Match List-I with List-II: List-I (Applications) List-II (Batteries/Cell) (A) Transistors (I) Anode - Zn/Hg; Cathode - HgO + C (B) Hearing aids (II) Hydrogen fuel cell (C) Invertors (III) Anode - Zn; Cathode - Carbon (D) Apollo space ship (IV) Anode - Pb; Cathode - Pb | PbO2 Choose the correct answer from the options given below:

Matching applications to their respective electrochemical cells: * Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with MnO2 ightarrow (III). * Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C ightarrow (I). * Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO2 ightarrow (IV). * Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells ightarrow (II). Space missions universally trigger fuel cell pairs in standard test patterns due to the secondary requirement of gathering pure drinking water byproduct. Class 12 Chemistry: Electrochemistry

Match List-I with List-II: List-I (Applications) List-II (Batteries/Cell) (A) Transistors (I) Anode - Zn/Hg; Cathode - HgO + C (B) Hearing aids (II) Hydrogen fuel cell (C) Invertors (III) Anode - Zn;

Matching applications to their respective electrochemical cells: * Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with MnO2 ightarrow (III). * Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C ightarrow (I). * Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO2 ightarrow (IV). * Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells ig

Match List-I with List-II: List-I (Applications) List-II (Batteries/Cell) (A) Transistors (I) Anode - Zn/Hg; Cathode - HgO + C (B) Hearing aids (II) Hydrogen fuel cell (C) Invertors (III) Anode - Zn; Cathode - Carbon (D) Apollo space ship (IV) Anode - Pb; Cathode - Pb | PbO2 Choose the correct answer from the options given below:

Matching applications to their respective electrochemical cells: * Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with MnO2 ightarrow (III). * Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C ightarrow (I). * Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO2 ightarrow (IV). * Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells ightarrow (II). Space missions universally trigger fuel cell pairs in standard test patterns due to the secondary requirement of gathering pure drinking water byproduct. Class 12 Chemistry: Electrochemistry

Match List-I with List-II: List-I (Applications) List-II (Batteries/Cell) (A) Transistors (I) Anode - Zn/Hg; Cathode - HgO + C (B) Hearing aids (II) Hydrogen fuel cell (C) Invertors (III) Anode - Zn;

Matching applications to their respective electrochemical cells: * Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with MnO2 ightarrow (III). * Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C ightarrow (I). * Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO2 ightarrow (IV). * Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells ig

Match List-I with List-II: List-I (Applications) List-II (Batteries/Cell) (A) Transistors (I) Anode

Year	2026	2025	2024	Total
Questions	13	19	8	40

List-I (Applications)	List-II (Batteries/Cell)
(A) Transistors	(I) Anode - Zn/Hg; Cathode - HgO + C
(B) Hearing aids	(II) Hydrogen fuel cell
(C) Invertors	(III) Anode - Zn; Cathode - Carbon
(D) Apollo space ship	(IV) Anode - Pb; Cathode - Pb \| $PbO₂$ $PbO_2$

Q26 jee_main_2025_03_april_evening Conductometric Titrations

40~mL

of a mixture of

CH₃COOH

and

HCl

(aqueous solution) is titrated against

0.1~M~NaOH

solution conductometrically. Which of the following statements is correct?

Conductometric titration curve for Q26 - JEE Main 2025 Evening — Conductance vs Volume of NaOH added curve showing two equivalence points at 2.0 mL and 5.0 mL.

A. The concentration of $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$ in the original mixture is $0.005~M$ $0.005\mathrm{~M}$
B. The concentration of $HCl$ $\mathrm{HCl}$ in the original mixture is $0.005~M$ $0.005\mathrm{~M}$
C. $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$ is neutralised first followed by neutralisation of $HCl$ $\mathrm{HCl}$
D. Point 'C' indicates the complete neutralisation of $HCl$ $\mathrm{HCl}$

Solution

Related Formula

At the equivalence point during titration:

Macid Vacid = Mbase Vbase

Core Logic

In a mixture of a strong acid ( $HCl$ $\mathrm{HCl}$) and a weak acid ( $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$):

