NEET ChemistryUpdated 19 Oct 2026

Electrochemistry: NEET PYQs and weightage

Electrochemistry carries 4.2 of 100 Chemistry marks in NEET by past-paper share: 30 of the 707 tagged NEET Chemistry PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 8 of 26 Chemistry chapters.

Take the 25-question Electrochemistry test Formula sheet ↓All NEET Chemistry chapters

4.2/100 Weight in Chemistry

Exam importance (3/5, by past-paper share)

30 NEET PYQs

453 Practice questions

427 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Chemistry PYQs in the bank (30 of 707), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Electrochemistry: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Standard cell emf
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = E_{right}^{\circ} - E_{left}^{\circ} (both taken as reduction potentials)$
Use it when: you have two half-cells from the table and need the cell voltage; positive $E_{cell}^{\circ}$ gives $Δ_{r} G^{\circ} = - n F E_{cell}^{\circ} < 0$ , so the reaction is spontaneous under standard conditions (unit concentration of every species: 1 M for ions, 1 bar for gases such as $H_{2}$ in the SHE, pure solids and liquids as they are); at other concentrations, the sign of $E_{cell}$ from the Nernst equation decides
Nernst equation (cell)
$E_{cell} = E_{cell}^{\circ} - \frac{R T}{n F} ln Q = E_{cell}^{\circ} - \frac{0.059}{n} lo g Q (at 298 K)$
Use it when: concentrations are not 1 M; $Q = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ , products over reactants, with pure solids taken as 1
Nernst equation (single electrode)
$E_{M^{n +} /M} = E_{M^{n +} /M}^{\circ} - \frac{0.059}{n} lo g \frac{1}{[ M ^{n +} ]} (at 298 K)$
Use it when: one electrode is not standard, e.g. a hydrogen electrode at a given pH
Gibbs energy from emf
$Δ_{r} G = - n F E_{cell}$
$Δ_{r} G^{\circ} = - n F E_{cell}^{\circ}$
Use it when: $E$ is given and the question asks for $Δ G$ ; put $E$ in V and $F$ in $C mol^{- 1}$ to get $J mol^{- 1}$ , then divide by 1000 for kJ
Equilibrium constant from emf
$E_{cell}^{\circ} = \frac{2.303 R T}{n F} lo g K_{c} = \frac{0.059}{n} lo g K_{c} (at 298 K)$
Use it when: $E_{cell}^{\circ}$ is given and the question asks for $K$ (at equilibrium $E_{cell} = 0$ )
Gibbs energy and K
$Δ_{r} G^{\circ} = - R T ln K = - 2.303 R T lo g K$
Use it when: you need to go from $Δ G^{\circ}$ to $K$ or back, with $R = 8.314 J K^{- 1} mol^{- 1}$
Faraday's first law
$Q = I t$
$w = \frac{M \times I \times t}{n \times F}$
$w$ = mass deposited
Use it when: a current flows for a time and the question asks for the mass, the moles or the charge; $t$ must be in seconds, $F \approx 96500 C mol^{- 1}$
Faraday's second law
$\frac{w _{1}}{w _{2}} = \frac{E _{1}}{E _{2}}$
$E = \frac{atomic mass}{electrons needed per ion}$
$E$ = equivalent weight
Use it when: the same charge passes through cells connected in series
Resistance and conductance
$R = ρ \frac{l}{A}$
$G = \frac{1}{R} = κ \frac{A}{l}$
$κ = \frac{1}{ρ}$
Use it when: the dimensions of the solution column are given; $1 S cm^{- 1} = 100 S m^{- 1}$
Cell constant
$G^{*} = \frac{l}{A} = R \times κ$
$κ = \frac{G ^{*}}{R}$
Use it when: a cell is calibrated with $KCl$ of known $κ$ and then reused with another solution
Molar conductivity
$Λ_{m} = \frac{κ}{c}$
$Λ_{m} (S cm^{2} mol^{- 1}) = \frac{κ ( S cm ^{- 1} ) \times 1000}{molarity ( mol L ^{- 1} )}$
Use it when: you have $κ$ and the concentration; $1 S m^{2} mol^{- 1} = 1 0^{4} S cm^{2} mol^{- 1}$
Kohlrausch's law
$Λ_{m}^{\circ} = ν_{+} λ_{+}^{\circ} + ν_{-} λ_{-}^{\circ} (example: Λ_{m}^{\circ} (CaC l_{2}) = λ^{\circ} (C a^{2 +}) + 2 λ^{\circ} (C l^{-}))$
Use it when: you need $Λ_{m}^{\circ}$ of any electrolyte, e.g. $Λ_{m}^{\circ} (HAc) = Λ_{m}^{\circ} (HCl) + Λ_{m}^{\circ} (NaAc) - Λ_{m}^{\circ} (NaCl)$
Weak electrolyte
$α = \frac{Λ _{m}}{Λ _{m}^{\circ}}$
$K_{a} = \frac{c α ^{2}}{1 - α} = \frac{c Λ _{m}^{2}}{Λ _{m}^{\circ} ( Λ _{m}^{\circ} - Λ _{m} )}$
Use it when: a weak acid's $Λ_{m}$ at concentration $c$ and its $Λ_{m}^{\circ}$ are given and the question asks for $α$ or $K_{a}$
Strong electrolyte
$Λ_{m} = Λ_{m}^{\circ} - A c^{1/2} (plot of Λ_{m} against c is a straight line, intercept Λ_{m}^{\circ}, slope - A)$
Use it when: you need $Λ_{m}^{\circ}$ of a strong electrolyte by extrapolating to $c = 0$ ; this does not work for weak electrolytes

