Formula
ΔG and EMF
ΔG = -nFE; ΔG° = -nFE° = -RT ln K. n = electrons transferred, F = Faraday's constant (96485 C/mol).
-- NCERT Class 12 Chemistry, Ch. 2, p. 40Cell Potential Gibbs
8 MCQs6 revision cards9-step worked example
Source: NCERT Redox Reactions and ElectrochemistryPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 25 Sep 2026
Cell Potential Gibbs, explained for NEET
The relationship ΔG° = −nFE° is where thermodynamics meets electrochemistry — and the most common leak here is a sign error. A positive E°_cell means the reaction is spontaneous (ΔG° < 0), and a negative E°_cell means non-spontaneous (ΔG° > 0). The minus sign in the formula enforces this: forget it, and every spontaneity prediction flips.
The core formula (NCERT Class 12 Chemistry Chapter 2, page 44):
ΔG° = −nFE°_cell
where n = number of electrons transferred in the balanced redox equation, F = 96485 C/mol (Faraday's constant), and E°_cell = standard cell EMF in volts.
How to get E°_cell: Both half-cell potentials must be written as reductions. Then E°_cell = E°_cathode − E°_anode. If E°_cell comes out positive, the cell works spontaneously; if negative, you need external energy.
Unit check. ΔG° comes out in joules (C/mol × V = J/mol). NEET options sometimes list kJ/mol — divide by 1000. A common error is reporting the answer in joules when options are in kilojoules, or vice versa.
The equilibrium link. At equilibrium, ΔG = 0 and therefore E = 0. This connects to the Nernst equation (covered in its own lesson) but the takeaway here is: when a cell's potential drops to zero, the reaction has reached equilibrium and no further net work is possible.
What to watch for in NEET questions: Problems typically give two standard reduction potentials and ask for ΔG°. The sequence is always: (1) identify cathode and anode, (2) compute E°_cell, (3) determine n from the balanced equation, (4) plug into ΔG° = −nFE°. Errors cluster at steps 2 and 3 — reversing cathode/anode or miscounting electrons.
Can you answer these Cell Potential Gibbs 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
For a galvanic cell operating under standard conditions, what is the relationship between Gibbs energy change (ΔG°) and the standard cell EMF (E°_cell)?
Show answer and why every option is right or wrong
Answer: A. The thermodynamic relationship is ΔG° = −nFE°_cell, as given in NCERT Class 12 Chemistry Chapter 2, page 44. The negative sign ensures that a positive E°_cell yields a negative ΔG°, consistent with spontaneity.
Why B is wrong: B omits the crucial negative sign. Without it, a spontaneous cell (positive E°_cell) would give positive ΔG°, contradicting the spontaneity criterion.
Why C is wrong: C places E°_cell in the denominator, which has no thermodynamic basis. The relationship is multiplicative, not inversely proportional.
Why D is wrong: D both omits the negative sign and places E°_cell in the denominator — a doubly incorrect formula.
MCQ 2Easy RecallPractice
If a cell has E°_cell > 0, what can be said about ΔG° for the cell reaction?
Show answer and why every option is right or wrong
Answer: C. Since ΔG° = −nFE°_cell, and n and F are always positive, a positive E°_cell makes ΔG° negative. This means the reaction is spontaneous under standard conditions (NCERT Class 12 Chemistry Chapter 2, page 44).
Why A is wrong: A reverses the sign relationship. A positive E°_cell gives negative ΔG°, not positive.
Why B is wrong: B applies only when E_cell = 0, which is the equilibrium condition, not when E°_cell > 0.
Why D is wrong: D is incorrect because n is always a positive integer and F is a positive constant; their product does not change the sign — only E°_cell determines whether ΔG° is positive or negative.
MCQ 3Easy RecallPractice
The value of one Faraday constant (F) is:
Show answer and why every option is right or wrong
Answer: B. The Faraday constant is the charge carried by one mole of electrons: F = 96485 C/mol (NCERT Class 12 Chemistry Chapter 2, page 44).
Why A is wrong: A gives the wrong unit. Joules per mole is a unit of energy, not charge. F is a charge-per-mole quantity.
Why C is wrong: C gives volts per mole. Voltage is electric potential difference (J/C), not charge.
Why D is wrong: D uses J·mol, which is dimensionally meaningless in this context. F has dimensions of charge per amount of substance.
