JEE Main 2025 (Online) 7th April Evening Shift

MCQ (Single Correct Answer)

-1

Given below are two statements :

1 M aqueous solutions of each of Cu(NO₃)₂, AgNO₃, Hg₂(NO₃)₂, Mg(NO₃)₂ are electrolysed using inert electrodes. Given: E⁰_Ag⁺/Ag = 0.80 V, E⁰_{Hg₂²⁺/Hg} = 0.79 V, E⁰_Cu²⁺/Cu = 0.24 V and E⁰_Mg²⁺/Mg = -2.37 V.

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 the cathode instead oxygen gas will be evolved at the cathode.

In the light of the above statements, choose the most appropriate answer from the options given below :

Both Statement I and Statement II are incorrect

Statement I is incorrect but Statement II is correct

Statement I is correct but Statement II is incorrect

Both Statement I and Statement II are correct

JEE Main 2025 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

On charging the lead storage battery, the oxidation state of lead changes from $x_1$ to $y_1$ at the anode and from $x_2$ to $y_2$ at the cathode. The values of $x_1, y_1, x_2, y_2$ are respectively :

$0,+2,+4,+2$

$+2,0,0,+4$

$+4,+2,0,+2$

$+2,0,+2,+4$

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

The standard cell potential $\left(\mathrm{E}_{\text {cell }}^{\ominus}\right)$ 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 $\mathrm{O}_2\left(\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\circ}\right)$ is 1.229 V .

Choose the correct statement:

Reactants are fed at one go to each electrode.

Oxygen is formed at the anode.

The standard half cell reduction potential for the reduction of $\mathrm{CO}_2\left(\mathrm{E}_{\mathrm{CO}_2 / \mathrm{CH}_3 \mathrm{OH}}^{\circ}\right)$ is 19 mV

Reduction of methanol takes place at the cathode.

JEE Main 2025 (Online) 3rd April Morning Shift

MCQ (Single Correct Answer)

-1

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

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

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

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

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

JEE Main Subjects