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. 14Cell Potential Gibbs
8 MCQs2 revision cards9-step worked example
Source: NCERT Redox ReactionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 14):
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.
Practice 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)?
MCQ 2Easy RecallPractice
If a cell has E°_cell > 0, what can be said about ΔG° for the cell reaction?
MCQ 3Easy RecallPractice
The value of one Faraday constant (F) is:
MCQ 4Direct ApplicationPractice
Given the standard reduction potentials: E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V. What is E°_cell for the Daniell cell (Zn anode, Cu cathode)?
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)
MCQ 6Direct ApplicationPractice
A cell reaction has ΔG° = −193.0 kJ/mol and n = 2. What is E°_cell? (F = 96485 C/mol)
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)
MCQ 8Concept TrapPractice
When a galvanic cell operates and the reaction reaches equilibrium, what are the values of E_cell and ΔG?
Quick recall before you leave
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 14). 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. ---
Before solving, remember these
Formulas
Cell EMF
Both as reduction potentials. E°_cell > 0 → spontaneous.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E°_cell | standard cell EMF | V |
| E°_red | reduction potential | V |
Valid when
- Standard conditions (1 M, 1 bar, 298 K)
- Both half-reactions as reductions
Faraday's law of electrolysis
Mass deposited at electrode. M = molar mass; I = current; t = time; n = electrons per ion.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m | mass deposited | g |
| M | molar mass | g/mol |
| I | current | A |
| t | time | s |
| n | electrons per ion | - |
| F | Faraday | C/mol |
Valid when
- Steady current
- Single product
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
Molar conductivity
Molar conductivity from specific conductance. Increases with dilution as more ions are free.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Λ_m | molar conductivity | S cm^2/mol |
| κ | specific conductance | S/cm |
| C | molarity | mol/L |
Valid when
- Aqueous solution
- Single electrolyte
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
Exam Traps & Common Mistakes
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: concept gap
Correction
n = number of electrons per ion to deposit. Cu²⁺ + 2e⁻ → Cu: n=2. Al³⁺ + 3e⁻: n=3. m = MIt/(nF).
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.
Past Year Questions
17 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
121.0 s
269.3 s
323.1 s
4210 s
NTA Answer: Option 2(final)
NEET 2025
1+4, –4, and +6
2+1, –1, and +6
3+2, –2, and +6
4+1, –2, and +4
NTA Answer: Option 2(final)
NEET 2025
1A-III, B-II, C-I, D-IV
2A-II, B-IV, C-I,D-III
3A-II, B-I, C-IV, D-III
4A-III, B-I, C-IV, D-II
NTA Answer: Option 3(final)
NEET 2024Revised key
1Both Statement I and Statement II are correct.
2Both Statement I and Statement II are incorrect.
3Statement I is correct but Statement II is incorrect.
4Statement I is incorrect but Statement II is correct.
NTA Answer: Option 1(revised_final)
NEET 2024Revised key
1A-II, B-IV, C-I, D-III
2A-III, B-IV, C-I, D-II
3A-II, B-III, C-I, D-IV
4A-III, B-IV, C-II, D-I
NTA Answer: Option 1(revised_final)
NEET 2024Revised key
13.15 g
20.315 g
331.5 g
40.0315 g
NTA Answer: Option 2(revised_final)
NEET 2024Revised key
1POCl and H PO 3 3 3
2POCl and H PO 3 3 4
3H PO and POCl 3 4 3
4H PO and POCl 3 3 3
NTA Answer: Option 4(revised_final)
NEET 2023
13.28 cm–1
21.26 cm–1
33.34 cm–1
41.34 cm–1
NTA Answer: Option 2(final)
NEET 2023
The number of bonds, bonds and lone pair of electrons in pyridine, respectively are:
112, 3, 0
211, 3, 1
312, 2, 1
411, 2, 0
NTA Answer: Option 2(final)
NEET 2022
12CuSO (aq) + 2Ag(s) → 2Cu(s) + Ag SO (aq) 4 2 4
2CuSO (aq) + Zn(s) → ZnSO (aq) + Cu(s) 4 4
3CuSO (aq) + Fe(s) → FeSO (aq) + Cu(s) 4 4
4FeSO (aq) + Zn(s) → ZnSO (aq) + Fe(s) 4 4
NTA Answer: Option 1(final)
NEET 2022
Which of the following statement is not correct about diborane?
1Both the Boron atoms are sp2 hybridised.
2There are two 3-centre-2-electron bonds.
3The four terminal B-H bonds are two centre two electron bonds.
4The four terminal Hydrogen atoms and the two Boron atoms lie in one plane.
NTA Answer: Option 1(final)
NEET 2022
1No, because E cell =−2.733V
2Yes, because E c e ll = + 0 .2 8 7 V
3No, because E c e ll = – 0 .2 8 7 V
4Yes, because E c e ll = + 2 .7 3 3 V
NTA Answer: Option 2(final)
NEET 2022
1158.7 Å
2158.7 pm
315.87 pm
41.587 pm
NTA Answer: Option 2(final)
NEET 2021
1C = RC V P
2C + C = R P V
3C – C = R P V
4C = RC P V
NTA Answer: Option 3(final)
NEET 2021
1540.48 S cm2 mol–1
2201.28 S cm2 mol–1
3390.71 S cm2 mol–1
4698.28 S
NTA Answer: Option 3(final)
NEET 2021
Which one of the following polymers is prepared by addition polymerisation?
1Dacron
2Teflon
3Nylon-66
4Novolac
NTA Answer: Option 2(final)
NEET 2021
11.75×10−4 molL−1
22.50×10−4 molL−1
31.75×10−5 molL−1
4Answer
NTA Answer: Option 4(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Compute Λ_m from κ and C. Or compute Λ°_m of weak electrolyte from strong electrolyte values via Kohlrausch.
Direct ApplicationMedium
Common distractors
inverts c multiplier
Uses Λ = κC instead of κ/C
forgets 1000 factor
Drops 1000 in unit conversion
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.14
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