- [Instructor] We already know that in an electrolytic cell,
current or movement of electrons
is used to drive a redox reaction.
And if we look at a generic reduction half-reaction,
the stoichiometry of the half reaction
shows how many electrons are needed
to reduce a generic metal ion.
For example, if we say that our generic metal ion is M2+,
it takes two electrons to turn M2+ into the solid metal.
So one mole of M2+ ions requires two moles of electrons
to form one mole of the metal.
So we could use this mole ratio of one mole of metal
to two moles of electrons to figure out,
say, how much of a metal is produced
in an electrolytic reaction.
So this is an example
of a quantitative electrolysis problem.
Moles of electrons are related to the amount of charge
that passes through the electrolytic cell
because one mole of electrons carries 96,485 coulombs.
So we could set this up as a constant
called Faraday's constant.
So Faraday's constant is symbolized by F,
and it's equal to 96,485 coulombs per mole of electrons.
And charge is related to current and time by this equation.
So I is equal to Q over t;
where I is the current; Q, capital Q is the charge;
and t is the time.
Sometimes, you see a lower case Q for charge.
Charge is measured in coulombs,
and time is in seconds.
And one coulomb per second is equal to one ampere,
so current is measured in amperes.
Next, let's look at a conversion chart that shows
how to approach quantitative electrolysis problems.
And let's start over here with this box on the far left,
which says that charge is equal to current times time.
This comes from our equation for current,
I is equal to Q over t.
So solving for charge, solving for Q,
Q is equal to I times t
or charge is equal to current times time.
Charge is related to moles of electrons
by Faraday's constant.
So we can convert back and forth between the two
using Faraday's constant.
And moles of electrons are related to the moles
of the substance that we're interested in the problem
by the balanced half-reaction.
So the balanced half-reaction allows us
to convert back and forth between moles of electrons
and moles of substance.
And from stoichiometry,
we know that the moles of a substance
is related to the grams of the substance by the molar mass.
So the molar mass allows us to convert back and forth
between these quantities.
So it helps to look at a conversion chart
when doing these sorts of problems.
However, if you don't remember the chart,
you can always use dimensional analysis and units
to figure out the answer to the question.
Next, let's use our conversion chart
to figure out a quantitative electrolysis problem.
Let's say we have an aqueous solution of silver nitrate
and a constant current of 2.40 amperes is applied
for 1,225 seconds.
And our goal is to figure out the mass of silver
that has formed.
As a quick reminder, an aqueous solution of silver nitrate
contains Ag+ ions and nitrate anions in solution.
So the applied current is used to reduce the Ag+ ions
to form solid silver.
And since our goal is to figure out the mass
of the solid silver that's formed,
we're trying to find grams of substance.
And in the problem we're given the current and the time,
so we're starting over here in the first box
with the current and the time.
So let's think about the steps that we need
to solve this problem.
The first step is to find the charge,
which we can get from the current and the time.
The second step is to convert charge
into moles of electrons.
From moles of electrons,
we can convert that into the moles of silver in this case.
And in the fourth step for this problem,
we can convert moles of silver into grams of silver.
So step one is to find the charge.
And Q, the charge, is equal to the current times the time.
So the current is equal to 2.40 amperes.
And remember,
one ampere is equal to one coulomb per second.
So we plug in 2.40 coulombs per second.
And the time is 1,225 seconds,
so we plug that in, seconds cancels out,
and we get that Q is equal to 2,940 coulombs.
So that's the charge that's transferred in that time period.
Now that we have the charge,
in the second step,
our goal is to find the moles of electrons.
And charge is related to moles of electrons
by Faraday's constant.
So dividing the charge by Faraday's constant
causes coulombs to cancel out,
and we get 0.0305 moles of electrons.
And now that we have moles of electrons,
in step three, we can find the moles of silver.
Looking at the balanced equation for our half reaction
showing the reduction of silver ions,
one mole of silver ions requires one mole of electrons
to form one mole of solid silver.
So our mole ratio of silver to moles of electrons
is one-to-one,
which gives us this conversion factor
of one mole of silver per one mole of electrons.
So if we multiply our moles of electrons
by our conversion factor, moles of electrons cancels out,
and that gives us 0.0305 moles of silver.
In this case, our mole ratio was one-to-one,
but if we were reducing a different metal ion,
our mole ratio might not be one-to-one.
So we have to be extra careful on the step
to make sure we get the right mole ratio.
And since we have moles of silver,
in our fourth and final step,
we can convert moles of silver into grams of silver.
So multiplying moles of silver by the molar mass of silver,
which is 107.87 grams per mole,
moles cancel and this gives us 3.29 grams of silver.
It's also possible to do all four steps at once
in a dimensional analysis approach.
So looking at our units,
if we multiply 1,225 seconds by 2.40 coulombs per second,
seconds cancels out and gives us coulombs.
Multiplying that by one mole of electrons
per 96,485 coulombs, cancels out coulombs.
Next, we have our conversion factor of one mole of silver
per one mole of electrons,
so moles of electrons cancels out.
And multiplying that by the molar mass
of 107.87 grams per one mole,
moles of silver will cancel out
and give us a final answer of 3.29 grams.
So we get the same answer, no matter which approach we take.