What is Nernst Equation: Derivation, Applications and Examples

For a cell reaction, the Nernst equation is expressed as: $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{RT}{nF}\mathrm{l}\mathrm{n}Q$

where:

• $E_{\text{cell}}$ : Cell potential under non-standard conditions.
• $E_{\text{cell}}^{\circ }$ : Standard cell potential (at ${25}^{\circ }\mathrm{C},1\mathrm{M},1\mathrm{a}\mathrm{t}\mathrm{m}$ ).
• $R$ : Gas constant ( $8.314\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}$ ).
• $T$ : Absolute temperature (K).
• $n$ : Number of electrons transferred.
• F: Faraday's constant (96485 C/mol).
• $Q$ : Reaction quotient.

At ${25}^{\circ }\mathrm{C}$ ( 298 K ), using $\mathrm{l}\mathrm{n}x=2.303\mathrm{l}\mathrm{o}\mathrm{g}x$ , the equation simplifies to: $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\mathrm{l}\mathrm{o}\mathrm{g}Q$

This form is commonly used for JEE Main and NEET problems which is why students must learn them in detail.

The Nernst equation is derived from the Gibbs free energy change ( $\mathrm{\Delta }G$ ) for a cell reaction: $\mathrm{\Delta }G=\mathrm{\Delta }{G}^{\circ }+RT\mathrm{l}\mathrm{n}Q$

Since $\mathrm{\Delta }G=-nF{E}_{\text{cell}}$ and $\mathrm{\Delta }{G}^{\circ }=-nF{E}_{\text{cell}}^{\circ }$ , substituting gives: $-nF{E}_{\text{cell}}=-nF{E}_{\text{cell}}^{\circ }+RT\mathrm{l}\mathrm{n}Q$

Dividing by $-nF$ : $E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{RT}{nF}\mathrm{l}\mathrm{n}Q$

At $298\mathrm{K},\frac{RT}{F}\approx 0.0257$ , and with $2.303\mathrm{l}\mathrm{o}\mathrm{g}x=\mathrm{l}\mathrm{n}x$ , the constant becomes $\frac{2.303\times 0.0257}{n}\approx$ $\frac{0.0591}{n}$ . Nernst equation derivation is important for entrance exams such as IISER and IIT JAM since conceptual questions are asked based on the steps.

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The reaction quotient $Q$ is the ratio of product concentrations to reactant concentrations, each raised to their stoichiometric coefficients, similar to the equilibrium constant. This is crucial for Nernst equation calculations since these are asked in exams like CUET and even CBSE board exam students. For a general cell reaction:

$\begin{array}{c}aA+bB\to cC+dD\\ Q=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}\end{array}$

For gases, partial pressures are used. Solids and pure liquids have activity = 1. 

In galvanic cells, the Nernst equation formula predicts how cell potential varies with concentration. For a Daniell cell ( $\mathrm{Z}\mathrm{n}\left|{\mathrm{Z}\mathrm{n}}^{2+}\right|\left|{\mathrm{C}\mathrm{u}}^{2+}\right|\mathrm{C}\mathrm{u}$ ):

$\begin{array}{c}\mathrm{Z}\mathrm{n}\left(\mathrm{s}\right)+{\mathrm{C}\mathrm{u}}^{2+}\left(\mathrm{a}\mathrm{q}\right)\to {\mathrm{Z}\mathrm{n}}^{2+}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{C}\mathrm{u}\left(\mathrm{s}\right)\\ E_{\text{cell}}^{\circ }=E_{{\mathrm{C}\mathrm{u}}^{2+}/\mathrm{C}\mathrm{u}}^{\circ }-E_{{\mathrm{Z}\mathrm{n}}^{2+}/\mathrm{Z}\mathrm{n}}^{\circ }=0.34-(-0.76)=1.10\mathrm{V}\\ Q=\frac{\left[{\mathrm{Z}\mathrm{n}}^{2+}\right]}{\left[{\mathrm{C}\mathrm{u}}^{2+}\right]},\mathrm{}n=2\\ E_{\text{cell}}=1.10-\frac{0.0591}{2}\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[{\mathrm{Z}\mathrm{n}}^{2+}\right]}{\left[{\mathrm{C}\mathrm{u}}^{2+}\right]}\end{array}$

