Nernst equation

What is the Nernst equation?

Scope of the equations: the Nernst equation gives electrode potentials and the emf, the cell potential when no current is drawn (NCERT page 34), not the terminal voltage while a substantial current flows (reasoned). Beyond the NCERT text: strictly, the equation uses activities, effective concentrations, and a gas enters as its partial pressure divided by 1 bar (Wikipedia, “Standard hydrogen electrode”). Below, as in NCERT, brackets mean concentrations in mol L⁻¹, and liquid-junction potentials at the salt bridge are ignored (reasoned).

In plain words: a Daniell cell with 1 M solutions starts at 1.10 V, but as it runs, zinc ions pile up, copper ions are used up, and the voltage slowly falls. The Nernst equation gives the voltage at every stage. When it reaches zero, the reaction is at equilibrium. That is why the standard emf, read on a voltmeter, gives the equilibrium constant and the Gibbs energy of the reaction.

Nernst equation for an electrode

For the electrode reaction Mⁿ⁺(aq) + ne⁻ → M(s), the electrode potential at any concentration, measured against the standard hydrogen electrode, is E = E^⊖ − (RT/nF) ln([M]/[Mⁿ⁺]). The concentration of the solid M is taken as unity (strictly, a pure solid has unit activity; reasoned), which gives NCERT Eq. (2.8), page 36:

• R is the gas constant, 8.314 J K⁻¹ mol⁻¹; T is the temperature in kelvin; F is the Faraday constant, the charge on one mole of electrons (NCERT pages 36 and 52).
• NCERT gives F = 96 487 C mol⁻¹, an older value. Since 2019 it has an exact value, the elementary charge times the Avogadro constant: 96 485.33 C mol⁻¹ to two decimals (NIST, 2022 CODATA recommended values). The difference changes no answer on this page. For rough work NCERT uses 96 500 C mol⁻¹ (page 52).
• At 298 K, 2.303RT/F = 2.303 × 8.314 × 298/96 487 = 0.0591 V, rounded by NCERT to 0.059 V (NCERT page 38). So at 298 K, E = E^⊖ + (0.059/n) log [Mⁿ⁺] (reasoned from NCERT Eq. (2.8)).
• What it means. The more concentrated the metal ions, the more positive the electrode potential: more ions are pushing to be reduced. Each tenfold rise in [Mⁿ⁺] adds 0.059/n V.
• The hydrogen electrode. For H⁺(aq) + e⁻ → ½H₂(g) with hydrogen at 1 bar, n = 1 and E^⊖ = 0, so E = 0.059 log [H⁺] = −0.059 pH at 298 K (reasoned from NCERT Eq. (2.13); NCERT Intext Question 2.4 asks for it). If the hydrogen is not at 1 bar, each tenfold rise in its pressure lowers E by 0.0295 V (reasoned).

Nernst equation for a cell

• The Daniell cell. Writing NCERT Eq. (2.8) for the copper cathode and the zinc anode and subtracting, E_cell = E_cathode − E_anode, gives NCERT Eq. (2.11), page 38:

  E_cell = E^⊖_cell − (RT/2F) ln([Zn²⁺]/[Cu²⁺])

• So E_cell increases as [Cu²⁺] increases and as [Zn²⁺] decreases (NCERT page 38).

• At 298 K, with base-10 logarithms, this becomes E_cell = E^⊖_cell − (0.059/2) log([Zn²⁺]/[Cu²⁺]) (NCERT Eq. (2.12), page 38).

• NCERT slip (page 38): in the step just before NCERT Eq. (2.11), the bracket that should enclose both logarithms is missing, so as printed RT/2F multiplies only the first one. Eq. (2.11) itself is right.

• Balance the overall electron transfer. Each separate half-cell equation uses its own electron count. After balancing and combining half-reactions, use the total number cancelled for the whole cell: For Ni(s)|Ni²⁺(aq)||Ag⁺(aq)|Ag, the reaction is Ni + 2Ag⁺ → Ni²⁺ + 2Ag, so n = 2 and Q = [Ni²⁺]/[Ag⁺]² (NCERT page 38).

