Nernst equation

In 1887, Walther Nernst, a German chemist, developed the Nernst equation, which describes the relationship between an electrochemical reaction’s potential and its standard electrode potential under standard and non-standard conditions.

The derivation of the Nernst equation involves the following steps:

1. Derive the formula for the reversible open circuit potential , E^o, of an electrochemical cell at constant temperature and pressure under standard conditions.
2. Derive the expression for E^o in terms of chemical potentials of the reaction species at constant temperature and pressure.
3. Derive a ‘correction factor’ for E^oin step2. This final expression, the Nernst equation, describes how the resultant E^ovaries with activities of chemical species.

Step 1

From the definition of enthalpy, H = U + pV, we have

$dH=dU+d(pV)\; \; \; \; \; \; \; \; 1$

Substitute the definition of the internal energy of a system undergoing reversible change, dU = dq_rev+ dw_rev in eq1

$dH=dq_{rev}+dw_{rev}+d(pV)\; \; \; \; \; \; \; \;2$

Substitute the definition of the change in Gibbs energy of a system, dG = dH – d(TS) in eq2

$dG+TdS+SdT=dq_{rev}+dw_{rev}+pdV+Vdp$

At constant temperature and pressure, dT = dp = 0

$dG=dq_{rev}+dw_{rev}+pdV-TdS\; \; \; \; \; \; \; \; 3$

Substitute the definition of the change in entropy of a system, dS = dq_rev/T in eq3

$dG=dw_{rev}+pdV\; \; \; \; \; \; \; \; 4$

Substitute the definition of total reversible work, w_rev = w_ex+ w_add, where w_ex is reversible expansion work and w_add is reversible additional work other than reversible expansion work, in eq4

$dG=dw_{ex}+dw_{add}+pdV\; \; \; \; \; \; \; \; 5$

Substitute the definition of the change in reversible expansion work, dw_ex= –pdV in eq5

$dG=dw_{add}\; \; \; \; \; \; \; \; 6$

Substitute the definition of the reversible electrical work between two points in an electric circuit, dw_add= –nFEdξ in eq6

$\Delta _rG=-nFE\; \; \; \; \; \; \; \; 7$

where n is the number of moles of electrons, F is the Faraday constant, E is the open circuit potential of the electrochemical cell and Δ_rG = dG/dξ is the Gibbs energy of the electrochemical cell reaction.

When the open circuit potential of an electrochemical cell is measured at standard conditions (1 bar, 298K, 1M), we write eq7 as:

$\Delta _rG^{\, o}=-nFE^{\, o}\; \; \; \; \; \; \; \; 8$

Step 2

The chemical potential of the j-th species, in a multi-component system is

$\mu_j=\left (\frac{\partial G}{\partial n_j} \right )_{T,p,n'}\; \; \; \; \; \; \; 9$

where n‘ is all n other than nj.

The total differential of the Gibbs energy of the system at constant temperature and pressure is

$dG=\left (\frac{\partial G}{\partial n_a} \right )_{T,p,n'}dn_a+\left (\frac{\partial G}{\partial n_b} \right )_{T,p,n'}dn_b+...\; \; \; \; \; \; \; 10$

Substituting eq9 in eq10

$dG=\mu_adn_a+\mu_bdn_b+...=\sum_{j}\mu_jdn_j\; \; \; \; \; \; \; \; 11$

Eq11 can be expressed in another way by considering the reversible reaction

$v_aa+v_bb+...\rightleftharpoons v_{\alpha}\alpha+v_{\beta}\beta+...$

with the change in Gibbs energy at constant temperature and pressure of the system given by eq11. Next, let’s define

$dn_j=v_jd\xi\; \; \; \; \; \; \; \; 12$

where vj is the stoichiometric number of the j-th species in the reversible reaction and dξ is the amount of substance that is being changed in the reaction.

ξ is called the extent of a reaction and has units of amount in moles. Note that the stoichiometric numbers for reactants are negative by convention and those for products are positive. Substituting eq12 in eq11

$\left ( \frac{\partial G}{\partial \xi} \right )_{T,p}= \sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 13$

Substitute the definition of the Gibbs energy of an electrochemical cell reaction, $\Delta _rG=\left ( \frac{\partial G}{\partial \xi} \right )_{T,p}$ , into eq13

$\Delta _rG=\sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 14$

Substitute eq7 from Step1 in eq14

$E=-\frac{1}{nF}\sum_{j}\mu_jv_j\; \; \; \; \; \; \; \; 15$

When the open circuit potential of an electrochemical cell is measured at standard conditions (1 bar, 298K, 1M), we write eq15 as:

