DENA LEGGETT: Hi, and welcome to AP Daily practice session.
In this session, I'm going to be looking
at a long-free response.
My name is Dena, my kids call me Doc Dena
and I am so glad you joined me for this.
This long for your response that we're looking at
covers electrochemistry, Beer's law, and kinetics.
So many topics in one question.
So let's dive in.
First one gives us a couple of half reactions
and it asks us to write a balanced equation
for the reaction.
Now there's some key terms you want to underline here
or you would not get A correct.
The prompt tells us the student uses a galvanic cell.
So if I want to write a correct balance equation,
that's really important information.
Galvanic cell means that we're dealing
with a thermodynamically favorable situation.
In other words, we're going to produce
voltage and E not of the cell should come out to be positive
if we set it upright.
So to do that, what we want to do to have it thermodynamically
favorable is we want the most positive reduction potential
to be reduced.
So we're going to choose the second reaction
as our reduction.
A, and we want our first to be an oxidation.
So to make that an oxidation, I flipped it.
I took the products and I made them reactants
and the reactants and made them products,
everything else is still the same.
Now I have to multiply that first reaction by 3.
I have to exchange the same number of electrons,
I can't lose 4 and gain 12 or vice versa.
So we need to multiply the first reaction by 3 so that I
get 12 electrons in both.
Now let's take a look at what happens
once we've added these up.
My electrons cancel, some of the water is going to cancel,
you just bring things down.
Everything that's on the left, you
keep on the left of the arrow, everything that's
on the right in both equations you keep on the right.
And there's my balanced equation, done.
OK.
This time I want to use the galvanic cell to find
the concentration of ethanol.
And we have the half reactions below.
And first off, we're going to calculate the E
naught for the overall cell.
We've done half the work for this.
We have figured out our reduction
and we have its reduction potential.
Now the way I teach it for my kids
because it's so intuitive for them,
we've reversed the reaction, we now
have an oxidation half reaction.
So I'm going to change the sine of the E naught
to make it an oxidation potential.
So instead of negative 0.085, I have positive 0.085.
Now I can just add those puppies up
and I have my cell potential.
What this represents is the maximum voltage
I could get from that cell.
Now we're still dealing with the electrochemistry
but it's a little drift into using the current
and finding Coulomb's and finding
moles that can be produced.
So this time, I want to calculate the charge
in Coulomb's.
So I have a 10 mL sample, tells me I'm going to run--
add 0.1 amps for 20 seconds.
I want to show you two ways to do this.
One way is to use the formula that's
going to be at the beginning.
I don't like my kiddos to have to flip back and forth
the formula charts, so my suggestion
is if you like the formula approach memorize that formula.
So let's start with that one.
So I, that's amperage is q over t, that's charge over time.
So I have 0.1 amps is equal to q over 20 seconds.
It's actually pretty easy, especially
if they ask you to land right at that Coulomb's of charge
or maybe ask you to land at the amperage.
I like to use dimensional analysis
because you can string along a longer problem when
you do that.
And so that's what we would do.
The key to doing these types of problems
is to remember two things.
One is that an amp is simply a Coulomb's per second.
So I can just change amps to Coulomb's per second
and use that as a conversion factor.
The other we'll see later on is the use of Faraday's constant.
So I started with 20 seconds, 0.1 amp
means I have 0.1 Coulomb's for every one second,
set it up per second to cancel and I get that same answer.
Use what's fastest for your mind.
Now I have to calculate the initial concentration.
In order to do that, I need moles and I need liters.
I already can get to my leaders pretty quickly.
The tricky part here is going to get the moles of ethanol.
So I already got my Coulomb's of charge that goes through, now I
can use Faraday's constant.
Faraday's constant is my link between the world of chemistry
and the World of electricity is what I like to call it.
And that tells me that for every 1 mole of electron that is
transferred, I need 96,485 Coulomb's.
So I can get my Coulomb's to cancel.
Now I'm at moles of electrons.
But if I want molarity of ethanol,
I need to get moles of ethanol.
Now my balanced equation is what's going to give me that,
so I popped that back in for you.
You see that the balanced half reaction
says I have to consume 12 moles for every 1 mole of ethanol
I make.