• In general, for aA + bB → cC + dD with n electrons transferred, NCERT Eq. (2.13), page 38, is:

• Here Q = [C]^c[D]^d/([A]^a[B]^b), the reaction quotient, with products on top. Pure solids are left out, as their concentration is taken as unity (NCERT page 36). Water as the solvent is left out too, and a gas enters as its pressure in bar (reasoned; NCERT’s hydrogen electrode uses 1 bar, page 35; see the scope note above). Strictly, pure solids and liquids have unit activity, water can be left out only while its activity is close to one, as in dilute solution, and pressure in bar stands for the ratio to 1 bar (reasoned).
• At 298 K, 2.303RT/F = 0.059 V, so for the cell, with E = E_cell and E^⊖ = E^⊖_cell:

Equilibrium constant from Nernst equation

• Close the circuit of a Daniell cell and the reaction Zn + Cu²⁺ → Zn²⁺ + Cu runs: [Zn²⁺] rises, [Cu²⁺] falls, and the voltmeter reading falls. After some time the concentrations stop changing and the voltmeter reads zero (NCERT page 39). Strictly, zero emf means equilibrium only for a working cell measured with no current drawn; a real cell can also stop through depletion or a failed electrode (reasoned). In NCERT’s words:

This indicates that equilibrium has been attained.

NCERT Chemistry Part I (Class 12), Section 2.3.1, page 39

• At equilibrium E_cell = 0 and Q = K_c, so the Nernst equation gives NCERT Eq. (2.14), page 39:

• For the Daniell cell at 298 K: 1.1 V = (0.059 V/2) log K_c, so log K_c = (1.1 × 2)/0.059 = 37.288 and K_c = 2 × 10³⁷ (NCERT page 39). The reaction goes essentially to completion. With 0.0591 V, log K_c = 2.2/0.0591 = 37.225 and K_c = 1.7 × 10³⁷: a small rounding moves K a lot, so use the constants the question gives.
• So equilibrium constants that are difficult to measure otherwise can be calculated from E^⊖_cell (NCERT page 39). A big K comes from only a modest emf, because E^⊖ depends on log K: each 0.059/n V multiplies K by ten (reasoned).

Electrochemical cell and Gibbs energy of the reaction

• NCERT slip (page 40): the first sentence of NCERT Section 2.3.2 says the “electrical work done in one second” equals the potential multiplied by the total charge passed. That holds only if the charge is the charge passed in that one second; work per second is power, potential × current. The equation that follows, Δ_rG = −nFE_cell, uses the whole charge nF, so read the sentence as: the electrical work equals potential × total charge passed (reasoned). In a problem, multiply the emf by the total charge passed.
• In NCERT’s words:

If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.

NCERT Chemistry Part I (Class 12), Section 2.3.2, page 40

• At constant temperature and pressure, the maximum (reversible) electrical work a galvanic cell can do equals the decrease in its Gibbs energy. With emf E and charge nF passed per mole of reaction, Δ_rG = −nFE_cell (NCERT Eq. (2.15), page 40).
• E is intensive, Δ_rG is extensive. E_cell is an intensive parameter, but Δ_rG is an extensive thermodynamic property, and its value depends on n (NCERT page 40). Strictly, Δ_rG is Gibbs energy per mole of reaction as written: it scales when the written equation is multiplied, not when a larger sample is used, so it is not extensive in the usual sense: NCERT’s point is that its value depends on n, on how the equation is written (reasoned). In an exam, give NCERT’s pair: E intensive, Δ_rG extensive. For Zn + Cu²⁺ → Zn²⁺ + Cu, Δ_rG = −2FE_cell; written as 2Zn + 2Cu²⁺ → 2Zn²⁺ + 2Cu, Δ_rG = −4FE_cell, with the same E_cell (NCERT page 40).
• If the concentration of every reacting species is unity, E_cell = E^⊖_cell (NCERT Eq. (2.16), page 40):

• From Δ_rG^⊖ the equilibrium constant follows by Δ_rG^⊖ = −RT ln K (NCERT page 40). Putting the two together gives Eq. (2.14) again.
• Signs. A positive E^⊖_cell means a negative Δ_rG^⊖ and K > 1: the cell reaction is spontaneous as written under standard conditions (reasoned from NCERT Eqs. (2.14) and (2.16)).

Solved examples

Common mistakes to avoid

Quick revision summary

• Nernst equation: E(cell) = E⊖(cell) − (RT/nF) ln Q; at 298 K, E(cell) = E⊖(cell) − (0.059/n) log Q.
• For an electrode, E = E^⊖ − (RT/nF) ln(1/[Mⁿ⁺]): more ions, more positive potential.
• Daniell cell: E_cell rises with [Cu²⁺] and falls with [Zn²⁺].
• At equilibrium, E(cell) = 0 and Q = Kc, so E⊖(cell) = (0.059/n) log Kc at 298 K; for the Daniell cell Kc ≈ 2 × 10³⁷.
• Gibbs energy: ΔrG⊖ = −nFE⊖(cell); for the Daniell cell about −212 kJ mol⁻¹.
• E is intensive, Δ_rG extensive: E is unchanged by rescaling the equation; n, Δ_rG and ln K scale with it.

