O2 gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3 using silver electrodes. (B) an aqueous solution of AgNO3 using platinum electrodes. (C) a dilute solution of H2SO4 using platinum electrodes. (D) a high concentration solution of H2SO4 using platinum electrodes. Choose the correct answer from the options given below:

Analyzing anodic reactions during electrolysis: * Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag ightarrow Ag+ + e-). No oxygen is evolved. * Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO3^- ions: * Case (C): In dilute H2SO4, water oxidation takes place, releasing O2 gas at the anode. * Case (D): In concentrated H2SO4, oxidation of SO42- creates peroxodisulphate ions (S2O82-), inhibiting oxygen evolution. Remember that active electrodes participate directly in redox reactions, whereas inert electrodes (Pt, Graphite) yield oxygen gas when water is oxidized in the presence of oxoanions like NO3^- or dilute SO42-. Class 12 Chemistry: Electrochemistry

O2 gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3 using silver electrodes. (B) an aqueous solution of AgNO3 using platinum electrodes. (C) a dilute solution of H

Analyzing anodic reactions during electrolysis: * Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag ightarrow Ag+ + e-). No oxygen is evolved. * Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO3^- ions: * Case (C): In dilute H2SO4, water oxidation takes place, releasing O2 gas at the anode. * Case (D): In concentrated H2SO4, oxidation of SO42- creates peroxodisulphate ions (S2O82-), inhibiting oxygen evolution. Remember

O2 gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3 using silver electrodes. (B) an aqueous solution of AgNO3 using platinum electrodes. (C) a dilute solution of H2SO4 using platinum electrodes. (D) a high concentration solution of H2SO4 using platinum electrodes. Choose the correct answer from the options given below:

Analyzing anodic reactions during electrolysis: * Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag ightarrow Ag+ + e-). No oxygen is evolved. * Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO3^- ions: * Case (C): In dilute H2SO4, water oxidation takes place, releasing O2 gas at the anode. * Case (D): In concentrated H2SO4, oxidation of SO42- creates peroxodisulphate ions (S2O82-), inhibiting oxygen evolution. Remember that active electrodes participate directly in redox reactions, whereas inert electrodes (Pt, Graphite) yield oxygen gas when water is oxidized in the presence of oxoanions like NO3^- or dilute SO42-. Class 12 Chemistry: Electrochemistry

O2 gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3 using silver electrodes. (B) an aqueous solution of AgNO3 using platinum electrodes. (C) a dilute solution of H

Analyzing anodic reactions during electrolysis: * Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag ightarrow Ag+ + e-). No oxygen is evolved. * Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO3^- ions: * Case (C): In dilute H2SO4, water oxidation takes place, releasing O2 gas at the anode. * Case (D): In concentrated H2SO4, oxidation of SO42- creates peroxodisulphate ions (S2O82-), inhibiting oxygen evolution. Remember

O2 gas will be evolved as a product of electrolysis of: (A) an aqueous solution of AgNO3 using silver electrodes. (B)…

Year	2026	2025	2024	Total
Questions	13	19	8	40

Q72 jee_main_2026_21_jan_morning Conductance of Electrolytic Solutions

The pH and conductance of a weak acid (HX) was found to be 5 and

4 \times 10⁻⁵

S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of

1 cm²

were at a distance of

15 cm

apart. The value of the limiting molar conductivity is .....

S m² mol⁻¹

. (nearest integer) (Given: degree of dissociation of the weak acid

(α) ll 1)

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

κ = G \times (l)/(A)

Λm = (κ \times 1000)/(C)

α = ΛmΛm\infty

$[H^+] = Cα$ $[H^+] = C\alpha$

Core Logic

Given: $pH = 5$ $pH = 5$, so $[H^+] = 10⁻⁵ M$ $[H^+] = 10^{-5}\text{ M}$. Since $[H^+] = C \cdot α = C \cdot ΛmΛm\infty$ , we have $10⁻⁵ = C \cdot ΛmΛm\infty$ .

