Emf Galvanic

8 MCQs1 revision card9-step worked example

Source: NCERT Redox ReactionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The EMF of a galvanic cell is the potential difference between its two electrodes when no current flows. The single formula that governs this topic:

E°_cell = E°_cathode − E°_anode

Both E° values are standard reduction potentials (NCERT Class 12 Chemistry Chapter 2, page 8). This is the load-bearing detail: you subtract the anode's reduction potential from the cathode's reduction potential. You do NOT flip the sign of either value before subtraction — both stay as reduction potentials.

The trap that costs marks: confusing which electrode is cathode and which is anode. In a galvanic cell, the electrode with the higher reduction potential is the cathode (reduction occurs there), and the lower one is the anode (oxidation occurs there). If E°_cell comes out negative, the reaction is non-spontaneous as written — either you assigned cathode/anode backwards, or the cell genuinely doesn't work in that direction.

How NEET tests this: a question gives you two standard reduction potentials and asks for the cell EMF, or gives the cell notation and asks you to identify cathode/anode and compute E°_cell. The common wrong answer comes from subtracting in the wrong order (cathode − anode vs. anode − cathode) or from accidentally using an oxidation potential without flipping sign.

Watch-out: cell notation convention places the anode on the left and cathode on the right: Anode || Cathode. When you read Zn | Zn²⁺ || Cu²⁺ | Cu, zinc is the anode (left) and copper is the cathode (right). E°_cell = E°_Cu − E°_Zn. If a question gives oxidation potentials, convert them to reduction potentials (flip sign) before applying the formula.

A positive E°_cell confirms the cell reaction is spontaneous under standard conditions.

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

In the standard cell notation Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), which electrode is the cathode?

MCQ 2Easy RecallPractice

The EMF of a galvanic cell under standard conditions is defined as:

MCQ 3Easy RecallPractice

In a spontaneous galvanic cell, the standard cell EMF (E°_cell) is:

MCQ 4Direct ApplicationPractice

Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, the standard EMF of the Daniell cell (Zn–Cu cell) is:

MCQ 5Direct ApplicationPractice

E°(Ag⁺/Ag) = +0.80 V and E°(Mg²⁺/Mg) = −2.37 V. What is E°_cell for the galvanic cell Mg | Mg²⁺ || Ag⁺ | Ag?

MCQ 6Direct ApplicationPractice

A galvanic cell has E°_cell = +0.46 V. If the standard reduction potential of the cathode is +0.34 V, the standard reduction potential of the anode is:

MCQ 7Concept TrapPractice

In a galvanic cell, the electrode with the more negative standard reduction potential acts as the:

MCQ 8Concept TrapPractice

A student computes E°_cell for a proposed galvanic cell and obtains −0.15 V. What does this result indicate?

Quick recall before you leave

Worked Example

1
Given
A galvanic cell is set up with the following half-cells: - Fe³⁺/Fe²⁺ electrode: E°(Fe³⁺/Fe²⁺) = +0.77 V - Sn⁴⁺/Sn²⁺ electrode: E°(Sn⁴⁺/Sn²⁺) = +0.15 V Both are standard reduction potentials at 298 K.
2
Required
Calculate the standard EMF of the galvanic cell and write the cell notation.
3
Concept
In a galvanic cell, the electrode with the higher standard reduction potential is the cathode (reduction site), and the lower one is the anode (oxidation site). E°_cell = E°_cathode − E°_anode.
4
Formula
E°_cell = E°_cathode − E°_anode
5
Substitution
Fe³⁺/Fe²⁺ has the higher reduction potential (+0.77 V) → cathode. Sn⁴⁺/Sn²⁺ has the lower reduction potential (+0.15 V) → anode. E°_cell = (+0.77) − (+0.15)
6
Calculation
E°_cell = 0.77 − 0.15 = 0.62 V Note: All given values are reduction potentials with two significant figures. No exact-constant adjustment is needed here — both values participate as measured quantities.
7
Final answer
E°_cell = +0.62 V Cell notation (anode left, cathode right): Sn²⁺ | Sn⁴⁺ || Fe³⁺ | Fe²⁺ (with Pt electrodes implied for both solution-phase couples) The positive value confirms the reaction is spontaneous under standard conditions.
8
Common trap
A frequent error is subtracting in the wrong order: E°_anode − E°_cathode = 0.15 − 0.77 = −0.62 V. The negative sign would wrongly suggest the cell is non-spontaneous. Always subtract anode FROM cathode.
9
Similar NEET-style question
Given E°(Cr³⁺/Cr) = −0.74 V and E°(Ni²⁺/Ni) = −0.25 V, calculate E°_cell for the galvanic cell and identify the anode and cathode. (Answer: Cr is the anode, Ni is the cathode. E°_cell = (−0.25) − (−0.74) = +0.49 V.) ---

Before solving, remember these

Formula

Cell EMF

E°_cell = E°_cathode - E°_anode (both as reduction potentials). E_cell > 0 → spontaneous. Cell notation: anode | anion soln || cation soln | cathode.

