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### Course: Chemistry archive>Unit 9

Lesson 5: Cell potentials under nonstandard conditions

# Galvanic cells and changes in free energy

Relationship between Gibbs free energy, reaction quotient Q, and cell voltage.

## Want to join the conversation?

• Where does the K= 1.58*10^37 come from?
• You can use ΔG° = -RT ln(Q) to find it
since we have:
ΔG° = -212 x 10^3 J
R = 8.314 J/Kmol
T = 298K

Although i got K = 1.45 x 10^37
(probably has to do withe rounding of ΔG°, it cant be nicely 212 KJ)

Otherwise, K is typically determined experimentally.
(1 vote)
• at , how do I know the number of electrons transferred? I'd like to solve for E myself to see that it's 0.98V but I don't know where to get n (number of electrons) from.
• n is the number of electrons in your half-cell reactions. In this case, n=2
• I am missing a whole bunch! I didn't understand a thing about what's going on here. So I went to the next section, Cell Potentials, which seems to give the foundation for understanding this video. So I'll come back to this one, later.

But, is it out of sequence? Should it be in the next section?
• If I'm ever lost in the sequence of the MCAT tutorials, I switch over to the physics/chem/org sections and usually find the missing videos.
• Can we consider the free energy here as the energy released due to the redox reaction and that it's utilized in transferring the electrons?
I just want to relate this to voltage = work/ q.
That in delta G= -nFE°, nF gives us the quantity of charge and E° is voltage. When we multiply these we get energy or work done to transfer charges.
• So does Q = K, the equilibrium constant in the last videos?
(1 vote)
• K is the expression of [products]/[reactants] at equilibrium.
Q is the same expression but at whatever time point you are looking at, let's say 4 minutes into a reaction.
If Q=K, the reaction is at equilibrium.
If Q<K, there are more reactants than at equilibrium so the reaction toward products will be favored.
If Q>K, there are more products than at equilibrium so the reverse reaction will be favored.
• Why do you always do the math with a calculator? Aren't these for MCAT prep? Anyways, any help on calculating ln 10,000 w/o a calculator?
(1 vote)
• These lectures are prepared to help students from around the world who are preparing to sit a range of different examinations. They're not specifically aimed at MCAT.
• At , why do we leave copper and zinc?
• Is Delta G a state variable or a difference between the state variables of two states? If so, what states?
• At why did delta G change when K increased such that delta G was no longer equal to Delta G naught? Wasn't the temperature and pressure kept the same? What change was made such that we were no longer at standard conditions?
• You should differentiate btw ΔG°and ΔG:

ΔG° helps to determine the work/energy (or Voltage) provided by the reaction to reach equilibrium under standard conditions ([X]= 1M; T=25 °C; P=1 atm). It is not influenced by the state or at what moment is the reaction, but solely provides information on the maximum released free energy or the maximum potential difference between the initial and final states (equilibrium) of the reaction. This is why ΔG° is only dependent to Keq (but always under standard conditions mentioned above).

On the other side, ΔG helps to identify the work (or Voltage) that must be provided by the reaction at a specific moment. For instance, at t=0, ΔG is at its maximum because the spontaneous free energy that should be released by the reaction is still awaiting the reaction to commence. Therefore, ΔG refers here to this unliberated energy (Similarly, at t=0 there is a maximum voltage/potential between reactions and products and the system (reaction) is awaiting to start to decrease this potential difference and achieve equilibrium. ΔG is identified by this maximum potential difference at this point.

At an instant t<tequilibrium, the reaction progresses, and some products are formed, causing the potential difference to decrease. This is why when Q= 10 000 in the video, the potential energy drops from its maximum at the beginning of the reaction (E0=1.1v) to E= 0.98v at Qt<Keq; at this point, the energy supplied by the reaction to attain equilibrium is reduced, and ΔG also decreases.

Once the potential difference has disappeared, all the spontaneous free energy has been released, and the reaction no longer possesses any electromotive force to alter its state. At this point, the reaction has reached equilibrium, and ΔG simply equals 0.
Hope this helps to leave your ambiguity.
(1 vote)
• how can the Q be increased from 1 to 10000?