Questions students ask

How is the Nernst equation derived from Gibbs energy?

The Nernst equation follows from the Gibbs energy of a reaction mixture, Δ_rG = Δ_rG^⊖ + RT ln Q (NCERT Class 11 Chemistry, Eq. (6.21), page 184): putting in Δ_rG = −nFE_cell and Δ_rG^⊖ = −nFE^⊖_cell (NCERT Eq. (2.15) and Eq. (2.16), page 40) and dividing by −nF gives E_cell = E^⊖_cell − (RT/nF) ln Q (reasoned). That is why the equation is logarithmic: concentrations enter the Gibbs energy through ln Q. NCERT itself states the equation without this step (NCERT page 36).

What are the applications of the Nernst equation?

The Nernst equation gives the emf or electrode potential at any concentration, the pH from a hydrogen electrode, and, at equilibrium, the equilibrium constant, Δ_rG^⊖ and solubility products (NCERT pages 36–40). For AgCl(s) → Ag⁺ + Cl⁻, E^⊖_cell = 0.22 − 0.80 = −0.58 V (NCERT Table 2.1), so log K_sp = −0.58/0.059 = −9.83 and K_sp ≈ 1.5 × 10^⁻¹⁰, near the tabulated 1.8 × 10^⁻¹⁰. Beyond the NCERT text, it also gives the equilibrium potential of one ion, such as K⁺, across a nerve-cell membrane (Wikipedia, “Nernst equation”).

How do you find n in the Nernst equation?

To find n in the Nernst equation, write the two half-reactions, multiply them so that the electrons given up at the anode equal those taken up at the cathode, and add them: n is the number of electrons that cancel (NCERT page 38). For 3Zn + 2Au³⁺ → 3Zn²⁺ + 2Au, three zinc atoms give up two electrons each and two Au³⁺ ions take three each, so n = 6 (half-reactions from NCERT Table 2.1). If the whole equation is doubled, n becomes 12 but Q is squared, and (0.059/12) log Q^² = (0.059/6) log Q, so the emf comes out the same either way (reasoned).

Does the 0.059 in the Nernst equation change with temperature?

Yes, the 0.059 V in the Nernst equation is 2.303RT/F worked out at 298 K, so it is proportional to the kelvin temperature: 0.0591 V at 298 K but 0.0615 V at 310 K, body temperature (R = 8.314 J K^⁻¹ mol^⁻¹, F = 96 487 C mol^⁻¹, NCERT page 36; reasoned). Its unit is the volt, because RT/F is joules per mole divided by coulombs per mole, and T must be in kelvin. NCERT Table 2.1 gives E^⊖ values at 298 K (NCERT page 37), so at another temperature E^⊖ should also be taken for that temperature (reasoned).

Is the Nernst equation in the JEE Main and NEET syllabus?

Yes, the Nernst equation is in both syllabuses: Unit 7, Redox Reactions and Electrochemistry, of the JEE Main syllabus lists “Nernst equation and its applications, relationship between cell potential and Gibbs’ energy change”, and Unit 7 of the NEET (UG) 2026 syllabus lists “Nernst equation and its applications; Relationship between cell potential and Gibbs’ energy change”. NCERT Class 12 Chemistry Part I covers it in Section 2.3 (pages 36–41), with Examples 2.1–2.3 and Intext Questions 2.4–2.6.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. How does the emf of a Daniell cell change if [Cu²⁺] is increased? If [Zn²⁺] is increased?

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   It increases with [Cu²⁺] and decreases with [Zn²⁺], since Q = [Zn²⁺]/[Cu²⁺] (NCERT page 38).

2. Q2. Calculate the emf of the cell in which Ni(s) + 2Ag⁺(0.002 M) → Ni²⁺(0.160 M) + 2Ag(s) takes place, given E^⊖_cell = 1.05 V (NCERT Intext Question 2.5).

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   Q = 0.160/(0.002)² = 4 × 10⁴, so E_cell = 1.05 − (0.059/2) × 4.602 = 1.05 − 0.136 = 0.914 V (NCERT Eq. (2.13)).

3. Q3. What does zero reversible emf of a functioning Daniell cell imply, and why is a zero reading under load insufficient?

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   NCERT: when the concentrations stop changing and the voltmeter reads zero, equilibrium has been attained, and Q = K_c (NCERT page 39). Strictly, this needs the emf of a working cell with no current drawn; zero voltage under load, a short circuit or a failed electrode does not by itself prove chemical equilibrium (reasoned).