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### Course: Chemistry archive > Unit 9

Lesson 3: Standard cell potentials# Voltage as an intensive property

Demonstrating that voltage is an intensive property by calculating the voltage and change in Gibbs free energy for a half reaction.

## Want to join the conversation?

- But when you calculated the Standard Free Energy change for 2 moles of At you just multiplied it by 2! This is means that you actually considered the Voltage to be the same as before. After that you tried to prove that the Voltage was the same as before! Is that wrong? Because according to mathematics I don't find it to be correct.(19 votes)
- I was confused by this proof too because we considered voltage to be the same as before from the beginning, as you said (ie. multiplying the standard free energy by 2)

The math**is**right, but here's another way of "proving" voltage isn't affected by the number of mols: Since voltage is a change of potential, and potential is the amount of work done on a**single**point charge by an electric field, voltage won't change with number of point charges because it represents a**single**point charge in an electric field. Voltage is not affected by the number of e- moved because it represents only**one**point charge.

Hope I could help :D(23 votes)

- What if you have 2 moles of 1 substance and 1 mole of another in a voltaic cell and thus twice the number of electrons sent? Would the voltage double?(3 votes)
- Voltage is an intensive property. It doesn't depend on the amount of substance.

Changing the amount of substance doesn't change the voltage. It changes the current.(9 votes)

- How can you connect it with Gibbs free energy, if standard potential is something we decided by convention? It is just comparing with hydrogen electrode has this potential, with another it will be else.(4 votes)
- Yes it would be different with a different reference point, but would still not change with the amount of substance.

As you said, it is the convention to use a hydrogen electrode. It would also change if we altered the conditions (i.e. temp.)(1 vote)

- Where did ∆G=-nF€° came from? It doesn't seem that I missed any video or article . I'm following these electrochemistry tutorials from chemistry(science).(2 votes)
- The lessons are a bit messed up here. explanation of delta G is given in a latter video but in this topic, it has somehow come out of nowhere i guess.(1 vote)

- This is making it sound like if you were to put two batteries together, it would still be the same voltage as one battery. Is that correct?(1 vote)
- here you are saying that voltage is an intensive property but in nernst you considered the effect of concentration on potential. this is confusing .(1 vote)
- reduction potentials depend on temperature, pressure and concentration. the standard reduction potential of a substance is the reduction potential under the condition 298K, 100Pa and concentration of relevant ions 1mol/L . Is that the case?(1 vote)

- If voltage is an intensive property then why do we have nernst equation. Which relates the voltage in terms of concentrations?(1 vote)
- Ok, I have a lot of questions now.

I know that Gibbs free energy is used to calculate if a reaction happens spontaneously, but does it have any physical significance?

What is standard Gibbs free energy or Delta G naught? Is it related to the equilibrium constant in some way?

What is exactly the standard electrode potential? I wonder why is constant, because potential should be affected by position with respect to the charges. Is using a galvanic cell and a voltimeter the only way to measure it?

Is standard electrode potential dependent on electronegativity?

Why is standard Gibbs energy related to standard electric potential as is shown in the formula in this video?(0 votes)

## 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.