0:00:00.280,0:00:05.000
in this quiz we're looking at another 
electrochemical cell starting with the

0:00:05.000,0:00:10.880
shorthand and note that for this week what's been 
covered is non-standard electrochemical cells and

0:00:10.880,0:00:15.680
that may be relevant for further questions 
but to begin with all we're looking for is

0:00:15.680,0:00:21.240
the oxidation and reduction half reactions that 
occur and we want to label which is at the anode

0:00:21.240,0:00:26.600
and which is at the cathode and so breaking into 
half reactions is where this notation is really

0:00:26.600,0:00:31.920
helpful because on one side of the double line 
we have the oxidation and on the other side we

0:00:31.920,0:00:37.240
have the reduction and so that's exactly what we 
have here barium to Barium 2 this is on the left

0:00:37.240,0:00:42.080
side of the double line that is our oxidation 
barium is losing two electrons and this is the

0:00:42.080,0:00:49.160
anode on the other side we have copper 2 gaining 
two electrons to become solid copper and therefore

0:00:49.160,0:00:53.200
being reduced this is the reduction and it's 
at the cathode um that's all we need for the

0:00:53.200,0:00:57.800
first part the next part we want to determine the 
number of electrons that are transferred in the

0:00:57.800,0:01:02.880
balanced redox reaction and I think the best way 
to do this is first to State the balanced reduxx

0:01:02.880,0:01:08.200
reaction which Begins by combining these half 
reactions there's two electrons in the product

0:01:08.200,0:01:12.880
and the reactant so we don't need to modify 
anything like that remember that when we combine

0:01:12.880,0:01:18.280
half reactions what we need to cancel out is 
the number of electrons in the oxidation and

0:01:18.280,0:01:23.160
the number of electrons in the reduction therefore 
no electrons will appear in the balanced chemical

0:01:23.160,0:01:28.400
reaction but it's two on both sides so here is 
our balanced chemical reaction and sure enough

0:01:28.400,0:01:33.520
what this shows is that two electrons are moving 
from barium to copper and we can determine that

0:01:33.520,0:01:40.720
n equals 2 in this case as we write this chemical 
reaction what It's associated with is the transfer

0:01:40.720,0:01:46.480
of two electrons and we can see this from the 
balanced chemical reaction itself as barium

0:01:46.480,0:01:53.000
goes from having um zero extra electrons to two 
fewer copper goes from having two fewer to having

0:01:53.000,0:01:59.360
the same number as protons so it is implied within 
the balanced chemical reaction alone but it can be

0:01:59.360,0:02:06.560
really useful to see that number explicitly in the 
half reaction combination okay so now we want to

0:02:06.560,0:02:11.320
calculate the standard cell potential and indicate 
whether it's spontaneous or non-spontaneous we

0:02:11.320,0:02:16.760
have our barium reaction right from this 
shorthand barium goes to Barium 2 plus two

0:02:16.760,0:02:23.120
electrons and note that this is an oxidation so 
the value we take for the corresponding reduction

0:02:23.120,0:02:28.320
off of appendix L we're going to flip the sign of 
that because this is actually the reverse of that

0:02:28.320,0:02:33.240
reduction reaction that appears on theable table 
the copper reduction is exactly as it appears on

0:02:33.240,0:02:38.680
the table and so we just take that half reaction 
and half cell potential directly from appendix

0:02:38.680,0:02:44.120
L now when we combine we added the two half 
reactions before to get the balanced chemical

0:02:44.120,0:02:50.640
reaction and the same thing can be done with these 
cell potentials we've already reversed the sign of

0:02:50.640,0:02:56.400
the oxidation and therefore we've already done um 
you know one negative of that value and so when we

0:02:56.400,0:03:01.320
add the oxidation to the reduction in a way that 
cancels the electrons we simply need to add their

0:03:01.320,0:03:08.640
corresponding half cell potentials so 3.24 volts 
now we want to think about non-standard conditions

0:03:08.640,0:03:15.240
and so for Part D what we're saying is that we 
start with a standard cell and note that reactant

0:03:15.240,0:03:20.120
will be consumed and product will be formed just 
like every other chemical reaction and so as this

0:03:20.120,0:03:28.120
runs the copper concentration and the barium solid 
will be depleted while the barium 2 the barium

0:03:28.120,0:03:34.560
solution and the copper solid will be created 
so still reactants becoming products is part of

0:03:34.560,0:03:40.000
this and it'll happen slowly over time and after 
a while the copper concentration has continued to

0:03:40.000,0:03:45.280
drop and the barium concentration has continued 
to rise to the point where the barium is now 1.75

0:03:45.280,0:03:52.160
molar and the copper is now .25 mol and so after 
it's run for this time we want to calculate the

0:03:52.160,0:03:56.600
new cell potential and here's where we want 
to use the nerst equation it's non-standard

0:03:56.600,0:04:02.720
conditions and what this says is that the standard 
cell potential not cell minus this term which

0:04:02.720,0:04:08.440
contains Q will combine to be equal to this the 
spell potential that we're looking for right so

0:04:08.440,0:04:14.200
basically the composition matters that's what this 
is saying and by using the standard cell potential

0:04:14.200,0:04:18.760
and our composition we'll determine the new 
cell potential so let's break down the different

0:04:18.760,0:04:24.120
parts of this equation n is equal to two moles of 
electrons we've determined this in Part B the cell

0:04:24.120,0:04:29.840
potential is positive 3.24 volts this is from part 
C and again we've done these parts already but but

0:04:29.840,0:04:34.960
if you're doing a non-standard cell from the very 
beginning all these previous parts are things that

0:04:34.960,0:04:42.520
you need to do Q here is the barium concentration 
over the copper concentration this is where it's a

0:04:42.520,0:04:48.440
little deceptive to look at the shorthand notation 
because it may look that barium is a reactant and

0:04:48.440,0:04:53.720
copper is a product but actually copper 2 is a 
reactant and barium is a product so these come

0:04:53.720,0:04:58.800
from the balanced chemical reaction and breaking 
the shorthand into that balanced chemical reaction

0:04:58.800,0:05:05.000
is really important we determine that Q is equal 
to 7 in this case from that ratio and from there

0:05:05.000,0:05:09.520
we simply use nerz calculate the non-standard 
potential it's based on that standard potential

0:05:09.520,0:05:16.520
of 3.24 volts minus the constant over Nal 2 log 
of 7 that describes the composition that we're

0:05:16.520,0:05:22.160
looking at right how it shifted from products to 
reactants based on uh you know how it changes from

0:05:22.160,0:05:26.840
standard conditions and when we evaluate that 
we get our non-standard potential is slightly

0:05:26.840,0:05:34.360
lower slightly more products than reactant so it 
will be slightly less forward pushing compared to

0:05:34.360,0:05:39.040
before slightly less spontaneous and we see that 
in the cell potential as well the value is still

0:05:39.040,0:05:42.800
positive but a little bit smaller than what 
we calculated when everything was one molar
