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 \| $PbO₂$ $PbO_2$

Q36 jee_main_2025_04_april_morning Batteries and Fuel Cells

On charging the lead storage battery, the oxidation state of lead changes from

x₁

to

y₁

at the anode and from

x₂

to

y₂

at the cathode. The values of

x₁, y₁, x₂, y₂

are respectively:

A. $+4, +2, 0, +2$ $+4, +2, 0, +2$
B. $+2, 0, +2, +4$ $+2, 0, +2, +4$
C. $0, +2, +4, +2$ $0, +2, +4, +2$
D. $+2, 0, 0, +4$ $+2, 0, 0, +4$

Solution

Related Formula

Net Charging Reaction: 2PbSO₄(s) + 2H₂O(l) arrow Pb(s) + PbO₂(s) + 2H₂SO₄(aq)

Core Logic

During the charging cycle, the discharge chemical reactions are driven in reverse:

At Anode: Lead sulfate ( $PbSO₄$ $PbSO_4$, where Lead is $+2$ $+2$) is reduced back to metallic lead ( $Pb$ $Pb$, oxidation state $0$ $0$):

x₁ = +2 arrow y₁ = 0

At Cathode: Lead sulfate ( $PbSO₄$ $PbSO_4$, where Lead is $+2$ $+2$) is oxidized back into lead dioxide ( $PbO₂$ $PbO_2$, where Lead is $+4$ $+4$):

x₂ = +2 arrow y₂ = +4

Thus, the values are $+2, 0, +2, +4$ $+2, 0, +2, +4$.

Pattern Recognition

Be careful with wording! Discharging consumes $Pb$ $Pb$ and $PbO₂$ $PbO_2$ to create $PbSO₄$ $PbSO_4$. Charging does the exact opposite, converting $PbSO₄$ $PbSO_4$ ( $+2$ $+2$) back into its parent elements ( $0$ $0$ and $+4$ $+4$).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q45 jee_main_2025_07_april_evening Electrolysis and Discharge Potential

Given below are two statements:

1 M

aqueous solution of each of

Cu(NO₃)₂

,

AgNO₃

,

Hg₂(NO₃)₂

;

Mg(NO₃)₂

are electrolysed using inert electrodes, Given:

EAg⁺/Agθ = 0.80V

,

EHg₂²⁺/Hgθ = 0.79V

,

ECu²⁺/Cuθ = 0.24V

and

EMg²⁺/Mgθ = -2.37V

Statement (I): With increasing voltage, the sequence of deposition of metals on the cathode will be

Ag

,

Hg

and

Cu

Statement (II): Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode. In the light of the above statement, choose the most appropriate answer from the options given below [cite: 426, 427]

A. $Both statement I and statement II are incorrect$
B. $Statement I is correct but statement II is incorrect$
C. $Both statement I and statement II are correct$
D. $Statement I is incorrect but statement II is correct$

Solution

Related Formula

Ease of discharge at Cathode \propto Standard Reduction Potential (E⁰)

Core Logic

At the cathode, the metal ion with the highest standard reduction potential ( $E⁰$ $E^0$) gets reduced and deposited first. Arranging the given potentials:

E⁰Ag^+/Ag (0.80V) > E⁰Hg₂²⁺/Hg (0.79V) > E⁰Cu²⁺/Cu (0.24V)

Thus, deposition follows the order $Ag arrow Hg arrow Cu$ as voltage is steadily increased, confirming Statement I.

For $Mg²⁺$ $\text{Mg}^{2+}$, its reduction potential is highly negative ( $-2.37 V$ $-2.37\text{ V}$), much lower than that of water ( $-0.83 V$ $-0.83\text{ V}$). Consequently, water undergoes reduction at the cathode instead of magnesium:

2H₂O + 2e^- arrow H₂(g) + 2OH^-

This results in the evolution of Hydrogen gas at the cathode, not oxygen gas. Oxygen gas is evolved at the anode via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044]

Step 1: Conclusion Match

Since Statement I is correct and Statement II is incorrect, we select option (2).

Pattern Recognition

Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and $Al$ $\text{Al}$), $H₂$ $\text{H}_2$ gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas ( $O₂$ $\text{O}_2$) is an anodic product generated by water oxidation.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q26 jee_main_2025_24_jan_morning Galvanic Cells and Standard Cell Potential

For the given cell:

Fe²⁺(aq) + Ag⁺(aq) arrow Fe³⁺(aq) + Ag(s)

The standard cell potential of the above reaction is given by:

Ag⁺ + e⁻ arrow Ag E⁰ = x V

Fe²⁺ + 2e⁻ arrow Fe E⁰ = y V

Fe³⁺ + 3e⁻ arrow Fe E⁰ = z V

A. $x + y - z$ $x + y - z$
B. $x + 2y - 3z$ $x + 2y - 3z$
C. $y - 2x$ $y - 2x$
D. $x + 2y$ $x + 2y$

Solution

Related Formula

Δ G⁰ = -nFE⁰

Core Logic

Using Gibbs free energy changes for individual steps to find the target reduction potential:

$Ag⁺ + e⁻ arrow Ag Δ G₁⁰ = -1Fx$
$Fe²⁺ + 2e⁻ arrow Fe Δ G₂⁰ = -2Fy$
$Fe³⁺ + 3e⁻ arrow Fe Δ G₃⁰ = -3Fz$

For the conversion of $Fe²⁺ arrow Fe³⁺ + e⁻$ $$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$$, we compute the free energy change as:

Δ G⁰ = Δ G₂⁰ - Δ G₃⁰ = -2Fy - (-3Fz) = 3Fz - 2Fy

Thus, $E⁰Fe²⁺/Fe³⁺ = 2y - 3z$ $$E^{0}_{Fe^{2+}/Fe^{3+}} = 2y - 3z$$

Combining with silver reduction:

E⁰cell = E⁰Ag⁺/Ag + E⁰Fe²⁺/Fe³⁺ = x + 2y - 3z

Step 1: Final Calculation

The overall potential equals $x + 2y - 3z$ $x + 2y - 3z$.

Galvanic Cells and Standard Cell Potential diagram for Q26 - JEE Main 2025 Morning

Pattern Recognition

Direct application of $Δ G⁰$ $\Delta G^0$ summation. Remember that standard cell potentials cannot be added directly unless the number of electrons involved is identical.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q48 jee_main_2025_28_jan_evening Faraday's Laws of Electrolysis

Electrolysis of

600~mL

aqueous solution of NaCl for

5

changes the

pH

of the solution to 12. The current in Amperes used for the given electrolysis is ______ (Nearest integer).

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Faraday's law of electrolysis equation:

Moles of electrons (equivalents) = (I \cdot t)/(F)

Water ion product relation:

pH + pOH = 14

Core Logic

During the electrolysis of brine ( $NaCl(aq)$ $NaCl(aq)$), hydroxide ions ( $OH^-$ $OH^-$) are generated at the cathode:

2H₂O + 2e^- arrow H₂ + 2OH^-

Given metrics:

Final $pH = 12 pOH = 14 - 12 = 2$
$[OH^-] = 10⁻² M$ $[OH^-] = 10^{-2}\mathrm{\ M}$
$Volume = 600 mL = 0.6 L$
$Time = 5 = 300 s$

Step 1: Calculate Moles of Hydroxide Produced

Find the absolute moles of $OH^-$ $OH^-$ ions generated:

Moles = Molarity \times Volume (L) = 10⁻² \times 0.6 = 6 \times 10⁻³ moles

Step 2: Relate to Electrical Current

Since $1$ $1$ mole of electrons produces $1$ $1$ mole of $OH^-$ $OH^-$, the moles of charge equals $6 \times 10⁻³$ $6 \times 10^{-3}$. Applying Faraday's equation:

6 \times 10⁻³ = (I \times 300)/(96500)

I = 6 \times 10⁻³ \times 96500300 = 1.93 A

Rounding to the nearest integer gives $2$ $2$.

Pattern Recognition

Always convert a given $pH$ $\text{pH}$ value into $[OH^-]$ $[OH^-]$ concentration when dealing with cathodic water reduction. Tracking the relationship where $1 e^- \equiv 1 OH^-$ $1\ e^- \equiv 1\ OH^-$ provides a direct shortcut to link $pH$ $\text{pH}$ changes to current flow.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q jee_main_2025_29_jan_morning Standard Reduction Potential and Oxidising Power

The standard reduction potential values of some of the p-block ions are given below. Predict the one with the strongest oxidising capacity.

A. $ESn⁴⁺/Sn²⁺ = +1.15V$
B. $ETl³⁺/Tl = +1.26V$
C. $EAl³⁺/Al = -1.66V$
D. $EPb⁴⁺/Pb²⁺ = +1.67V$

Solution

Related Formula

Oxidising Capacity \propto Standard Reduction Potential (E)

Core Logic

A higher positive value of standard reduction potential ( $E$ $E^{\ominus}$) indicates a stronger tendency to undergo reduction, hence behaving as a stronger oxidising agent. Comparing the given values:

$ESn⁴⁺/Sn²⁺ = +1.15V$
$ETl³⁺/Tl = +1.26V$
$EAl³⁺/Al = -1.66V$
$EPb⁴⁺/Pb²⁺ = +1.67V$

Since $+1.67V$ $+1.67\mathrm{V}$ is the highest value, $Pb⁴⁺$ $\mathrm{Pb}^{4+}$ possesses the strongest oxidising capacity.

Pattern Recognition

Strongest oxidising agent = Most positive reduction potential. Weakest oxidising agent = Most negative reduction potential.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: p-Block Elements

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Moles of Hydroxide Produced

Step 2: Relate to Electrical Current

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Electrochemistry Questions — jee_main_2025_29_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back