$HCl$ $\mathrm{HCl}$ is a strong acid and is completely ionized. When $NaOH$ $\mathrm{NaOH}$ is added, highly mobile $H^+$ $\mathrm{H}^+$ ions are replaced by less mobile $Na^+$ $\mathrm{Na}^+$ ions, causing a sharp drop in conductance (segment AB).
At point B ( $2.0~mL$ $2.0\mathrm{~mL}$), $HCl$ $\mathrm{HCl}$ is completely neutralized.
Segment BC represents the neutralization of the weak acid $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$ to form highly conducting sodium acetate, causing a moderate rise in conductance up to point C ( $5.0~mL$ $5.0\mathrm{~mL}$).
Beyond point C, excess $OH^-$ $\mathrm{OH}^-$ ions cause a rapid rise in conductance (segment CD).

Step 1: Calculate concentration of $HCl$ $\mathrm{HCl}$

Volume of $NaOH$ $\mathrm{NaOH}$ used to neutralize $HCl$ $\mathrm{HCl}$ is $V₁ = 2.0~mL$ $V_1 = 2.0\mathrm{~mL}$:

MHCl \times 40~mL = 0.1~M \times 2.0~mL

MHCl = (0.2)/(40) = 0.005~M

Step 2: Calculate concentration of $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$

Volume of $NaOH$ $\mathrm{NaOH}$ used to neutralize $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$ is $V₂ = 5.0~mL - 2.0~mL = 3.0~mL$ :

MCH₃COOH \times 40~mL = 0.1~M \times 3.0~mL

MCH₃COOH = (0.3)/(40) = 0.0075~M

Pattern Recognition

Conductometric titration curves are analyzed sequentially: the strongest electrolyte is always neutralized first. A steep drop in conductance always signals the neutralization of a strong acid ( $H^+$ $\mathrm{H}^+$ depletion). A weak acid titration shows a gentle upward slope due to salt formation.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Q34 jee_main_2025_03_april_evening Fuel Cells and Standard Cell Potential

The standard cell potential

(Ecell)

of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as

1.21~V

. The standard half cell reduction potential for

O₂

(EO₂/H₂O°)

is

1.229~V

. Choose the correct statement:

A. The standard half cell reduction potential for the reduction of $CO₂$ $\mathrm{CO}_2$ $(ECO₂/CH₃OH°)$ is $19~mV$ $19\mathrm{~mV}$
B. Oxygen is formed at the anode.
C. Reactants are fed at one go to each electrode.
D. Reduction of methanol takes place at the cathode.

Solution

Related Formula

Standard cell EMF is related to standard reduction potentials:

Ecell° = Ecathode° - Eanode°

Core Logic

In a methanol-oxygen fuel cell:

Anode reaction (Oxidation): Methanol is oxidized to carbon dioxide:

CH₃OH + H₂O arrow CO₂ + 6H^+ + 6e^-

Cathode reaction (Reduction): Oxygen is reduced to water:

O₂ + 4H^+ + 4e^- arrow 2H₂O

Hence, cathode is the oxygen electrode, and anode is the methanol electrode.

Step 1: Calculate Standard Reduction Potential of Anode

Using the EMF equation:

1.21~V = 1.229~V - Eanode°

Eanode° = 1.229 - 1.21 = 0.019~V = 19~mV

The standard half-cell reduction potential for the $CO₂/CH₃OH$ couple is $19~mV$ $19\mathrm{~mV}$, matching Option (1).