 

Concentration cells generate potential due to differences in electrolyte concentrations. For a Zn $\mid {\mathrm{Z}\mathrm{n}}^{2+}\left(C_1\right)‖Z{n}^{2+}\left(C_2\right)‖$ Zncell : $E_{\text{cell}}^{\circ }=0$ (same electrodes)

$\begin{array}{c}{E}_{\text{cell}}=-\frac{0.0591}{n}\mathrm{l}\mathrm{o}\mathrm{g}\frac{C_2}{C_1}\mathrm{}\left(C_2>C_1,n=2\right)\\ E_{\text{cell}}=\frac{0.0591}{2}\mathrm{l}\mathrm{o}\mathrm{g}\frac{C_1}{C_2}\end{array}$

Electrons flow from the lower concentration (anode) to the higher concentration (cathode). 

At equilibrium, $E_{\text{cell}}=0$ , and $Q=K$ (equilibrium constant). The Nernst equation becomes:

$\begin{array}{c}0=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\mathrm{l}\mathrm{o}\mathrm{g}K\\ \mathrm{l}\mathrm{o}\mathrm{g}K=\frac{n{E}_{\text{cell}}^{\circ }}{0.0591}\end{array}$

This relates cell potential to equilibrium constants which is a key concept for JEE Main.

Problem: Calculate the potential of the cell $\mathrm{A}\mathrm{g}\left(\mathrm{s}\right)\left|{\mathrm{A}\mathrm{g}}^{+}\left(0.01\mathrm{M}\right)\right|\left|{\mathrm{C}\mathrm{u}}^{2+}\left(0.1\mathrm{M}\right)\right|\mathrm{C}\mathrm{u}\left(\mathrm{s}\right)$ at ${25}^{\circ }\mathrm{C}$ . Given ${\mathrm{E}}_{{\mathrm{A}\mathrm{g}}^{+}/\mathrm{A}\mathrm{g}}^{\circ }=+0.80\mathrm{V},{\mathrm{E}}_{{\mathrm{C}\mathrm{u}}^{2+}/\mathrm{C}\mathrm{u}}^{\circ }=+0.34\mathrm{V}$ .

Solution: $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=0.34-0.80=-0.46\mathrm{V}$

Since $E_{\text{cell}}^{\circ }$ is negative, reverse the cell: $\mathrm{C}\mathrm{u}\left|{\mathrm{C}\mathrm{u}}^{2+}\left(0.1\mathrm{M}\right)\right|\left|{\mathrm{A}\mathrm{g}}^{+}\left(0.01\mathrm{M}\right)\right|\mathrm{A}\mathrm{g}$ .

$E_{\text{cell}}^{\circ }=0.80-0.34=0.46\mathrm{V}$

Reaction: $\mathrm{C}\mathrm{u}\left(\mathrm{s}\right)+2{\mathrm{A}\mathrm{g}}^{+}\left(\mathrm{a}\mathrm{q}\right)\to {\mathrm{C}\mathrm{u}}^{2+}\left(\mathrm{a}\mathrm{q}\right)+2\mathrm{A}\mathrm{g}\left(\mathrm{s}\right),n=2$ .

$\begin{array}{c}Q=\frac{\left[{\mathrm{C}\mathrm{u}}^{2+}\right]}{{\left[{\mathrm{A}\mathrm{g}}^{+}\right]}^2}=\frac{0.1}{(0.01{)}^2}=1000\\ E_{\text{cell}}=0.46-\frac{0.0591}{2}\mathrm{l}\mathrm{o}\mathrm{g}1000=0.46-0.0886=0.3714\mathrm{V}\end{array}$

This tests Nernst equation and cell spontaneity, common in JEE Main.