First, calculate conductivity ( $κ$ $\kappa$): Conductance $G = 4 \times 10⁻⁵ S$ $G = 4 \times 10^{-5}\text{ S}$. Cell constant $G^* = (l)/(A) = 15 cm1 cm² = 15 cm⁻¹$ .

κ = G \cdot G^* = (4 \times 10⁻⁵) \times 15 = 6 \times 10⁻⁴ S cm⁻¹

Molar conductivity ( $Λm$ $\Lambda_m$):

Λm = (κ \times 1000)/(C) = 6 \times 10⁻⁴ \times 1000C = (0.6)/(C)

Substitute $Λm$ $\Lambda_m$ into the proton concentration formula:

[H^+] = 10⁻⁵ = C \cdot 0.6 / CΛm\infty

10⁻⁵ = 0.6Λm\infty

Λm\infty = 0.610⁻⁵ = 60000 S cm² mol⁻¹

Convert units to $S m² mol⁻¹$ $\text{S m}^2\text{ mol}^{-1}$: Since $1 m² = 10⁴ cm²$ $1\text{ m}^2 = 10^4\text{ cm}^2$,

Λm\infty = 60000 \times 10⁻⁴ S m² mol⁻¹ = 6 S m² mol⁻¹

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Q67 jee_main_2026_21_jan_evening Daniell Cell and Cell Potential Variation

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?

A. $(1) Plot 1$ $(1) \text{ Plot 1}$
B. $(2) Plot 2$ $(2) \text{ Plot 2}$
C. $(3) Plot 3$ $(3) \text{ Plot 3}$
D. $(4) Plot 4$ $(4) \text{ Plot 4}$

Solution

Core Logic

For a closed circuit Daniell cell operating under given conditions, standard cell potential $Ecell°$ $E_{\text{cell}}^{\circ}$ remains constant with time during discharge until equilibrium or noticeable concentration shifts occur in a specific manner, or cell potential versus time plots reflect constancy of standard electromotive force terms before polarization effects take over.

Step 1: Final Conclusion

Plot 2 accurately depicts the expected trend where $Ecell°$ $E_{\text{cell}}^{\circ}$ remains constant with time.

Pattern Recognition

Sees: graphical representation of electrochemical cell potentials over time. Trap: Confusing standard cell potential $E°$ $E^{\circ}$ with operating cell potential $Ecell$ $E_{\text{cell}}$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q75 jee_main_2026_21_jan_evening Solubility Product and Electrode Potential

MX

is a sparingly soluble salt that follows the given solubility equilibrium at 298 K:

MX(s) leftharpoons M^+(aq) + X^-(aq); Kₛₚ = 10⁻¹⁰

If the standard reduction potential for

M^+(aq) arrow M(s)

is

(EM^+/M) = 0.79 V

, then the value of the standard reduction potential for the metal/metal insoluble salt electrode

EX^-/MX(s)/M

is

____ mV

. (nearest integer) [Given:

(2.303RT)/(F) = 0.059 V

]

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

$EX^-/MX/M° = EM^+/M° + (0.059)/(n) Kₛₚ$ r

Core Logic

Substituting the given values:

EX^-/MX/M° = 0.79 + (0.059)/(1) (10⁻¹⁰)

EX^-/MX/M° = 0.79 + 0.059(-10) = 0.79 - 0.59 = 0.20 V

Step 1: Converting to Millivolts

0.20 V = 200 mV

Pattern Recognition

Sees: relationship between standard reduction potential, solubility product, and metal-insoluble salt electrodes. Trap: Sign or logarithmic base calculation errors.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q72 jee_main_2026_22_january_morning Nernst Equation

Consider the following electrochemical cell at 298K

Pt | HSnO₂⁻(aq) | Sn(OH)₆²⁻(aq) || Bi₂O₃(s) | Bi(s)

If the reaction quotient at a given time is

10⁶

, then the cell EMF

(Ecell)

is

x \times 10⁻¹ V

(Nearest integer). Given the standard half-cell reduction potential as

E_Bi₂O₃/Bi,OH⁻⁰ = -0.44 V and E_Sn(OH)₆²⁻/HSnO₂⁻,OH⁻⁰ = -0.90 V

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Ecell = Ecell° - (0.0591)/(n) Q

(Using 0.06 for approximation if needed, as per standard practice unless specified).

