Formula
Nernst equation
E = E° - (RT/nF) ln Q = E° - (0.0591/n) log Q at 298 K. Q = reaction quotient. At equilibrium E=0 and Q=K, giving E° = (0.0591/n) log K.
-- NCERT Class 12 Chemistry, Ch. 2, p. 39Nernst Equation
8 MCQs5 revision cards9-step worked example
Source: NCERT Redox Reactions and ElectrochemistryOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Nernst Equation, explained for NEET
The Nernst equation tells you what happens to a cell's EMF when concentrations deviate from the standard 1 M. It is also where a large fraction of NEET negative marks originate — not from the equation itself, but from plugging in the wrong value of n.
The equation (at 298 K):
E = E° − (0.0591/n) × log₁₀ Q
where E° is the standard cell EMF, n is the number of electrons transferred in the balanced redox equation, and Q is the reaction quotient ([products]/[reactants], each raised to stoichiometric powers; pure solids and liquids excluded).
Standard cell EMF recap: E°_cell = E°_cathode − E°_anode, with both values taken as reduction potentials (NCERT Class 12 Chemistry Chapter 2, page 34).
The high-frequency trap: wrong n. The electron count must come from the balanced overall equation, not from a single half-reaction in isolation. For the Daniel cell (Zn | Zn²⁺ || Cu²⁺ | Cu), each half-reaction involves 2 electrons, so n = 2. But if KMnO₄ acts as oxidising agent in acidic medium, Mn goes from +7 to +2, transferring 5 electrons — and that number changes to 3 (→ MnO₂) in neutral medium or 1 (→ MnO₄²⁻) in strongly basic medium. The medium dictates n, not just the reagent.
Key boundary conditions to memorise:
- At standard conditions: Q = 1, log Q = 0, so E = E°.
- At equilibrium: E = 0, so E° = (0.0591/n) × log₁₀ K. This links electrochemistry to the equilibrium constant.
Watch-out: NEET distractors routinely offer an answer that uses n = 1 where the balanced equation demands n = 2 (or vice versa). Before substituting into the Nernst equation, write out the balanced redox equation and explicitly count electrons. Every time.
Can you answer these Nernst Equation MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the Nernst equation E = E° − (0.0591/n) log Q at 298 K, what does 'n' represent?
Show answer and why every option is right or wrong
Answer: C. In the Nernst equation, n is the total number of electrons transferred per balanced redox equation (NCERT Class 12 Chemistry Chapter 2, page 38).
Why A is wrong: A is wrong because n counts electrons transferred, not moles of any particular reactant. The stoichiometric coefficients of reactants and n are related but not identical.
Why B is wrong: B is wrong because n is not the number of ions — it is the electron count from the balanced equation. A cell with Cu²⁺ has many ions in solution, but n = 2 for the Cu²⁺/Cu half-cell.
Why D is wrong: D is wrong because Avogadro's number divided by Faraday constant has no direct role here; n is a simple integer from the balanced equation.
MCQ 2Easy RecallPractice
At equilibrium, the cell potential E in the Nernst equation equals:
Show answer and why every option is right or wrong
Answer: A. At equilibrium, no net reaction occurs and the driving force is zero: E = 0. Setting E = 0 in the Nernst equation gives E° = (0.0591/n) log K (NCERT Class 12 Chemistry Chapter 2).
Why B is wrong: B is wrong because E = E° only when Q = 1 (standard conditions), not at equilibrium where Q = K.
Why C is wrong: C is wrong because 0.0591/n is the pre-logarithmic factor, not a cell potential value. It has units of V but is only meaningful when multiplied by log Q.
Why D is wrong: D is wrong because −E° would imply the cell has reversed to exactly the standard magnitude, which has no thermodynamic basis at equilibrium.
MCQ 3Easy RecallPractice
When all species in a galvanic cell are at their standard states (1 M concentration, 298 K), the value of the reaction quotient Q is:
Show answer and why every option is right or wrong
Answer: C. At standard conditions all concentrations are 1 M, so Q = (1)^products / (1)^reactants = 1, making log Q = 0 and E = E° (NCERT Class 12 Chemistry Chapter 2).
Why A is wrong: A is wrong because Q = 0 would require zero product concentration, which is a different physical scenario (initial mixing, not standard state). log 0 is undefined.
Why B is wrong: B is wrong because Q = 10 would require specific non-standard concentrations. At standard state, every activity is 1 by definition.
Why D is wrong: D is wrong because at standard states, every concentration term is 1 M regardless of the specific reaction, so Q = 1 universally.
MCQ 4Direct ApplicationPractice
For the cell reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E° = +1.10 V, what is the cell EMF when [Cu²⁺] = 0.01 M and [Zn²⁺] = 1.0 M at 298 K?
Show answer and why every option is right or wrong
Answer: D. n = 2 (Zn → Zn²⁺ + 2e⁻). Q = [Zn²⁺]/[Cu²⁺] = 1.0/0.01 = 100. E = 1.10 − (0.0591/2) × log 100 = 1.10 − 0.02955 × 2 = 1.10 − 0.0591 = 1.04 V (NCERT Class 12 Chemistry Chapter 2, page 39).
