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What is the Nernst equation?

Class 12 ChemistryNCERT chapter: Electrochemistryइसे हिंदी में पढ़ें

Exam-ready answer

The Nernst equation gives the electrode potential (or cell emf) at any concentration in terms of the standard potential. For Mⁿ⁺(aq) + ne⁻ → M(s), E = E° − (RT/nF) ln(1/[Mⁿ⁺]), where R is the gas constant, T the temperature in kelvin, n the number of electrons and F the Faraday constant. At 298 K, E = E° − (0.059 V/n) log(1/[Mⁿ⁺]).

Explanation

Standard electrode potentials E° are measured with all ions at 1 mol L⁻¹ and 298 K. Real cells rarely have these concentrations, so Nernst showed how potential depends on them. For the reduction Mⁿ⁺(aq) + ne⁻ → M(s), E(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − (RT/nF) ln(1/[Mⁿ⁺]). The concentration of the pure solid M is taken as unity. Here R = 8.314 J K⁻¹ mol⁻¹ and F is about 96487 C mol⁻¹.

For the Daniell cell, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with n = 2, combining the two electrode equations gives E_cell = E°_cell − (RT/2F) ln([Zn²⁺]/[Cu²⁺]). Converting to base-10 logarithms at 298 K, E_cell = E°_cell − (0.059 V/2) log([Zn²⁺]/[Cu²⁺]). So increasing [Cu²⁺] or decreasing [Zn²⁺] raises the cell potential. For a general reaction, E_cell = E°_cell − (RT/nF) ln Q, where Q is the reaction quotient.

As a cell discharges, reactant ions are used up and product ions build up, so Q rises and E_cell falls. This is one reason a battery's voltage drops with use. When the reaction reaches equilibrium, the cell can do no more work: E_cell = 0 and Q = K_c. Then E°_cell = (2.303RT/nF) log K_c, which lets us calculate equilibrium constants from E° values. Also, ΔrG° = −nFE°_cell links cell potential to Gibbs energy.

Points to remember

  1. Electrode: E(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − (RT/nF) ln(1/[Mⁿ⁺]); pure solids have concentration taken as 1.
  2. Cell: E_cell = E°_cell − (RT/nF) ln Q; at 298 K, E_cell = E°_cell − (0.059 V/n) log Q.
  3. Daniell cell: E_cell = E°_cell − (0.059 V/2) log([Zn²⁺]/[Cu²⁺]).
  4. R = 8.314 J K⁻¹ mol⁻¹, F ≈ 96487 C mol⁻¹, T in kelvin, n = number of electrons transferred.
  5. At equilibrium E_cell = 0: E°_cell = (2.303RT/nF) log K_c; also ΔrG° = −nFE°_cell.

Exam tip

Write the cell reaction first and find n, then write the Nernst equation with products over reactants inside the log and omit pure solids. In numericals, show the step E°_cell = E°_cathode − E°_anode and state the answer in volts to correct significant figures.

Questions students also ask

Where does the value 0.059 V come from?
Converting natural log to base-10 log introduces 2.303. Then 2.303RT/F = 2.303 × 8.314 J K⁻¹ mol⁻¹ × 298 K ÷ 96487 C mol⁻¹, which is about 0.059 V. So 0.059 V/n replaces RT/nF ln only at 298 K; at other temperatures the full expression must be used.
How is the equilibrium constant found from the Nernst equation?
At equilibrium the cell reaction has no driving force, so E_cell = 0 and the reaction quotient equals K_c. Substituting gives E°_cell = (2.303RT/nF) log K_c. At 298 K this becomes log K_c = nE°_cell/0.059 V, so a known E°_cell gives K_c directly.
Why does a battery's voltage fall as it is used?
As the cell reaction proceeds, reactant concentrations decrease and product concentrations increase. The reaction quotient Q rises, so the term (RT/nF) ln Q grows and E_cell becomes smaller. When the reaction approaches equilibrium, the potential approaches zero and the cell is exhausted.

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