MCQ 4Direct ApplicationPractice
The standard emf of the Daniell cell, Zn | Zn²⁺ || Cu²⁺ | Cu, is 1.10 V. Taking F = 96 500 C mol⁻¹, the standard Gibbs energy change for the cell reaction is:
Show answer and why every option is right or wrong
Answer: C. C is correct. ΔG° = −nFE°cell. In Zn + Cu²⁺ → Zn²⁺ + Cu, two electrons are transferred, so n = 2: ΔG° = −2 × 96 500 × 1.10 = −212 300 J mol⁻¹ = −212.3 kJ mol⁻¹. The negative sign matches the positive emf: the cell reaction is spontaneous.
Why A is wrong: A is wrong because −106.2 kJ mol⁻¹ takes n = 1. Zinc goes from 0 to +2, so two electrons are transferred per formula unit.
Why B is wrong: B is wrong because +212.3 kJ mol⁻¹ drops the minus sign in ΔG° = −nFE°. A positive emf always gives a negative ΔG°.
Why D is wrong: D is wrong because −2.123 × 10⁵ is the value in joules, labelled kJ. Divide by 1000 to convert.
MCQ 5Direct ApplicationPractice
For a cell reaction where n = 2 and E°_cell = +0.34 V, the standard Gibbs energy change ΔG° is: (F = 96485 C/mol)
Show answer and why every option is right or wrong
Answer: D. ΔG° = −nFE° = −(2)(96485)(0.34) = −65610 J/mol ≈ −65.6 kJ/mol. The negative value confirms spontaneity, consistent with the positive E°_cell (NCERT Class 12 Chemistry Chapter 2, page 44).
Why A is wrong: A combines both errors: uses n = 1 and drops the negative sign.
Why B is wrong: B has the correct magnitude but wrong sign. This error arises from omitting the negative sign in ΔG° = −nFE°, giving +nFE° instead.
Why C is wrong: C uses n = 1 instead of n = 2: −(1)(96485)(0.34) = −32805 J ≈ −32.8 kJ. Always use the actual number of electrons transferred in the balanced equation.
MCQ 6Direct ApplicationPractice
A cell reaction has ΔG° = −193.0 kJ/mol and n = 2. What is E°_cell? (F = 96485 C/mol)
Show answer and why every option is right or wrong
Answer: D. Rearranging: E°_cell = −ΔG°/(nF) = −(−193000)/(2 × 96485) = 193000/192970 ≈ 1.00 V. The positive value is consistent with a spontaneous reaction (negative ΔG°) (NCERT Class 12 Chemistry Chapter 2, page 44).
Why A is wrong: A uses n = 4 instead of n = 2, halving the correct answer: 193000/(4 × 96485) ≈ 0.50 V.
Why B is wrong: B gets the correct magnitude but the wrong sign. This happens when the student forgets that E°_cell = −ΔG°/(nF) and instead computes ΔG°/(nF), yielding a negative value.
Why C is wrong: C uses n = 1 instead of n = 2, doubling the correct answer: 193000/(1 × 96485) ≈ 2.00 V.
MCQ 7CalculationPractice
Given: E°(Ag⁺/Ag) = +0.80 V, E°(Cu²⁺/Cu) = +0.34 V. For the cell Cu | Cu²⁺ || Ag⁺ | Ag, what is ΔG° for the overall reaction Cu + 2Ag⁺ → Cu²⁺ + 2Ag? (F = 96485 C/mol)
Show answer and why every option is right or wrong
Answer: A. Step 1: E°_cell = E°_cathode − E°_anode = 0.80 − 0.34 = +0.46 V. Step 2: The balanced equation Cu → Cu²⁺ + 2e⁻ and 2Ag⁺ + 2e⁻ → 2Ag shows n = 2. Step 3: ΔG° = −nFE° = −(2)(96485)(0.46) = −88766 J ≈ −88.8 kJ/mol (NCERT Class 12 Chemistry Chapter 2, page 44).
Why B is wrong: B uses n = 1 instead of n = 2. Although one Ag⁺ requires 1 electron, the balanced equation transfers 2 electrons overall: −(1)(96485)(0.46) ≈ −44.4 kJ.
Why C is wrong: C uses n = 4, perhaps double-counting electrons from both half-cells: −(4)(96485)(0.46) ≈ −177.5 kJ.
Why D is wrong: D has the correct magnitude but wrong sign, from dropping the negative in ΔG° = −nFE°.
MCQ 8Concept TrapPractice
When a galvanic cell operates and the reaction reaches equilibrium, what are the values of E_cell and ΔG?