Why A is wrong: A is wrong because 1.16 V results from inverting Q (using 0.01/1.0 = 0.01, log = −2), which would add the correction instead of subtracting. The trap: writing Q upside-down (trap: wrong Q expression).
Why B is wrong: B is wrong because 0.98 V results from using n = 1 instead of n = 2, giving (0.0591/1) × 2 = 0.1182, so E = 1.10 − 0.12 ≈ 0.98 V (trap: wrong n — trap: nernst n electrons negmark).
Why C is wrong: C is wrong because 1.10 V is the standard EMF — it ignores the non-standard concentrations entirely, i.e., forgetting to apply the Nernst correction.
MCQ 5Direct ApplicationPractice
Given E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V, the standard EMF of the cell Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s) is:
Show answer and why every option is right or wrong
Answer: D. E°_cell = E°_cathode − E°_anode = 0.80 − 0.34 = +0.46 V. Ag⁺/Ag is the cathode (higher reduction potential); Cu²⁺/Cu is the anode (NCERT Class 12 Chemistry Chapter 2).
Why A is wrong: A is wrong because 1.14 V results from adding both reduction potentials (0.80 + 0.34). The correct operation is subtraction: cathode minus anode (trap: adding instead of subtracting reduction potentials).
Why B is wrong: B is wrong because 0.80 V is simply the cathode reduction potential alone, ignoring the anode contribution entirely.
Why C is wrong: C is wrong because −0.46 V results from reversing cathode and anode (0.34 − 0.80). In the given cell notation, Cu is the anode (left) and Ag is the cathode (right), so E° must be positive for a spontaneous cell.
MCQ 6Direct ApplicationPractice
For the half-reaction MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O in acidic medium, what value of n should be used in the Nernst equation for a cell based on this half-reaction?
Show answer and why every option is right or wrong
Answer: A. In acidic medium, MnO₄⁻ is reduced to Mn²⁺ with a transfer of 5 electrons (Mn goes from +7 to +2). Therefore n = 5 in the Nernst equation (NCERT Class 12 Chemistry Chapter 2).
Why B is wrong: B is wrong because n = 3 applies when KMnO₄ is reduced in neutral or weakly basic medium to MnO₂ (Mn: +7 → +4). The question specifies acidic medium (trap: kmno4 medium dependence).
Why C is wrong: C is wrong because n = 1 applies only when KMnO₄ is reduced in strongly basic medium to MnO₄²⁻ (Mn: +7 → +6). The medium is specified as acidic here (trap: kmno4 medium dependence).
Why D is wrong: D is wrong because 7 is the oxidation state of Mn in MnO₄⁻, not the number of electrons transferred. The electron count is the change in oxidation state, not the absolute value.
MCQ 7CalculationPractice
For the cell Zn | Zn²⁺(0.1 M) || Ag⁺(0.1 M) | Ag, given E°(Zn²⁺/Zn) = −0.76 V and E°(Ag⁺/Ag) = +0.80 V, the cell EMF at 298 K is:
Show answer and why every option is right or wrong
Answer: B. E° = 0.80 − (−0.76) = 1.56 V. The cell reaction is Zn + 2Ag⁺ → Zn²⁺ + 2Ag, so n = 2. Q = [Zn²⁺]/[Ag⁺]² = 0.1/(0.1)² = 10. E = 1.56 − (0.0591/2) × log 10 = 1.56 − 0.02955 = 1.53 V (NCERT Class 12 Chemistry Chapter 2, page 39).
Why A is wrong: A is wrong because 1.56 V is the standard EMF (E°). The question gives non-standard concentrations, so the Nernst correction must be applied. This error comes from ignoring non-standard conditions.
Why C is wrong: C is wrong because 1.50 V results from using n = 1 instead of n = 2: correction = (0.0591/1) × 1 = 0.0591, giving 1.56 − 0.0591 ≈ 1.50 V (trap: wrong n — mistake: nernst wrong n).
Why D is wrong: D is wrong because 1.59 V results from inverting Q (using [Ag⁺]²/[Zn²⁺] = 0.01/0.1 = 0.1, log = −1), which adds the correction instead of subtracting. The trap: writing Q with products in the denominator.
MCQ 8Concept TrapPractice
In neutral medium KMnO₄ is reduced to MnO₂, so n = 3 for the cell reaction. A student uses the correct E°_cell and the correct reaction quotient Q, with Q > 1, but substitutes n = 5 in E = E° − (0.0591/n) log Q. Compared with the true EMF, the student's calculated value will be:
Show answer and why every option is right or wrong
Answer: A. A is correct. In neutral medium Mn goes +7 → +4, so n = 3. The Nernst correction (0.0591/n) log Q is SUBTRACTED from E°, and with Q > 1 the log is positive, so it is a genuine subtraction. Dividing by 5 instead of 3 makes that subtraction smaller than it should be, and a smaller subtraction leaves a larger E. The student's value is therefore higher than the true EMF (NCERT Class 12 Chemistry Chapter 2, page 38, Eq. 2.13; trap: kmno4 medium dependence).
