JEE Main 2025 (Online) 3rd April Morning Shift | Electrochemistry Question 6 | Chemistry

1

JEE Main 2025 (Online) 3rd April Morning Shift

MCQ (Single Correct Answer)

+4

-1

Correct order of limiting molar conductivity for cations in water at 298 K is :

A

$\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}$

B

$\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$

C

$\mathrm{Mg}^{2+}>\mathrm{H}^{+}>\mathrm{Ca}^{2+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$

D

$\mathrm{H}^{+}>\mathrm{Na}^{+}>\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{K}^{+}$

2

JEE Main 2025 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

+4

-1

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) Inverters	(III) Anode - Zn; Cathode - Carbon
(D) Apollo space ship	(IV) Anode - Pb; Cathode - Pb \| PbO₂

Choose the correct answer from the options given below :

A

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

B

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

C

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

D

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

3

JEE Main 2025 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

+4

-1

$\mathrm{O}_2$ gas will be evolved as a product of electrolysis of :

(A) an aqueous solution of $\mathrm{AgNO}_3$ using silver electrodes.

(B) an aqueous solution of $\mathrm{AgNO}_3$ using platinum electrodes.

(C) a dilute solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.

(D) a high concentration solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes.

Choose the correct answer from the options given below :

A

(B) and (C) only

B

(B) and (D) only

C

(A) and (D) only

D

$(A)$ and $(C)$ only

4

JEE Main 2025 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

+4

-1

For a Mg | Mg²⁺ (aq) || Ag⁺ (aq) | Ag the correct Nernst Equation is :

A

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}$

B

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]}$

C

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Mg}^{2+}\right]}$

D

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}+\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Ag}^{+}\right]^2}{\left[\mathrm{Mg}^{2+}\right]}$