So that gives me my moles of ethanol.
To calculate molarity, I'm going to put my moles over my volume
and I need to convert that volume to liters,
so I showed that to very explicitly here.
Some of you might do that step without showing it.
Here's a warning.
If you show your work and you have that conversion set up
correctly and your answer is wrong,
that becomes a math error.
If you convert milliliters to liters in your head
and you have it wrong, that can become a conceptual error.
So do math, show math.
Lesson today, do math, show math, show your work.
OK.
Now we're still on the same long, long question
and now they've given me a whole bucket load of information
that I have to wade through.
So you want to focus on some key information here.
They gave me an initial molarity of dichromate
and initial concentration of ethanol.
That's the link here, is that ethanol
the link between the two parts.
Then it gives me a path length and a wavelength.
And so we're going to be dealing with Beer's law here.
So let's focus on that as to calculate my concentration
at 1.5 minutes.
Now it does tell us that the absorbance
is for the dichromate here.
So when we use Beer's law, we have
absorbance as molar absorptivity times the path length times
molarity.
Now there are two approaches to this as well.
I'm going to take the slightly longer one so I can reinforce
that equation for you.
At 0 minutes, I know my molarity was 0.5.
I know my absorbance was 0.782.
And it gives me my-- excuse me, it gives me my molarity here
and it gives me my path length, I misspoke.
This is my molarity, this is my path length.
So I can use this to calculate my molar absorptivity
activity here.
Otherwise I'd have two unknowns at 1.5.
At 1.5 I don't know my concentration,
so I have to find my absorptivity
to use this equation.
Now I can hop to that 1.5 timeline.
I can plug in my new absorbance.
I have the same path length of 0.5,
now I can calculate the concentration of the dichromate
at 1.5 minutes.
And c is 7.1 times 10 to the minus 4.
Now, what you could have done here
is taken the ratio of absorbance to concentration.
So I could have had known concentration over 0.782
is equal to unknown concentration over the 0.553.
Would have been a little faster but this gave me an opportunity
to talk about that path length and that molar absorptivity.
OK.
Now I have a second trial, but this time I changed my life.
And I want to keep my same absorbance.
So let's evaluate that equation that we took time with.
This prompt is telling me I want constant absorbance.
I have a constant molar absorptivity,
that depends on the absorbing species.
I've now doubled my path length.
All right.
So the only way I can keep that constant absorbance
with doubling my path length is I have to have my molarity.
So it's going to-- the light's going
to go through more distance, so I
need to have a less similarity along the way,
or I would increase.
So just mathematically if I double the path length,
I have to have my molarity.
Now we're hopping into the kinetics,
this time it's giving me a rate low expression, which
is first order with dichromate.
And that ethanol molarity that we saw
is wrapped into the rate constant there.
So k observed is the k of the reaction
along with that molarity.
So what we want to do is explain how the data shows us
that it really is first order with respect
to the dichromate ion.
We've got to agree with that rate low expression.
Now, whenever you are given time and concentration or time
and molarity data and you're asked for order,
the first thing I encourage you to look at is half lives.
That's going to give you your main evidence,
is the absorbance is halved in 3 minutes and then I look again
and it's halved in another 3 minutes.
That means I have a constant half life.
I want to give you a warning.
I graded papers and so many papers said it has a half life.
All of these reactions have a half life.
What differentiates first order is
that it has a constant half life that
is independent on my starting molarity and hence independent
on my starting absorbance.
So you have to put constant in there or you won't get credit.
Now another thing people do here is they make the claim, oh, I
have my data, the linear--
if I take the natural log of absorbance versus time,
it's going to be linear therefore it's first order.
You can't just make that claim, you have to prove it.
You have to graph this data to prove
that the natural log of the absorbance versus time
is linear therefore supporting first order.
You can't just make the statement.
So you're going to have to graph it somehow, someway.
OK.
Well, that was another fast review.
There's a lot more information on the AP
College for YouTube channel and I gave you
the QR code for that.
And so I recommend you hop over there
if you need some more help.
In the meantime, well, thanks for letting me
be here on your journey of learning even more chemistry.