Pattern Recognition

Fuel cells are galvanic cells where reactants (like fuels and oxidants) are fed continuously to the electrodes, not at one go. Oxidation always occurs at the anode (methanol) and reduction at the cathode (oxygen).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q35 jee_main_2025_07_april_morning Kohlrausch's Law

Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate. Statement II: If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are

x₁

,

x₂

and

x₃

S cm² mol⁻¹

, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by

x₁ + x₂ + 2x₃

. In the light of the given statements, choose the correct answer from the options given below:

A. $Both Statement I and Statement II are false$
B. $Statement I is false but Statement II is true$
C. $Statement I is true but Statement II is false$
D. $Both Statement I and Statement II are true$

Solution

Related Formula

λm\infty = ν_+ λ_+\infty + ν_- λ_-\infty

Core Logic

Statement I: Mohr's salt is a double salt with chemical formula:

FeSO₄ \cdot (NH₄)₂SO₄ \cdot 6H₂O

When dissolved in water, it completely dissociates into three distinct ionic species:

Fe²⁺ (ferrous), NH₄^+ (ammonium), and SO₄²⁻ (sulphate)

Thus, Statement I is true.

Statement II: According to Kohlrausch's law of independent migration of ions:

λm\infty(Mohr's Salt) = 1 \cdot λm\infty(Fe²⁺) + 2 \cdot λm\infty(NH₄^+) + 2 \cdot λm\infty(SO₄²⁻)

λm\infty = x₁ + 2x₂ + 2x₃

Statement II claims the expression is $x₁ + x₂ + 2x₃$ $x_1 + x_2 + 2x_3$ (missing the coefficient $2$ $2$ for ammonium). Thus, Statement II is false.

Pattern Recognition

Kohlrausch's law matches stoichiometric coefficients directly to the ion quantities released. Mohr's salt formula contains $(NH₄)₂$ $(NH_4)_2$, requiring a multiplier of $2$ $2$ for ammonium ion conductance.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: d- and f-Block Elements

Q48 jee_main_2025_07_april_morning Nernst Equation

1 Faraday electricity was passed through

Cu²⁺

(1.5 M, 1 L)/Cu and 0.1 Faraday was passed through

Ag⁺

(0.2 M, 1 L)/Ag electrolytic cells. After this, the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is ______ V.

Galvanic cell assembly diagram with salt bridge for Q48 — The cell assembly combines Cu and Ag half cells after individual initial electrolysis modifications.

Given:

ECu²⁺/Cu° = 0.34 ~V

EAg⁺/Ag° = 0.8 ~V

(2.303RT)/(F) = 0.06 ~V

Numerical Answer. Answer: 0.4 to 0.4

Solution

Related Formula

Ecell = E°cell - (0.06)/(n) Q

Core Logic

First, analyze the electrolysis step to determine final ionic concentrations:

For $Cu²⁺/Cu$ $\mathrm{Cu}^{2+}/\mathrm{Cu}$ half-cell:
Initial moles of $Cu²⁺ = 1.5 M \times 1 L = 1.5 mol$ .
Reductive half-reaction: $Cu²⁺ + 2e^- arrow Cu$ .
Passing $1 Faraday$ $1 \text{ Faraday}$ converts: $(1)/(2) = 0.5 mol$ $\frac{1}{2} = 0.5 \text{ mol}$ of $Cu²⁺$ $\mathrm{Cu}^{2+}$.
Remaining moles of $Cu²⁺ = 1.5 - 0.5 = 1.0 mol$ .
Final concentration $[Cu²⁺] = 1.0 M$ $[\mathrm{Cu}^{2+}] = 1.0 \text{ M}$.
For $Ag⁺/Ag$ $\mathrm{Ag}^{+}/\mathrm{Ag}$ half-cell:
Initial moles of $Ag^+ = 0.2 M \times 1 L = 0.2 mol$ .
Reductive half-reaction: $Ag^+ + e^- arrow Ag$ .
Passing $0.1 Faraday$ $0.1 \text{ Faraday}$ converts: $0.1 mol$ $0.1 \text{ mol}$ of $Ag^+$ $\mathrm{Ag}^+$.
Remaining moles of $Ag^+ = 0.2 - 0.1 = 0.1 mol$ .
Final concentration $[Ag^+] = 0.1 M$ $[\mathrm{Ag}^+] = 0.1 \text{ M}$.

