .36, and that will be 1 .53.
02:36
So this means that zirconium is capable of acting to reduce water.
03:05
Okay, there's my first question.
03:13
Let's go to my second question.
03:21
B.
03:24
Write the balanced equation for reduction of water.
03:28
Hmm.
03:30
Balanced equation for the reduction of water.
03:42
I.
03:43
Zirconium.
03:45
Ok.
03:49
And let's get right here.
03:51
So i will have h2o and i will have let's get over here back to my equation so i can see it.
04:16
I'll have zirconium plus 4 oh, and this will be zro2, 4 e minus.
04:43
Then i'm going to take this equation times 2 to get 4, 4, 2, and 4.
04:57
So my ohs will cross out, and i will end up with 1 zirconium plus one two three four three h2o and i get a zr o2 dot h2o plus two h2 i think this was an s this is an l okay that one wasn't so hard c back to c find e cell didn't i already do this delta g and k.
06:10
My e cell, i already did one problem, it was positive 1 .53 volts.
06:17
We did that in part a.
06:21
And then our delta g will be equal to negative nfe minus.
06:29
So my delta g will be equal to, remember from up here we had four moles of electrons times faraday's which is 96485 c over moles of electrons times 1 .53.
06:54
4 times 96485 times 1 .353.
07:01
And then i like to multiply this by 1000 joules per kilojoule.
07:13
And for my volts, i'm going to put my 1 .53 j over c there.
07:23
So my delta g would be equal to negative 590 .5 kilojoules.
07:34
And now my k is equal to the antilog of my, let me see here, my k is equal to my antilog of 4 times 1 .53 divided by 0 .0592.
08:23
Let me try this.
08:27
Second e.
08:28
4, this should be minus, times 1 .53 divided by 0 .0592.
08:37
4 times 1 .53 divided by 0 .0592 is 103 .4.
08:59
Let's try the anti -log of the second answer...