What is the meaning of thermodynamic tendency?

While giving a test, I encountered a question which was as follows

Lead reduces $$\ce{NO_3^-}$$ into $$\ce{NO}$$ and $$\ce{NO_2}$$ depending on the concentration of $$\ce{HNO_3}$$ in solution. Assuming that $$\ce{[Pb^{2+}]} = \pu{0.1 M}$$ and $$p_\ce{NO} = p_\ce{NO2} = \pu{0.001 bar}$$ and use the following data:

\begin{align} E^\circ_\ce{Pb^2+/Pb} &= \pu{-0.13 V}\\ E^\circ_\ce{NO3^-/NO} &= \pu{0.96 V}\\ E^\circ_\ce{NO3^-/NO2} &= \pu{0.79 V}\\ \end{align} and at $$\pu{298K}$$ $$\frac{(2.303)RT}{F} = \pu{0.06 V}$$, find the concentration of $$\ce{HNO3}$$ at which thermodynamic tendency for reduction of $$\ce{NO3^-}$$ into $$\ce{NO}$$ and $$\ce{NO2}$$ by lead is the same.

In this question what is the meaning of thermodynamic tendency? Is it represented by $$E_\mathrm{cell}$$ or $$\Delta G$$? Quoting a reliable source would help.

The answer is $$\pu{10^{0.625} M}$$ which comes by equating the $$E_\mathrm{cell}$$ of the reactions. My intuition told me to equate $$\Delta G$$ but it seems I am wrong.

• How is equating Ecell different from equating ΔG? Aug 7, 2021 at 9:56
• The number of electron that are involved in the reaction are different hence equating $\ce{E_{cell}}$ and equating $\ce{\Delta{G}}$ are two different things. Aug 7, 2021 at 10:32
• Guys have tendency to be attracted to ( in their eyes ) the most beautiful girl regardless of the success. Similarly, TD systems have tendency to reach the state with the minimal Gibbs energy, regardless of if they reach it ( e.g. due kinetic reasons ). Diamonds have TD tendency to become graphite, staying being diamonds for billions years. Aug 7, 2021 at 15:25
• I am unable to understand that what are you wanting to imply? Aug 7, 2021 at 19:52
• The Hess law is applicable on $\Delta G_\mathrm{halfr}$ ( implicitly referring to SHE as the 2nd half reaction), or on its equivalent $-nF.E_\mathrm{redox}$. // As $\Delta U = q \cdot \Delta E$ (energy, charge, potential difference ) Aug 12, 2021 at 9:53

For solving this problem, the $$\Delta G$$ values cannot be compared, or equalized, because they do not correspond to the same equation. Both reactions start from the same products, $$\ce{Pb + NO3^-}$$. But the first reaction leads to $$\ce{NO}$$ and the other one to $$\ce{NO2}$$, and these two nitrogen oxides do not have the same $$G^\circ_\mathrm{form}$$ values. So the Gibbs energy values are no use for solving this problem. It is much better to compare potentials through the Nernst equation.
• I am not able to understand that if the $\Delta{G^o_form}$ values are different then why can't we compare the $\Delta{G}$ of both the reactions? Also then, under what circumstances the $\Delta{G}$ values of two reactions can be compared? Aug 7, 2021 at 13:04