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### Course: Class 12 Chemistry (India) > Unit 2

Lesson 2: Standard cell potentials- Standard reduction potentials
- Using reduction potentials
- Voltage as an intensive property
- Worked example: Identifying the best oxidising agent using Eo(red.) values.
- Worked example: Constructing a galvanic cell using reduction potential values
- Spontaneity and redox reactions
- Free energy and cell potential
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential

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# Free energy and cell potential

Relationship between Gibbs free energy and standard cell potential. Created by Jay.

## Want to join the conversation?

- Can you have negative voltage?(8 votes)
- A spontaneously operating cell will always have a positive voltage.

A negative voltage just means that the cell will operate spontaneously in the opposite direction.(8 votes)

- I didn't catch why we have 2 moles of electrons at5:35?(6 votes)
- Zn(s) loses 2 electrons when it is oxidized to become Zn2+ (aqueous). Those 2 electrons travel the wire to the cathode and the Copper ions (Cu 2-) in solution pick them up (get reduced) to copper solid. So for every mole of Zn(s) that's oxidized, 2 moles of electrons are stripped to travel the wire and reduce one mole of copper ions.(9 votes)

- How do we get the relation between Gibbs free energy and standard cell potential??(9 votes)
- What is the role of the ZnSO4 solution in the left side of the cell, because the Zn eletrode is the one who will lose its electrons so why do we need a ZNSO4 solution?(5 votes)
- The role of the salt solutions is to maintain an electrogradient. The redox reaction of the two metals in the galvanic cells has electrons moving from a negative (high e- concentration) to a positive (low e- concentration area). For a battery to work, there always needs to be an electromotive force because of difference in potentials. When the two cells are equalized, there's no oomph for those electrons to move.

With the salt bridge, positive ions move towards the cathode because they're now attracted to the accumulating electrons, and anions move towards the anode because they're now attracted to the oxidized material (positive charges). This ensures that a gradient is maintained--it's seen often in cellular biology....basically a gradient needs to be set up and maintained to ensure that there is always a "downhill" for electrons to flow.(7 votes)

- Where did that 1.10V come from? Just from the experiment?(4 votes)
- The overall cell potential of 1.10V was obtained through the addition of the literature values given for each of the half-cell reactions (these can be found in chemistry textbooks etc) - have a look at the next video in this section (standard reduction potentials) where this is explained thoroughly.(4 votes)

- Why a reaction can be a spontaneous reaction only when the change in free energy is negative?(3 votes)
- Because that's how it is defined. It's a function of entropy, enthalpy and temperature:

delta_G = delta_H - T * delta_S(2 votes)

- But as the process progresses , the concentration of ions will change and tha leads to the E changing. obviously the equation doesn't hold anymore.....(2 votes)
- The equation does hold, but the E value and hence the G value would change. But since we are only looking at 'standard' values Go and Eo, we're not really concerned with what happens as the reaction progresses.(2 votes)

- Standard change in free energy just means Change in free energy at standard conditions i.e. stp(25degrees celcius) and standard pressure?(1 vote)
- Correct, except that standard state is 25 °C and 1 atm, while STP is 0 °C and 1 bar.(2 votes)

- Is there any way to get the standard change in free energy (ΔG naught) if the reaction is not redox?

Can I use the ΔG=-nF(emf) for a non galvanic cell redox reaction like the formation of water?(1 vote) - How do you figure out how many moles of electrons are transferred? or are you given the number of moles all the time?(1 vote)