$E^{\, o}=-\frac{1}{nF}\sum_{j}\mu^o_{\, j}v_j\; \; \; \; \; \; \; \; 16$

Step 3

Firstly, let’s consider a system with a single pure substance. Substitute the definition of the change in Gibbs energy of a system, dG = dH – d(TS), in the change of enthalpy of the system dH = dU + d(pV)

$dG+SdT+TdS=dU+pdV+Vdp\; \; \; \; \; \; \; \; 17$

The reversible change in internal energy of a pure substance (a system with a single pure substance only does expansion work) can be expressed as

$dU=dq_{rev}+dw_{ex}\; \; \; \; \; \; \; \; 18$

Substitute the definitions of entropy, dS = dq_rev/T, and expansion work, –pdV, in eq18

$dU=TdS-pdV\; \; \; \; \; \; \; \; 19$

Substitute eq19 in eq17

$dG=Vdp-SdT\; \; \; \; \; \; \; \; 20$

At constant temperature, dT = 0 and eq20 becomes dG = Vdp. Integrating this new expression on both sides from $p_i$ to $p_f$ ,

$G\left ( p_f \right )=G\left ( p_i \right )+nRTln\frac{p_f}{p_i}$

Let p_i = 1 bar = p^o(i.e. standard conditions) and p_f = p

$G=G^{\, o}+nRTln\frac{p}{p^{\, o}}$

Secondly, we extend the above working to a multi-component system of ideal gases. Since G is an extensive property, we add the Gibbs energies of all components to give the total Gibbs energy of the system:

$G_{total}=\left ( G_a^{\: o}+n_aRTln\frac{p_a}{p_a^{\, o}} \right )+\left ( G_b^{\: o}+n_bRTln\frac{p_b}{p_b^{\, o}} \right )+...$

Let’s assume the reference pressures of the component gases are the same since they are all ideal gases, i.e. p_a^o= p_b^o= p^o

$G_{total}=\left ( G_a^{\, o}+G_b^{\, o}+...\right )+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )$

$G_{total}=G_{total}^{\, o}+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )$

For simplicity, let G_Totaland G^o_Total be G and G^orespectively.

$G=G^{\, o}+\left ( n_aRTln\frac{p_a}{p^o}+n_bRTln\frac{p_b}{p^o}+...\right )$

Taking the partial derivative of the above with respect to n_j, at constant temperature, pressure and the amount of the other components, n‘, we have

$\left ( \frac{\partial G}{\partial n_j} \right )_{T,p,n'}=\left ( \frac{\partial G^{\, o}}{\partial n_j} \right )_{T,p,n'}+\left [ \frac{\partial \left ( n_aRTln\frac{p_a}{p^{\, o}}+ n_bRTln\frac{p_b}{p^{\, o}}+...\right )}{\partial n_j} \right ]_{T,p,n'}$

For the last partial derivative of the above equation, only the j-th term survives.

$\left ( \frac{\partial G}{\partial n_j} \right )_{T,p,n'}=\left ( \frac{\partial G^{\, o}}{\partial n_j} \right )_{T,p,n'}+RTln\frac{p_j}{p^{\, o}}\; \; \; \; \; \; \; \; 21$

Substitute eq9 from Step 2 in eq21

$\mu_j=\mu_j^{\, o}+RTln\frac{p_j}{p^{\, o}}\; \; \; \; \; \; \; \; 22$

Similarly, for a solute in a dilute solution satisfying Henry’s law, we can write

$\mu_j=\mu_j^{\, o}+RTln\frac{c_j}{c^{\, o}}\; \; \; \; \; \; \; \; 23$

where c_jand c^oare the j-th solute’s concentration and molar concentration respectively.

We can fine-tune eq22 and eq23 by replacing p_j/p^oand c_j/c^owith γ_j(p_j/p^o) and γ_j(c_j/c^o) where γ_j is a factor called the activity coefficient that can account for both ideal and non-ideal fluids (γ = 1 and γ < 1 for ideal and non-ideal fluids respectively). Next, we combine the modified equations into a single equation by letting a_j= γ_j(p_j/p^o) and a_j= γ_j(c_j/c^o), where a_j is the activity of species j.

$\mu_j=\mu_j^{\, o}+RTlna_j\; \; \; \; \; \; \; \; 24$

Substituting eq24 in eq15 from Step 2,

$E=-\frac{1}{nF}\sum_{j}\left ( \mu_j^{\, o}+RTlna_j \right )v_j=-\frac{1}{nF}\sum_{j} \mu_j^{\, o}v_j-\frac{RT}{nF}\sum_{j}v_jlna_j \; \; \; \; \; \; \; \; 25$