Class 12 Chemistry · Academic Guide

Nernst Equation Numericals for Class 12 Chemistry (Solved Step by Step)

How to set up and solve electrochemistry numericals using the Nernst equation — with two fully worked board-style examples.

Class 12 · Chemistry · Electrochemistry · Updated 22 August 2026

Quick summary: The Nernst equation links a cell’s actual electrode/cell potential to the standard potential and the concentrations of the species involved. For Class 12 boards and entrance exams, most numericals reduce to plugging values into E = E° − (0.0591/n) log Q at 298 K. Below we break the formula down and solve two typical questions.

The Nernst equation (what to memorise)

For a general electrode or cell reaction at temperature T, the Nernst equation is written as:

E_cell = E°_cell − (RT / nF) ln Q

At the standard exam temperature of 298 K (25 °C), substituting R, F and converting to base-10 logarithm gives the form you will actually use:

E_cell = E°_cell − (0.0591 / n) log Q

Here n is the number of electrons transferred in the balanced cell reaction and Q is the reaction quotient (products over reactants, each raised to its stoichiometric power; pure solids and liquids are taken as 1).

Symbol	Meaning	Note for numericals
E°_cell	Standard cell potential	= E°_cathode − E°_anode (both as reduction potentials)
n	Electrons transferred	From the balanced equation
Q	Reaction quotient	Ignore pure solids/liquids
0.0591	2.303RT/F at 298 K	Use 0.059 or 0.0591 as told

Worked example 1 — Daniell cell

Q. Calculate the EMF of the cell Zn | Zn²⁺ (0.1 M) || Cu²⁺ (1.0 M) | Cu at 298 K. Given E°_cell = 1.10 V.

Step 1 — Write the reaction and find n

Zn + Cu²⁺ → Zn²⁺ + Cu. Two electrons are transferred, so n = 2.

Step 2 — Write Q

Q = [Zn²⁺] / [Cu²⁺] = 0.1 / 1.0 = 0.1

Step 3 — Substitute

E = 1.10 − (0.0591/2) log(0.1) = 1.10 − (0.02955)(−1) = 1.13 V

Because the reactant Cu²⁺ is more concentrated than the product Zn²⁺, the cell potential rises slightly above E° — consistent with Le Chatelier’s principle.

Worked example 2 — concentration cell

Q. Find the EMF of the concentration cell Ni | Ni²⁺ (0.01 M) || Ni²⁺ (0.1 M) | Ni at 298 K.

In a concentration cell both electrodes are the same metal, so E°_cell = 0 and n = 2. The cell runs to equalise the two concentrations, so:

E = 0 − (0.0591/2) log(0.01/0.1) = −(0.02955)(−1) = 0.0295 V

A small positive EMF is expected: dilute solution acts as the anode, concentrated as the cathode.

Common mistakes to avoid

Using the wrong n — always balance electrons first.
Inverting Q (products go on top, reactants below).
Forgetting that a negative log makes the second term add to E°.
Mixing oxidation and reduction potentials when finding E°_cell.

Board tip: If the question gives EMF and asks for a concentration, rearrange the same equation for log Q — the method never changes, only the unknown does.

Frequently asked questions

Why is the value 0.0591 used?

It is 2.303RT/F evaluated at 298 K. If the exam specifies a different temperature, recompute 2.303RT/F.

Does the Nernst equation apply to a single electrode?

Yes. Replace E°_cell with the electrode’s standard reduction potential and use its half-reaction for Q.

What happens at equilibrium?

E_cell becomes 0 and Q equals the equilibrium constant K, giving E°_cell = (0.0591/n) log K.

How much weight does electrochemistry carry in Class 12 boards?

It is a core physical-chemistry unit that regularly appears in both short and long numerical questions. Confirm the exact weightage from the current CBSE syllabus for your session.

Struggling with Class 12 Chemistry numericals?

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