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## Standard cell potentials

Current time:0:00Total duration:5:22

# Voltage as an intensive property

## Video transcript

- [Voiceover] When you are working with standard reduction potentials, it's important to realize that voltage is an intensive property. And by the end of the
video you'll understand what I mean by the word "intensive." We're gonna use this first half
reaction here as an example. So this is the reduction of
silver ion to silver metal, so we have Ag+ plus an
electron gives us solid silver. The standard reduction potential is positive .80 volts
for this half reaction. And we're gonna start by calculating the standard change in free energy. So from an earlier video, we
know if we have the voltage, we can find the standard change
in free energy Delta G zero. So for this half reaction
to be equal to negative, remember that "n" is the
number of moles of electrons. And here we have one mole of electrons. So we have one mole, so we
plug that into our equation. Next, we have F which
is Faraday's constant. And from an earlier video we know that's 96,500 coulombs per mole, so it's the charge of
one mole of electrons. We need to multiply that by
the voltage which was .80. Instead of writing
"volts," I'm gonna write "joules per coulomb" here, so
we can see our units cancel. So moles of electrons cancels out charge, coulombs cancels out. And we should get our
answer in terms of joules. So this is equal to negative. And let's figure out what it is. This'd be 1 times 96,500 times .80, and it should be the negative of that. So -77,200 joules, which I'll just say is -77 kilojoules. So we get -77 kilojoules is
the change in free energy that accompanies the formation
of one mole of silver. What about if we wanted to
form two moles of silver? So we need to multiply everything in our half reaction by 2, so we get 2Ag+. So two moles of silver ions, plus two moles of electrons should give us two moles of solid silver. What would be the standard
change in free energy that accompanies the formation
of two moles of solid silver? Let me change colors here. If this is the standard
change in free energy for the formation of one
mole of solid silver, then we should just be able to multiply that number by two, right? So -77 kilojoules times 2 gives us -154 kilojoules. So that's the standard
change in free energy that accompanies the formation
of two moles of silver. So, free energy is what's
called an extensive property. So, it depends on how much
you're dealing with here. So, we're dealing with
two moles of electrons and the formation of two moles of silver. So, we have to change the standard change in free energy accordingly. What about the voltage? So, we have our standard
change in free energy. Let's use this equation again
to solve for the voltage. So we're gonna plug in -154 kilojoules in for our standard change in free energy, and we need to make that into joules, that's -154,000 joules. And that's equal to, let me
change colors again here. This would be equal to negative, so we have the negative sign. And then "n" is number
of moles of electrons. Well now we're saying
two moles of electrons, so this'd be -2 here. And then multiply that
by Faraday's constant. So we have Faraday's
constant which is 96,500. And then we can solve for the voltage. So we can solve for E zero. So let's plug that in on our calculator. We have -154,000, right? And we need to divide that by -2. And then divide that
by Faraday's constant. So 96,500 which gives us a voltage of .80. So if you round that we get a voltage of .80 volts. So it's the same voltage
to form two moles of silver as it was to form one mole of silver. Let me highlight that, so I'm
let me change colors here. So we had a voltage of .80 volts to form two moles of silver. That's the same voltage to
form one mole of silver. So, voltage is an intensive
property, it's the same. It doesn't matter how much
silver you're forming, how many moles of electrons you're using, the voltage is the same. And it's important to remember that when you're doing your
standard reduction potentials, because if you need to do
something like we did up here, multiply a half reaction by 2, you don't multiply the voltage by 2. because voltage is an intensive property.