Why B is wrong: B is wrong because 'lower' would need the subtracted correction to be too LARGE. A bigger denominator (5 rather than 3) makes it smaller, so with Q > 1 the computed EMF overshoots rather than falls short (trap: getting the direction right in words but inverting it in the arithmetic).
Why C is wrong: C is wrong because using the wrong n directly changes the magnitude of the correction term. The equation is sensitive to n — any incorrect value produces a different EMF.
Why D is wrong: D is wrong because the direction follows without knowing E°. E° is the same in both calculations and cancels out of the comparison; n appears only in the correction term, so a larger n always shrinks that term, and with Q > 1 the result is always on the high side.
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Nernst Equation: quick recall before you leave
How do you solve a Nernst Equation question? A worked example
Pattern: Nernst equation — compute non-standard cell EMF (NEET pattern: nernst equation problem).
- 1
Given
Cell: Zn(s) | Zn²⁺(aq, 0.001 M) || Cu²⁺(aq, 0.1 M) | Cu(s)
E°(Zn²⁺/Zn) = −0.76 V
E°(Cu²⁺/Cu) = +0.34 V
Temperature = 298 K - 2
Required
Calculate the cell EMF (E) at the given non-standard concentrations.
- 3
Concept
The Nernst equation adjusts the standard EMF for the effect of concentration. The reaction quotient Q depends on the balanced overall equation.
- 4
Formula
E°_cell = E°_cathode − E°_anode
E = E° − (0.0591/n) × log₁₀ Q - 5
Substitution setup
Balanced reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Electrons transferred: n = 2 (Zn → Zn²⁺ + 2e⁻)
E° = 0.34 − (−0.76) = 1.10 V
Q = [Zn²⁺]/[Cu²⁺] = 0.001/0.1 = 0.01 - 6
Calculation
log₁₀(0.01) = −2
E = 1.10 − (0.0591/2) × (−2)
E = 1.10 − (0.02955) × (−2)
E = 1.10 + 0.0591
E = 1.1591 V
Note on exact values: The coefficient 2 in n = 2 is a counting integer (electrons per balanced equation) and does not limit significant figures. - 7
Final answer
E = 1.16 V (rounded to 3 significant figures, matching the precision of the given E° values).
The factor 0.0591 is derived from (RT ln 10)/F at 298 K and is conventionally treated as exact for NEET calculations. - 8
Common trap
Using n = 1 instead of n = 2 would double the correction term: (0.0591/1) × (−2) = −0.1182, giving E = 1.10 + 0.1182 = 1.22 V — a distractor commonly seen in NEET options (trap: nernst n electrons negmark; mistake: mistake: nernst wrong n).
- 9
Similar NEET-style question
For the cell Fe(s) | Fe²⁺(0.01 M) || Ag⁺(1.0 M) | Ag(s), with E°(Fe²⁺/Fe) = −0.44 V and E°(Ag⁺/Ag) = +0.80 V, calculate E at 298 K. (Answer: n = 2, Q = 0.01/1² = 0.01, E = 1.24 + 0.0591 = 1.30 V.)
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What to remember before solving Nernst Equation questions
Which Nernst Equation formulas do you need for NEET?
Nernst equation
Cell potential at non-standard conditions. n = electrons transferred. At equilibrium E=0, Q=K.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | cell potential | V |
| E° | standard | V |
| n | electrons | - |
| Q | reaction quotient | - |
Valid when
- 298 K (else use RT/F)
- Single redox process
Where do students lose marks on Nernst Equation?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Negative Marking
Multi-step Nernst problem: identify electrons, write Q correctly, plug into 0.0591/n. Each sub-step has factor errors.
When it triggers
Cell EMF problem at non-standard conditions.
How to avoid
Step-by-step: (1) write balanced redox; (2) count n electrons; (3) compute Q from concentrations; (4) plug into Nernst. Verify by checking limits: at standard conditions Q=1, log Q=0, E=E°.
Category: Inorganic Exception
Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).
When it triggers
Question gives KMnO4 oxidation in unspecified or specific medium.
How to avoid
Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.
Root cause: formula misuse
Correction
n = electrons transferred per balanced redox equation. For Cu²⁺ + 2e⁻ → Cu: n=2. For Mn(VII) → Mn(II): n=5.
More in Redox Reactions and Electrochemistry: 1 exam trap or mistake · 4 formulas · 1 question pattern from its other lessons.
Nernst Equation questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 21 past-paper questions from Redox Reactions and Electrochemistry →
How does NEET ask about Nernst Equation?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Apply Nernst equation to compute non-standard cell EMF given Q. n = electrons transferred.
Multi StepMedium
Common distractors
wrong n electrons
Uses incorrect electron count from half-reactions
Sources
NCERT refs: Class 12 Chemistry Chapter 2, p.34
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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