Electrochemistry Numericals (Class 12): Nernst, Conductance and Faraday

Class 12 · Chemistry

Electrochemistry Numericals (Class 12): Nernst, Conductance and Faraday

Three formula families in one chapter, and marks are lost mainly by mixing up which one a question needs.

Class 12 · CBSE & ISC · Concept and numericals

The short answer: Cell potential questions use the Nernst equation with n from the balanced cell reaction. Conductance questions use molar conductivity and Kohlrausch’s law. Electrolysis questions use Faraday’s laws relating charge to amount deposited. Identifying which family a question belongs to is the first step.

Identifying the question type

If the question mentionsUse
Cell potential, concentrations, spontaneityNernst equation
Conductivity, dilution, degree of dissociationMolar conductivity and Kohlrausch
Current, time, mass depositedFaraday's laws
Equilibrium constant from a cellNernst at equilibrium, or the free energy relation

Cell potential and the Nernst equation

Write both half-reactions, identify cathode and anode, and compute

cell = E°cathode − E°anode

Then for non-standard concentrations at 298 K:

E = E° − (0.0591/n) log Q
Two errors account for most lost marks here. First, n is the number of electrons in the balanced overall cell reaction, not in a half-reaction as tabulated. Second, electrode potentials are not multiplied when a half-reaction is multiplied to balance electrons — the potential is intensive and stays the same. Writing the balanced cell reaction before touching the formula prevents both.

Conductance

Λm = κ / c     α = Λm / Λ°m

Kohlrausch's law lets the limiting molar conductivity of a weak electrolyte be built from ionic contributions, since it cannot be obtained by extrapolation. From the degree of dissociation, the dissociation constant follows through the equilibrium expression.

Units are the usual trap. Conductivity is per unit length; molar conductivity divides by concentration, so the units change. Converting concentration from mol per litre to mol per cubic metre where required is where errors creep in, so decide the unit system before substituting.

Faraday's laws of electrolysis

The amount of substance deposited is proportional to the charge passed, and for a given charge it is inversely proportional to the number of electrons required per ion.

Charge Q = I × t    Moles of electrons = Q / F    Moles deposited = (Q/F) / n

A worked order that avoids errors:

  1. Compute the charge from current and time, keeping time in seconds.
  2. Divide by the Faraday constant to get moles of electrons.
  3. Write the electrode half-reaction to find how many electrons each ion needs.
  4. Divide to get moles of substance, then multiply by molar mass if a mass is wanted.

Step three is the one skipped most often. A metal deposited from a divalent ion needs two electrons per atom, so half as much is deposited for the same charge as a monovalent one.

Connecting cell potential to equilibrium

ΔG° = −nFE° = −RT ln K

So a measured standard cell potential gives the equilibrium constant directly. At 298 K this can be written with the same 0.0591 factor, which is convenient but must not be used at other temperatures.

Sanity checks

  • A positive cell potential means the reaction as written is spontaneous.
  • Molar conductivity must increase on dilution, never decrease.
  • Degree of dissociation must lie between zero and one.
  • Mass deposited must be positive and consistent with the current and time given.

Frequently asked questions

Why is the electrode potential not multiplied when balancing?

Because potential is energy per unit charge, which does not change when the reaction is scaled. Only n changes.

Why does molar conductivity rise on dilution?

For a strong electrolyte because interionic interference decreases; for a weak one because the degree of dissociation increases. The observation is the same, the reason differs.

How do I get n for the Nernst equation?

From the balanced overall cell reaction, after the electrons cancel between the two half-reactions.

Why is the 0.0591 value temperature specific?

Because it comes from RT/F converted to base-10 logarithms at 298 K. At any other temperature the general form must be used.

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