Core Logic

First, calculate $E°cell$ $E^{\circ}_{cell}$:

E°cell = E°cathode - E°anode

E°cell = (-0.44 V) - (-0.90 V) = +0.46 V

Identify the number of transferred electrons ( $n$ $n$) for the overall reaction. Anode reaction: $Sn²⁺ arrow Sn⁴⁺ + 2e⁻$ $Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Tin goes from +2 in $HSnO₂^-$ $HSnO_2^-$ to +4 in $Sn(OH)₆²⁻$ $Sn(OH)_6^{2-}$). Cathode reaction: $Bi³⁺ + 3e⁻ arrow Bi$ $Bi^{3+} + 3e^{-} \rightarrow Bi$ (Bismuth goes from +3 in $Bi₂O₃$ $Bi_2O_3$ to 0). To balance the electrons, we multiply anode by 3 and cathode by 2, yielding $n = 6$ $n = 6$ total electrons transferred.

Step 1: Apply Nernst Equation

Applying the Nernst equation (approximating 0.0591 to 0.06 for simplicity):

Ecell = E°cell - (0.06)/(n) ₁₀ Q

Ecell = 0.46 - (0.06)/(6) ₁₀(10⁶)

Ecell = 0.46 - 0.01 \times 6

Ecell = 0.46 - 0.06 = 0.40 V

Step 2: Format Output

The problem asks for the answer in the form $x \times 10⁻¹ V$ $x \times 10^{-1}\text{ V}$.

0.40 = 4 \times 10⁻¹

So, $x = 4$ $x = 4$.

Pattern Recognition

Be extremely careful finding 'n'. You must balance the half-reactions. A +2 to +4 change (n=2) and a +3 to 0 change (n=3) resolves to a common multiple of 6.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q67 jee_main_2026_22_january_evening Standard Reduction Potentials and Reducing Strength

Consider the following reduction processes:

Al³⁺ + 3e^- arrow Al(s)

,

E⁰ = -1.66 V

Fe³⁺ + e^- arrow Fe²⁺

,

E⁰ = +0.77 V

Co³⁺ + e^- arrow Co²⁺

,

E⁰ = +1.81 V

Cr³⁺ + 3e^- arrow Cr(s)

,

E⁰ = -0.74 V

The tendency to act as reducing agent decreases in the order:

A. $Al > Cr > Fe²⁺ > Co²⁺$
B. $Al > Fe²⁺ > Cr > Co²⁺$
C. $Al > Cr > Co²⁺ > Fe²⁺$
D. $Cr > Fe²⁺ > Al > Co²⁺$

Solution

Related Formula

Reducing Power \propto 1Standard Reduction Potential (E⁰red)

Core Logic

Step 1: List standard reduction potential values:

$E⁰(Al³⁺/Al) = -1.66 V$
$E⁰(Cr³⁺/Cr) = -0.74 V$
$E⁰(Fe³⁺/Fe²⁺) = +0.77 V$
$E⁰(Co³⁺/Co²⁺) = +1.81 V$

Step 2: Arrange species in increasing order of $E⁰red$ $E^0_{red}$ (decreasing reducing power):

Al > Cr > Fe²⁺ > Co²⁺

Pattern Recognition

Sees: Reduction potential values for reducing strength ordering. Shortcut: More negative standard reduction potential $\implies$ stronger reducing agent.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Converting to Millivolts

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Nernst Equation

Step 2: Format Output

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