Now, connect the two components into a galvanic cell:

Anode reaction: $Cu(s) arrow Cu²⁺(aq) + 2e^-$
Cathode reaction: $2Ag⁺(aq) + 2e^- arrow 2Ag(s)$
Net cell reaction: $Cu(s) + 2Ag⁺(aq) arrow Cu²⁺(aq) + 2Ag(s)$
$n = 2$ $n = 2$

Calculate standard cell potential:

E°cell = E°Ag^+/Ag - E°Cu²⁺/Cu = 0.80 - 0.34 = 0.46 V

Applying Nernst Equation:

Ecell = E°cell - (0.06)/(2) ( [Cu²⁺][Ag^+]² )

Ecell = 0.46 - 0.03 ( (1)/((0.1)²) ) = 0.46 - 0.03 (100)

Ecell = 0.46 - 0.03(2) = 0.46 - 0.06 = 0.40 V

(Note: The potential is $0.4 V$ $0.4\text{ V}$ or $400 mV$ $400\text{ mV}$).

Pattern Recognition

Electrolysis modifies the bulk concentrations. First, use Faraday's laws to get the new concentration values ( $[Cu²⁺] = 1.0 M$ $[Cu^{2+}] = 1.0\text{ M}$, $[Ag^+] = 0.1 M$ $[Ag^+] = 0.1\text{ M}$). Then plug these straight into standard Nernst equations.

Evaluation Rubric / Model Answer

Requires complete calculations showing concentrations updated by electrolysis, followed by a double-transfer Nernst equation calculation.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q50 jee_main_2025_08_april_evening Nernst Equation

Consider the following half-cell reduction reaction:

Cr₂O₇²⁻(aq) + 6e^- + 14H^+(aq) 2Cr³⁺(aq) + 7H₂O(l)

The process is conducted with a concentration ratio of

[Cr³⁺]²[Cr₂O₇²⁻] = 10⁻⁶

. The specific pH value at which the EMF (

E

) of this reduction half-cell becomes exactly zero is _________ (as the nearest integer value). Given parameters:

E^°Cr₂O₇²⁻/Cr³⁺ = 1.33 V

and

(2.303RT)/(F) = 0.059 V

.

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

The Nernst equation for a reduction half-cell is:

E = E^° - (2.303RT)/(nF) Q

For this reaction, the reaction quotient $Q$ $Q$ is:

Q = [Cr³⁺]²[Cr₂O₇²⁻] \cdot [H^+]¹⁴

Execution

Step 1: Identify the number of transferred electrons ( $n = 6$ $n = 6$) and substitute the condition $E = 0$ $E = 0$:

0 = 1.33 - (0.059)/(6) ( 10⁻⁶[H^+]¹⁴ )

Step 2: Isolate the logarithmic term:

1.33 = (0.059)/(6) [ (10⁻⁶) - ([H^+]¹⁴) ]

(1.33 \times 6)/(0.059) = -6 - 14 [H^+]

Step 3: Perform the arithmetic division:

135.254 = -6 - 14 [H^+]

Step 4: Rearrange the terms using the definition of pH ( $- [H^+] = pH$ $-\log[\text{H}^+] = \text{pH}$):

135.254 + 6 = 14 \cdot pH

141.254 = 14 \cdot pH

pH = (141.254)/(14) = 10.089

Rounding to the nearest integer value gives 10.

Pattern Recognition

The exponent of the hydrogen ion concentration ( $[H^+]¹⁴$ $[\text{H}^+]^{14}$) heavily influences the cell potential. A small shift in pH causes a large change in EMF due to this factor of 14, which explains why the potential drops to zero even in a highly basic environment ( $pH \approx 10$ $\text{pH} \approx 10$).

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Ionic Equilibrium

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculate concentration of HCl$\mathrm{HCl}$

Step 2: Calculate concentration of CH₃COOH$\mathrm{CH}_3\mathrm{COOH}$

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Calculate Standard Reduction Potential of Anode

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

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Step 1: Calculate concentration of $HCl$ $\mathrm{HCl}$

Step 2: Calculate concentration of $CH₃COOH$ $\mathrm{CH}_3\mathrm{COOH}$