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For example let us take the reaction:

Zn(s) ---> Zn^2+(aq) + 2e^-(aq) ;ΔG°, which we are considering as our half cell

And now suppose, by taking the value of E° of SHE(Standard Hydrogen Electrode) as zero and using it as reference, we have measured the E° of the zinc electrode. But clearly this value is relative to SHE and not the absolute value.

Now we have the equation ΔG°= -nFE° , relating the change in Gibbs energy of the half cell reaction and the value of E°. The change in standard Gibbs energy of a reaction is constant at a constant temperature and has nothing to do with the chosen reference electrode so how can we relate both the values?

My another doubt is why even bother defining a reference electrode if we can find the electrode potentials of a half cell with its change in gibbs energy. Or is it the other way around, that we calculate the value of ΔG based on the measured value of E.

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    Commented Oct 2, 2023 at 11:21
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2 Answers 2

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If we use the tabelated reduction potential $E°$ of a half-reaction in the equation $ΔG° = -nFE°$ the we implicitly assume the reaction $\ce{Ox(aq) + $\frac n2$ H2(g) -> Red(aq) + n H+(aq)}$.

If other than SHE half reaction had been chosen as the conventional potential reference, the respective implicit reaction would have been different, with different $ΔG°$ and different $E°$ for both the reaction and the zinc half reaction.

If we consider two general half-reactions then $Δ_\text{r}G{^\circ} = -nF(E{^\circ}_\text{cathode} - E{^\circ}_\text{anode} )$

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  • $\begingroup$ So does that mean when i refer to the value of ΔG° of the reaction that i mentioned above , what is actually being implied is that its the ΔG° value for the reaction Zn(s) + 2H+(aq) ---> Zn2+(aq) + H2(g) ? $\endgroup$
    – Anish reddy
    Commented Oct 2, 2023 at 10:53
  • $\begingroup$ It is, if the E° used in the equation is the E° for Zn^2+ + 2- -> Zn(s) half-reaction. $\endgroup$
    – Poutnik
    Commented Oct 2, 2023 at 11:05
  • $\begingroup$ Thank you! that fully cleared my doubt $\endgroup$
    – Anish reddy
    Commented Oct 2, 2023 at 11:40
  • $\begingroup$ Oops, I have overlooks the direction you have used. It is the opposite direction, H2 + Zn^2+ -> 2 H+ + Zn. E°Zn/Zn^2+ < 0, ΔG > 0 so your directions is not spontaneous reaction. The opposite one is. $\endgroup$
    – Poutnik
    Commented Oct 2, 2023 at 11:43
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You can’t calculate the Gibbs energy for a half cell, just like you can’t do a redox reaction with a half cell, only. In the formula $$ΔG°= -n F E°$$ the $E°$ refers to the cell potential, not the half cell reduction potential. The former is independent of a reference electrode, while the latter is not.

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