Trap

E° does not change when you multiply a half-reaction by a number. Cell potential is intensive, so never multiply it by stoichiometric coefficients. ΔG is extensive: writing the reaction twice doubles n and therefore doubles ΔG, but E stays the same.

How does NEET ask Electrochemistry?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1ChemistryNEET UG PYQFaraday's Laws of Electrolysis

Try right here · No sign-in

The weight of Silver (At. Wt. = 108) displaced by a quantity of electricity which displaces 5600 ml of O

_{2}

at STP will be

Answer key (B) & worked solution

Correct answer: Option B

eqts of Ag = eqts of O

_{2}

\frac{w}{108} = \frac{5.6}{5.6}

w = 108 g

Q2ChemistryNEET UG PYQNernst Equation and Its Applications

In the electrochemical cell: Zn|ZnSO

_{4}

(0.01 M) ||CuSO

_{4}

(1.0M) | Cu, the emf of this Daniel cell is E

_{1}

. When the concentration of ZnSO

_{4}

is changed to 1.0 M and that of CuSO

_{4}

changed to 0.01 M, the emf changes to E

_{2}

. From the followings, which one is the relationship between E

_{1}

and E

_{2}

(G i v e n, \frac{R T}{F} = 0.059)

_{1}

> E

_{2}

E_{2} = 0 \neq = E_{1}

_{1}

= E

_{2}

_{1}

< E

_{2}

Answer key (A) & worked solution

Correct answer: Option A

Z n ∣ Z n S O_{4} (0.01 M) ∣ ∣ C u S O_{4} (1.0 M) ∣ C u

∴ E_{1} = E_{ce l l}^{o} - \frac{2.303 R T}{2 \times F} \times lo g \frac{( 0.01 )}{1}

When concentrations are changed

∴ E_{2} = E_{ce l l}^{o} - \frac{2.303}{2 F} \times lo g \frac{1}{0.01}

i.e.,

E_{1} > E_{2}

Q3ChemistryNEET UG PYQFaraday's Laws of Electrolysis

Match List-I with List-II List-I (Conversion)List-II (No. of Faraday required)A. 1 mol of H

_{2}

O to O

_{2}

I. 3FB. 1 mol of MnO

_{4}

– to Mn

^{2 +}

II. 2FC. 1.5 mol of Ca from molten CaCl

_{2}

III. 1FD. 1 mol of FeO to Fe

_{2}

_{3}

IV. 5F Choose the correct answer from the options given below:

AA-III, B-IV, C-I, D-II

BA-II, B-III, C-I, D-IV

CA-III, B-IV, C-II, D-I

DA-II, B-IV, C-I, D-III

Answer key (D) & worked solution

Correct answer: Option D

(d) A-II, B-IV, C-I, D-III

Q4ChemistryNEET UG PYQNernst Equation and Its Applications

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be

A0.059 V

B0.59 V

C0.118 V

D1.18 V

Answer key (B) & worked solution

Correct answer: Option B

The reaction can be represented as

\frac{1}{2} H_{2} (g) \to H^{+} + e^{-}

E_{o x} = E_{o x i}^{0} - \frac{0.059}{n} lo g \frac{[ H ^{+} ]}{( P _{H_{2}} ) ^{1/2}}

Since p H = 10, [H^{+}] = 1 0^{- 10} M

S o, E_{o x} = 0 - \frac{0.059}{1} lo g \frac{10 ^{- 10}}{( 1 ) ^{1/2}}

E_{o x} = 0.59 V

Q5ChemistryNEET UG PYQNernst Equation and Its Applications

The pressure of H

_{2}

required to make the potential of H

_{2}

– electrode zero in pure water at 298 K is :