-- NCERT Class 12 Chemistry, Ch. 2, p. 8

Formulas

Cell EMF

E_{cell}^{\circ} = E_{cath}^{\circ} - E_{anode}^{\circ}

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

m = \frac{M I t}{n F}

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

Δ G = - n F E

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

Λ_{m} = \frac{1000 κ}{C}

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

E = E^{\circ} - \frac{0.0591}{n} lo g Q

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.

Trap

Nernst n electrons cumulative error

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°.

Trap

KMnO4 oxidation product depends on medium

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.

Mistake

Uses n=1 in Faraday's law for ions like Cu²⁺ (actual n=2) or Al³⁺ (n=3).

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).

Mistake

Uses incorrect electron count n in Nernst equation.

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

If the rate constant of a reaction is 0.03 s–1, how much time does it take for 7.2 mol L–1 concentration of the reactant to get reduced to 0.9 mol L–1? (Given: log 2 = 0.301)

121.0 s

269.3 s

323.1 s

4210 s

NTA Answer: Option 2(final)

NEET 2025

Consider the following compounds : KO , H O and H SO 2 2 2 2 4 The oxidation state of the underlined elements in them are, respectively,

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

Match List-I with List-II List-I List-II (Example) (Type of Solution) A. Humidity I. Solid in solid B. Alloys II. Liquid in gas C. Amalgams III. Solid in gas D. Smoke IV. Liquid in solid Choose the correct answer from the options given below:

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

Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

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

Match List I with List II. List I List II (Conversion) (Number of Faraday required) A. 1 mol of H O to O I. 3F 2 2 B. 1 mol of MnO− 4 to Mn2+ II. 2F C. 1.5 mol of Ca from molten CaCl III. 1F 2 D. 1 mol of FeO to Fe O IV. 5F 2 3 Choose the correct answer from the options given below:

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

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol–1, 1 F = 96487 C)

13.15 g

20.315 g

331.5 g

40.0315 g

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

The products A and B obtained in the following reactions, respectively, are 3ROH + PCl → RCl + A 3 ROH + PCl → RCl + HCl + B 5

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

The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm–1 cm–1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of cell constant is

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

At 298 K, the standard electrode potentials of Cu2+ / Cu, Zn2+ / Zn, Fe2+ / Fe and Ag+ / Ag are 0.34 V, –0.76 V, –0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?

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

Given below are half cell reactions: - 13 - M n O −4 + 8 H + + 5 e − → M n 2 + + 4 H 2 O E ºM n 2 + /M n O −4 = − 1 .5 1 0 V 1 2 O 2 + 2 H + + 2 e − → H 2 O Eº =+1.223V O2/H2O Will the permanganate ion, M n O –4 liberate O from water in the presence of an acid? 2

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

If radius of second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of third Bohr orbit of Li2+ ion?

1158.7 Å

2158.7 pm

315.87 pm

41.587 pm

NTA Answer: Option 2(final)

NEET 2021

Which one among the following is the correct option for right relationship between C and C for one mole P V of ideal gas?

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

The molar conductance of NaCl, HCl and CH COONa 3 at infinite dilution are 126.45, 426.16 and cm2 mol–1 91.0 S respectively. The molar conductance of CH COOH at infinite dilution is. 3 Choose the right option for your answer. cm2 mol–1

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

The molar conductivity of 0.007 M acetic acid is 20 cm2 mol–1. S What is the dissociation constant of acetic acid? Choose the correct option. Λ° =350Scm2 mol−1  H+   Λ° mol−1 =50Scm2  CHCOO−  3 2.50×10−5 molL−1

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.8

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Emf Galvanic

Lesson

Practice MCQs

Quick recall before you leave

Cell EMF formula

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Sources

Free NEET study resources