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According to Chemistry Libretexts, the Gibbs free energy change of a cell reaction is related to the cell voltage as

$$ \Delta G = -nFE,$$

where $n$ is the number of moles of electrons passed and $F$ is the charge on a mole of electrons.

But $E$ decreases in value through the course of the cell reaction, so shouldn't $$\Delta G = -\int_{q_\mathrm{i}}^{q_\mathrm{f}} E\,\mathrm dq,$$

where $q_\mathrm{f} - q_\mathrm{i} = nF?$

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    $\begingroup$ It's strange to see a question like that from Michael Faraday. $\endgroup$
    – andselisk
    Jul 17, 2020 at 10:34

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According to Nernst's equation,

$$E = E^o - \frac{RT}{nF}\ln Q$$

As you can see, $E$ is independent of $q$, therefore your proposed integral becomes,

$$\Delta G = - \int_{q_i}^{q_f}{E\cdot dq} = -E\int_{q_i}^{q_f}{dq} = -E\Delta q = -nFE$$


Alternatively, you can prove the the relation between Gibbs energy and cell potential as,

  • Multiply Nernst's equation by $nF$, we get,

$$nFE = nFE^o - RT\ln Q$$

  • Swapping the sides, we get,

$$-nFE = -nFE^o + RT\ln Q$$

  • As we know,

$$\Delta G = \Delta G^o + RT\ln Q$$

$$\Rightarrow \Delta G = -nFE$$

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  • $\begingroup$ so you are saying that the concentration "Q" is completely unrelated to the charge? $\endgroup$
    – Babu
    Jul 17, 2020 at 16:43
  • $\begingroup$ But 'q' is the amount of charged passed from the cathode to the anode of the electrolytic cell. And for charge to pass, a species must be reduced at the cathode and a species must be oxidized at the anode. The concentration of the species in the chemical equation changes, and hence Q is related with 'q'.. And if Q is related with q, then E is also related with q (from Nernst equation).. Correct me if I'm wrong.. $\endgroup$ Jul 17, 2020 at 16:52
  • $\begingroup$ Any chemical reaction follows some "nth order kinetics", so the conc comes out to be a function of time, therefore Q should also be a function of time. Now, the integral is with respect to charge passed, hence E comes out. $\endgroup$ Jul 17, 2020 at 17:25
  • $\begingroup$ @Rahul Verma if Q is a function of time, then by nernst equation, E is a function of time. Since it requires finite time to pass charge, E would vary as the charge is being passed. So E cannot be taken out of the integral since it's not constant. $\endgroup$ Jul 17, 2020 at 18:35
  • $\begingroup$ @MichaelFaraday: as you know, ∆G is a state fn, so I think the integral isn't required. It is because ∆H, ∆S and T are state / point fn $\endgroup$ Jul 18, 2020 at 6:45

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