## Video transcript

- [Voiceover] When you're
dealing with a voltaic cell, it's important to relate
the potential of the cell to the free energy of the redox reaction. And here's the equation that relates free energy to the cell potential. So delta G, delta G is
the change in free energy. And we know for a spontaneous reaction delta G is negative. So let me go ahead and
write that down here. So delta G is negative for our spontaneous redox reaction that we
have in our voltaic cell. And we're going to calculate
the value for delta G at the end of this video. E over here, this represents
our cell potential, or our cell voltage. And this is easy to measure. Just hook up a voltmeter and for this cell the cell potential is positive 1.10 volts. So very easy to measure the voltage associated with a voltaic cell. Next, the n, the n represents the moles of electrons that are transferred
in the redox reaction. In this example we're
talking about two moles of electrons are transferred
in our redox reaction. And finally, let's talk about F, which represents Faraday's constant. And Faraday's constant is
the magnitude of charge that's carried by one mole of electrons. So we can calculate Faraday's constant, let's go ahead and do that up here. So one electron has a charge of 1.6 times 10 to the negative 19 coulombs. Coulombs is the unit for charge. So let's go ahead and write that. 1.6, actually it's 1.602, times 10 to the negative 19 coulombs, for every one electron. So one electron has that charge. If we want to find the magnitude of charge carried by one mole of electrons, we would need to multiply
by Avogadro's number. So if we multiply by Avogadro's number, which we know is 6.022 times 10 to the 23rd. That's how many electrons there are in one mole of electrons. So that's how many electrons
in one mole of electrons. If you multiply these
out, you're gonna end up with charge per one mole of electrons. Let's get out the calculator and let's find Faraday's constant. So we have 1.602 times 10 to the negative 19. And we're going to multiply
that by Avogadro's number, 6.022 times 10 to the 23rd. And we get 96,472, so this is 96,472. The units would be, this is coulombs per mole of electrons. So coulombs per mole. And if you do a more careful
calculation, you'll get 96,485 coulombs per mole. And so sometimes you'll see textbooks use this number for
different calculations. Most of the time you can just round this to 96,500. So 96,500 coulombs per mole. And that's good enough for almost all the calculations that you will do. So that explains each
term of our equation here. So this cell potential
that we talked about, 1.10 volts, this is actually
the standard cell potential. Which is the voltage measured when the cell operates under
standard conditions. Which is defined as, all of
your solids are in pure form and that's the case for us here, because we're dealing with zinc metal. So that's a solid in its pure form. And copper metal, a
solid in its pure form. Also defined as the solutions are at one molar concentrations. We have a one molar concentration solution of zinc sulfate, which gives us one molar concentration of zinc two plus ions in solution. And a one molar concentration
of copper sulfate, which gives us one molar concentration of copper two plus ions in solution. And our temperature is 25 degrees. So this is 25 degrees C. So if those are your conditions
and you hook up a voltmeter, this is the voltage that you will get. So we can actually modify the equation that we've talked about by
adding in a superscript here. So we could write E
zero instead of just E. And E zero is our standard cell potential. It just means that everything
is under the conditions, under standard conditions. And then this would be delta G zero. So that's delta G zero now, the standard change in free energy. So let's go ahead and
solve for delta G zero. We already know it's gonna be negative, because this is a
spontaneous redox reaction. But let's go ahead and plug in everything. So let's get a little
bit more room down here. And let's plug in our numbers. So we're trying to solve for delta G zero, the standard change in free energy. So we have a negative sign here, I'll talk about what that
means a little bit later. So next we have n. Remember what n represented here. So n is the moles of
electrons that are transferred in our redox reaction. Which would be two. So this would be two moles of electrons. And we put these in parentheses here. Next we have Faraday's constant, F. And so we've already seen that we can use 96,500 coulombs per mole for Faraday's constant. So this is 96,500 coulombs per mole. Next we plug in the value for
our standard cell potential. So E zero is 1.10 volts for this voltaic cell. And, when you're thinking about units, so a volt is how many joules per coulomb? So instead of plugging in volts here, we're gonna plug in joules over coulombs so our units will work out properly. So this is positive 1.10 jules per coulomb. Let's do this math. So let's look our units first and let's see what cancels out here. So our moles cancels out. Our charge, coulombs, cancels out to give us joules for our answer. So let's go ahead and
calculate how many joules. Let's do the calculations, we have two times 96,500. And then we're gonna
multiply that by 1.10. So that would give us, this would give us, 212,300 joules. So remember this was in joules. And let's go ahead and
convert that to kilojoules. So 212,000 joules would be 212 kilojoules. And remember we have a negative sign here. So don't forget about this negative sign. This would be negative 212 kilojoules for our value for delta G zero. So the standard change in free energy for this voltaic cell is
negative 212 kilojoules. And we already know that
when delta G is negative, that's a spontaneous reaction. So this is a spontaneous
redox reaction here. Notice that delta G zero and
E zero have opposite signs. We had a positive sign for our voltage and we ended up with a
negative sign for delta G. And that's why we have this
negative in our equation here. So a spontaneous reaction
in a voltaic cell has a positive cell potential, so a positive value for the voltage, but a negative change in free energy, because that indicates
a spontaneous reaction.