Show answer and why every option is right or wrong
Answer: B. At equilibrium, no net reaction occurs, so the driving force (cell potential) drops to zero: E_cell = 0. Since ΔG = −nFE_cell, when E_cell = 0, ΔG = 0 as well. This is the thermodynamic definition of equilibrium (NCERT Class 12 Chemistry Chapter 2, page 44).
Why A is wrong: A describes standard-state conditions, not equilibrium. At standard conditions, E = E° and ΔG = ΔG°; at equilibrium, both drop to zero.
Why C is wrong: C correctly states E_cell = 0 but incorrectly retains ΔG = ΔG°. Once E_cell = 0, the formula ΔG = −nFE gives ΔG = 0, not the standard value.
Why D is wrong: D incorrectly retains E_cell = E°_cell at equilibrium. As the reaction proceeds, concentrations change and E_cell decreases from E° toward zero.
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Cell Potential Gibbs: quick recall before you leave
How do you solve a Cell Potential Gibbs question? A worked example
- 1
Given
• E°(Zn²⁺/Zn) = −0.76 V (reduction potential)• E°(Cu²⁺/Cu) = +0.34 V (reduction potential)• F = 96485 C/mol
- 2
Required
ΔG° for the overall cell reaction, in kJ/mol.
- 3
Concept
The Gibbs energy change is related to cell EMF by ΔG° = −nFE°_cell (NCERT Class 12 Chemistry Chapter 2, page 44). We first need E°_cell from the two reduction potentials, then determine n from the balanced equation.
- 4
Formula
E°_cell = E°_cathode − E°_anode
ΔG° = −nFE°_cell - 5
Substitution
Zn is oxidized (anode), Cu²⁺ is reduced (cathode).
E°_cell = (+0.34) − (−0.76) = +1.10 V
Balanced half-reactions:• Zn → Zn²⁺ + 2e⁻ (oxidation)• Cu²⁺ + 2e⁻ → Cu (reduction)
Therefore n = 2.
ΔG° = −(2)(96485)(1.10) - 6
Calculation
ΔG° = −(2)(96485)(1.10)
ΔG° = −(2)(106133.5)
ΔG° = −212267 J/mol
ΔG° = −212.3 kJ/mol
Note: n = 2 is an exact integer (electrons counted from the balanced equation) and F = 96485 C/mol is a defined constant. Neither limits significant figures. The precision is governed by the given reduction potentials (3 significant figures), so the answer is reported to 4 significant figures as −212.3 kJ/mol. - 7
Final answer
ΔG° = −212.3 kJ/mol
The negative value confirms the Daniell cell reaction is spontaneous under standard conditions, consistent with the positive E°_cell. - 8
Common trap
The most frequent error is reversing cathode and anode: computing E°_cell = (−0.76) − (0.34) = −1.10 V. This gives ΔG° = +212.3 kJ/mol, flipping the spontaneity prediction. Always identify the more positive reduction potential as the cathode.
A second error: using n = 1 (per Ag-type half-reactions) when the balanced equation clearly transfers 2 electrons. Always write out both half-reactions and count electrons explicitly. - 9
Similar NEET-style question
"Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V, calculate ΔG° for the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. (F = 96485 C/mol)."
Approach: E°_cell = 0.77 − 0.54 = +0.23 V. n = 2 (each Fe³⁺ gains 1e⁻, two Fe³⁺ ions transfer 2e⁻ total). ΔG° = −(2)(96485)(0.23) = −44383 J ≈ −44.4 kJ/mol.
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What to remember before solving Cell Potential Gibbs questions
Which Cell Potential Gibbs formulas do you need for NEET?
ΔG from EMF
Connection between thermodynamics and electrochemistry. F = 96485 C/mol.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔG | Gibbs energy change | J |
| n | electrons transferred | - |
| F | Faraday 96485 | C/mol |
| E | cell EMF | V |
Valid when
- Single redox process
More in Redox Reactions and Electrochemistry: 4 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
Cell Potential Gibbs questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys.
NEET 2023
1Both A and R are true and R is NOT the correct explanation of A
2A is true but R is false
3A is false but R is true
4Both A and R are true and R is the correct explanation of A
NTA Answer: Option 4(final)
All 21 past-paper questions from Redox Reactions and Electrochemistry →
Sources
NCERT refs: Class 12 Chemistry Chapter 2, p.44
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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