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Nernst Equation

The Nernst equation relates the electrode potential or cell EMF to concentration, pressure, temperature, and reaction quotient for a redox reaction.

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Student-friendly explanation

Standard EMF applies only under standard conditions. When ion concentrations change, the reaction quotient changes and the cell EMF also changes. At 298 K, the logarithmic form commonly used in Class 12 is E cell = E degree cell - (0.0591/n) log Q, where n is the number of electrons transferred and Q is the reaction quotient written from the balanced cell reaction. For pure solids and liquids, activity is taken as one, so they are not included in Q.

How to write this in exams

  1. 1

    Start with the exact idea

    The Nernst equation relates the electrode potential or cell EMF to concentration, pressure, temperature, and reaction quotient for a redox reaction.

  2. 2

    Then show how to use it

    Balance the cell reaction. Count total electrons transferred to get n. Write Q using aqueous and gaseous species only. Substitute concentrations with units converted consistently. Calculate the logarithm and interpret whether EMF rises or falls compared with standard EMF.

  3. 3

    Add one concrete example

    For Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s), Q = [Zn2+]/[Cu2+]. Therefore E cell = E degree cell - (0.0591/2) log([Zn2+]/[Cu2+]).

  4. 4

    Avoid this incomplete answer

    A common wrong answer is using Q = [Cu2+]/[Zn2+] for the Daniell cell, which changes the sign of the correction term.

Definition

The Nernst equation relates the electrode potential or cell EMF to concentration, pressure, temperature, and reaction quotient for a redox reaction.

Example

For Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s), Q = [Zn2+]/[Cu2+]. Therefore E cell = E degree cell - (0.0591/2) log([Zn2+]/[Cu2+]).

Rule to remember

At 298 K: E cell = E degree cell - (0.0591/n) log Q. For a half-cell Mn+ + ne- M, E = E degree - (0.0591/n) log(1/[Mn+]), which can also be written as E = E degree + (0.0591/n) log[Mn+].

Memory hook

Nernst asks: balanced reaction first, quotient second, EMF last.

Examples and method

Worked example

For Daniell cell at 298 K, E degree cell = 1.10 V, [Zn2+] = 0.10 M, [Cu2+] = 1.0 M, and n = 2. Q = 0.10/1.0 = 0.10. E cell = 1.10 - (0.0591/2) log(0.10). Since log(0.10) = -1, E cell = 1.10 + 0.02955 = 1.12955 V. The cell EMF increases because product ion concentration is lower relative to reactant ion concentration.

Method to apply

Balance the cell reaction. Count total electrons transferred to get n. Write Q using aqueous and gaseous species only. Substitute concentrations with units converted consistently. Calculate the logarithm and interpret whether EMF rises or falls compared with standard EMF.

Diagram support

A diagram is not required. A flow chart from balanced reaction to Q to substitution is more useful for numerical accuracy.

How CBSE asks it

Questions ask for EMF under non-standard concentration, equilibrium constant relation, concentration cell EMF, or effect of dilution on electrode potential.

Avoid common mistakes

Common confusion

Students often put the concentration ratio upside down or use n as the coefficient of metal instead of the number of electrons transferred.

Common wrong answer

A common wrong answer is using Q = [Cu2+]/[Zn2+] for the Daniell cell, which changes the sign of the correction term.

Exam tip

Write the balanced cell reaction before writing Q. This prevents most concentration-ratio and electron-number errors.

Quick check

For Zn + Cu2+ Zn2+ + Cu, what concentration ratio appears in Q?

Q = [Zn2+]/[Cu2+], because solids Zn and Cu are omitted.

Answer writing and exam use

1-mark answer

The Nernst equation relates the electrode potential or cell EMF to concentration, pressure, temperature, and reaction quotient for a redox reaction.

2-mark answer

The Nernst equation relates the electrode potential or cell EMF to concentration, pressure, temperature, and reaction quotient for a redox reaction. At 298 K: E cell = E degree cell - (0.0591/n) log Q. For a half-cell Mn+ + ne- M, E = E degree - (0.0591/n) log(1/[Mn+]), which can also be written as E = E degree + (0.0591/n) log[Mn+]. For Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s), Q = [Zn2+]/[Cu2+]. Therefore E cell = E degree cell - (0.0591/2) log([Zn2+]/[Cu2+]).

3-mark answer

Standard EMF applies only under standard conditions. When ion concentrations change, the reaction quotient changes and the cell EMF also changes. At 298 K, the logarithmic form commonly used in Class 12 is E cell = E degree cell - (0.0591/n) log Q, where n is the number of electrons transferred and Q is the reaction quotient written from the balanced cell reaction. For pure solids and liquids, activity is taken as one, so they are not included in Q. At 298 K: E cell = E degree cell - (0.0591/n) log Q. For a half-cell Mn+ + ne- M, E = E degree - (0.0591/n) log(1/[Mn+]), which can also be written as E = E degree + (0.0591/n) log[Mn+]. For Daniell cell at 298 K, E degree cell = 1.10 V, [Zn2+] = 0.10 M, [Cu2+] = 1.0 M, and n = 2. Q = 0.10/1.0 = 0.10. E cell = 1.10 - (0.0591/2) log(0.10). Since log(0.10) = -1, E cell = 1.10 + 0.02955 = 1.12955 V. The cell EMF increases because product ion concentration is lower relative to reactant ion concentration. Questions ask for EMF under non-standard concentration, equilibrium constant relation, concentration cell EMF, or effect of dilution on electrode potential. A common wrong answer is using Q = [Cu2+]/[Zn2+] for the Daniell cell, which changes the sign of the correction term.
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