A10

^{- 4}

atm

B10

^{- 14}

atm

C10

^{- 12}

atm

D10

^{- 10}

atm

Answer key (B) & worked solution

Correct answer: Option B

2 H_{(a q .)}^{+}

+ 2e

^{-}

→ H

_{2}

(reduction reaction) As pure water has [H

^{+}

] = [

^{-}

OH] = 10

^{- 7}

E = E^{0} - \frac{0.059}{2} lo g \frac{P _{H_{2}}}{[ 10 ^{- 7} ] ^{2}}

= 0 - \frac{0.059}{2} lo g \frac{P _{H_{2}}}{[ 10 ^{- 7} ] ^{2}}

As pure water has [H

^{+}

] = [

^{-}

OH] = 10

^{- 7}

So as to make E = 0,

lo g \frac{P _{H_{2}}}{[ H ^{+} ] ^{2}} = 0

Which is only possible when P

_{H 2}

= 10

^{-}

^{14}

So log

lo g \frac{P _{H_{2}}}{[ H ^{+} ] ^{2}} = lo g \frac{10 ^{- 14}}{10 ^{- 14}} = lo g 1 = 0

Q6ChemistryNEET UG PYQFaraday's Laws of Electrolysis

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 : 63g mol

^{- 1}

, 1F = 96487C)

A0.315 g

B31.5g

C0.0315 g

D3.15 g

Answer key (A) & worked solution

Correct answer: Option A

The reaction is Cu

^{+ 2}

+ 2e– → Cu Using Faraday’ law Moles × nf =

\frac{i t}{F}

\frac{W}{63} \times 2 = \frac{9.6487 \times 100}{96487}

W = \frac{0.63}{2} = 0.315 g

Hence, correct answer is (a)

Q7ChemistryNEET UG PYQFaraday's Laws of Electrolysis

When 0.1 mol

M n O_{4}^{2 -}

is oxidised, the quantity of electricity required to completely oxidise

M n O_{4}^{2 -}

M n O_{4}^{-}

A9650 C

B96.50 C

C96500 C

D2 × 96500 C

Answer key (A) & worked solution

Correct answer: Option A

M + 6 n O_{4}^{- 2} \to M + 7 n O_{4}^{-} + e^{-}

1 equivalent requires = 1F Valency factor = 1 ⇒ 1 mol requires = 1F = 96500 C 0.1 mol requires = 0.1 F = 9650 C

Q8ChemistryNEET UG PYQRelationship Between Cell Potential and Gibbs' Energy Change

If the

E_{ce l l}^{\circ}

for a given reaction has a negative value, which of the following gives the correct reactionships for the values of

Δ G^{\circ}

and

K_{e q}

Δ G^{\circ} > 0; K_{e q} > 1

Δ G^{\circ} 1

Δ G^{\circ} < 0; K_{e q} < 1

Δ G^{\circ} > 0; K_{e q} < 1

Answer key (D) & worked solution

Correct answer: Option D

E_{ce l l}^{\circ}

∴ Δ G^{\circ} = - n F E_{ce l l}^{\circ}

Δ G^{\circ} = + v e \Rightarrow Δ G^{\circ} > 0

∵ Δ G^{\circ} = - 2.303 R T lo g K_{e q}

∴ K_{e q} < 1

Which topics in Electrochemistry carry the most questions?

The 12 busiest of 20 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Faraday's Laws of Electrolysis	8	60
Nernst Equation and Its Applications	6	47
Relationship Between Cell Potential and Gibbs' Energy Change	4	27
Conductance and Conductivity	3	63
Electrochemical Series and Its Applications	3	53
Kohlrausch's Law and Its Applications	2	40
Electrode Potentials, Including Standard Electrode Potential	1	30
Electrolysis and Its Applications	1	19
Emf of a Galvanic Cell and Its Measurement	1	13
Fuel Cells	1	8
Galvanic Cell and Its Construction	0	18
Batteries	0	16

Why does Electrochemistry show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Electrochemistry, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Electrochemistry?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Electrochemistry test

Chapters either side of Electrochemistry by weight

Chemical Kinetics Weight 4.5/100 d- and f-Block Elements Weight 4.4/100 Thermodynamics Weight 4.2/100 Amines Weight 4.1/100

Electrochemistry: NEET PYQs and weightage

Electrochemistry: formula sheet for NEET

Standard cell emf

Nernst equation (cell)

Nernst equation (single electrode)

Gibbs energy from emf

Equilibrium constant from emf

Gibbs energy and K

Faraday's first law

Faraday's second law

Resistance and conductance

Cell constant

Molar conductivity

Kohlrausch's law

Weak electrolyte

Strong electrolyte

How does NEET ask Electrochemistry?

Which topics in Electrochemistry carry the most questions?

Why does Electrochemistry show "–" until you answer 3 questions?

What happens when you test Electrochemistry?

Chapters either side of Electrochemistry by weight