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 :)

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    $\begingroup$ In the nernst equation by design all gases are described with partial pressures and aqueous ions by concentrations, it's mixed $\endgroup$ Commented Jan 22 at 19:15
  • $\begingroup$ @GauravSaiMaddipati any proofs for the aqueous ions using concentrations ?! $\endgroup$ Commented Jan 22 at 21:50
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    $\begingroup$ A substance is its chemical activity. Two factors are involved1. the internal energy of the substance This is contained in the number of molecules and the temperature. The second factor is the entropy or dispersion of the substance . This is best measured in mole fraction in gases, liquids and solids, no contradictions there. In liquids molarity is a reasonable substitute for mole fraction simply substituting a volume or a mass. The stand activity is defined as 1. Molar. For pure solids, liquids and gases the standard is mole fraction of one. At high concentrations things get complicated. $\endgroup$
    – jimchmst
    Commented Jan 22 at 23:20
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    $\begingroup$ Im sure there is a proof, but i have asked this question to my prof and this is what he replied, i didn't pursue it further. Im sure someone much more knowledgeable than I could give a better answer hence i just left a comment $\endgroup$ Commented Jan 23 at 6:55

1 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.

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.
  • $\begingroup$ I went through the book you referred and found the Fugacity Equation [Equation $(2)$ in your answer] is derived for Ideal Gases in ($2-37$) and generalised for all systems in ($2-40$) in the book. But I can't find the proofs for Equations $(3)$ to $(5)$ in your answer. Kindly tell me where to find! $\endgroup$ Commented Jan 24 at 8:16
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    $\begingroup$ @AsmitKarmakar You can't prove those equations, although the first one seems to be one. They are definitions. For example, look in Introduction to Chemical Engineering Thermodynamics (Smith, Van Ness, and Abbott, 7th edition) in Chapter 11.8. It says explicitly that $\mu_j$ for the ideal liquid solution model is defined (in bold). Then, he defines the Eq. (11.90) and arrives at Eq. (11.92) which is the Eq. (4) in this post. $\endgroup$ Commented Jan 24 at 10:30
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    $\begingroup$ For Eq. (5) you may see the part named "Activities" in Physical Chemistry (Peter Atkins and Julio de Paula, 9th edition). It says "It proves convenient to write", and then pops out my equation for the ideal case (where $\gamma_j = 1$). Then he defines the activity in Eq. (5.61) (they use brackets [] in the book for definitions), and arrives finally at my Eq. (5). $\endgroup$ Commented Jan 24 at 10:39
  • $\begingroup$ Since $\mu_{j}$ is the molar Gibbs Free Energy, definitions of $\mu_{j}$ are equivalently definitions of Gibbs Free Energy. But Gibbs Free Energy should be the same everywhere! Or is it not? Is $G=H-TS$ valid only for Gaseous State? Nevertheless, how can there be three alternative definitions of $G$ for solutions? (Equations $(3), (4) & (5)$) $\endgroup$ Commented Jan 24 at 11:51
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    $\begingroup$ The definition of $G$ is what you have written. We have several ways to write the chemical potential of a species in a solution $\mu_j$. The one you read in books is derived for a solution of ideal gases. Then, you add the fugacity coefficient (arbitrarily) to extend them to real solutions. All the other forms are "copies" of this one, which are more suitable for other states of aggregation. $\endgroup$ Commented Jan 24 at 12:01

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