Inconsistency in between Nernst Equation and Gibbs Free Energy Equation

Question

Part 1 - Derivation of the Gibbs Free Energy Equation: [copied from this]

Using the fundamental equations for the state function (and its natural variables): \begin{align} \mathrm{d}G &= -S\mathrm{d}T + V\mathrm{d}P\\ V &= \left(\frac{\partial G}{\partial P}\right)_T\\ \bar{G}(T,P_2) &= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p \end{align} Here $\bar{x}$ represents molar $x$, i.e. $x$ per mole \begin{align} \bar{V} &= \frac{RT}{P}\\ \bar{G}(T,P_2) &= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1} \end{align} Defining standard state as $P = \pu{1 bar}$ and $\bar{G}=\mu$ $$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$ consider the general gaseous reaction $\ce{a A + b B -> c C + d D}$ $$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$ for "unit progress" in reaction. Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$ \begin{align} \Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\ \Delta G &= \Delta G^\circ + RT\ln Q \end{align}

Here, $Q=\ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$

Part 2 - Derivation of the Nernst Equation [Copied from this]

Realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$. $$ $$The fact that $\Delta G = W_\text{non-exp}$ can be derived as under: \begin{align} \mathrm{d}S &= \frac{\delta q}{T} && \text{(reversible case)}\\ \mathrm{d}U &= \delta q + \delta W_\text{non-exp} + \delta W_\text{exp}\\ \delta W_\text{non-exp} &= \mathrm{d}U - \delta W_\text{exp} - \delta q \\ &= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S && \text{(const. $p$ and $T$)} \\ &= \mathrm{d}H - T\,\mathrm{d}S \\ &= \mathrm{d}G \end{align}

Combining the above two, we get Nernst Equation: $$E_{cell}=E^{\circ}_{cell}-\frac{RT}{nF}lnQ$$

The Q for this remains same i.e. $Q_p=\ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}$

But we always use $Q_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$ in the Nernst Equation in terms of molarity instead of $Q_p$ (in terms of partial pressure). How is this justified?

Thnx :)

Answer 1

Rather than being an inconsistency, it has to do with the freedom of choosing a mathematical expression for the chemical potential that is more convenient to you.

A general expression for the chemical potential is $$ \mu_j = \mu_j^0 + RT \ln\left(\gamma_j \frac{\xi}{\xi_0}\right) \tag{1} $$ where $\gamma_j$ is the activity coefficient of species $j$, $\xi$ is some measure of the concentration, and $\xi_0$ is that measure of the concentration for a defined standard state.

Across chemistry you will see the next variations of Eq. (1) \begin{align} \mu_j &= \mu_j^{\mathrm{(g)}} + RT\ln\left(\gamma_j^{\mathrm{(g)}} \frac{P_j}{P_0}\right) \tag{2} \\ \mu_j &= \mu_j^{\mathrm{(l)}} + RT\ln\left(\gamma_j^{\mathrm{(l)}} x_j\right) \tag{3} \\ \mu_j &= \mu_j^{\mathrm{(c)}} + RT\ln\left(\gamma_j^{\mathrm{(c)}} \frac{c_j}{c_0}\right) \tag{4} \\ \mu_j &= \mu_j^{\mathrm{(m)}} + RT\ln\left(\gamma_j^{\mathrm{(m)}} \frac{m_j}{m_0}\right) \tag{5} \\ \end{align}

Below is a description:

Standard state	Measure of concentration	Activity coefficient
Pure substance in the ideal-gas state at a pressure of $\pu{1 bar}$	Partial pressure	Fugacity coefficient
Hypothetical ideal liquid solution at a pressure of $\pu{1 bar}$	Molar fraction	Mole-fraction activity coefficient
Hypothetical ideal 1-molar solution at a pressure of $\pu{1 bar}$	Molar concentration	Molarity activity coefficient
Hypothetical ideal 1-molal solution at a pressure of $\pu{1 bar}$	Molality	Molality activity coefficient

You can use any of the Eqs. (2)-(5) for any compound to obtain an expression of $\Delta G$. In all of these cases, the activity coefficient is going to be different depending on the expression you use. Of course, the measure of concentration is going to be different also for every species.

The only precaution you must take is that once you select the 'scale' for each compound, you have to be consistent with it. For example, if an equilibrium is attained you can calculate the equilibrium constant $K = \exp(-\Delta G/RT)$. When explaining in words what $\Delta G$ points too, you tell the reactants and products it refers to and the states you have considered for all of them. As you can see, depending on the different scales, you will have different $K$'s.

When someone reads your $K$ and the scale you used for every compound, that person will put correctly the measure of concentration for further calculations. If you just say $K = 50$ with no additional information, I won't know which $\xi$ to choose for every compound.

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A discussion of this topic can be found at the start of chapter $9$ of:

• J. M. Prausnitz, R. N. Lichtenhaler, E. G. de Azevedo, "Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd ed., Prentice Hall PTR, 1999.