NEET (UG)

Practice Test 1 — Electrochemistry

12 questions • 18 minutes • auto-graded with full solutions

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Section A — MCQ (Single Correct & Statement-based)

Question 1

In a galvanic cell, reduction occurs at the:

Solution: Reduction is at the cathode.

Question 2

$E^{\circ}_{cell}$ is calculated as:

Solution: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.

Question 3

The Nernst factor at 298 K is:

Solution: $(0.059/n)\log Q$.

Question 4

Statements: (I) $\Delta G^{\circ} = -nFE^{\circ}_{cell}$. (II) A positive $E^{\circ}_{cell}$ gives a spontaneous reaction. Which are correct?

Solution: Both statements are correct.

Question 5

On dilution, molar conductivity of an electrolyte:

Solution: $\Lambda_m$ increases on dilution.

Question 6

To deposit 1 mole of $\text{Al}$ from $\text{Al}^{3+}$, the charge needed is:

Solution: $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ needs $3\ F$.

Question 7

A hydrogen–oxygen fuel cell converts:

Solution: Fuel cells convert fuel energy directly to electricity.

Question 8

When a cell reaches equilibrium ($Q=K$), the EMF is:

Solution: EMF = 0 at equilibrium (dead cell).

Section B — Assertion & Reason

Question 9

A: A cell with a positive standard EMF drives a spontaneous reaction.
R: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$, so a positive $E^{\circ}_{cell}$ gives a negative $\Delta G^{\circ}$.

Solution: A negative $\Delta G^{\circ}$ is the condition for spontaneity — R explains A.

Question 10

A: The molar conductivity of a solution increases on dilution.
R: On dilution, the degree of dissociation and the freedom of ion movement increase.

Solution: Greater dissociation and ion mobility raise molar conductivity — R explains A.

Question 11

A: In a galvanic cell the anode is the positive electrode.
R: Oxidation occurs at the anode.

Solution: In a galvanic cell the anode is negative (so A is false); oxidation does occur at the anode (R true).

Question 12

A: Depositing one mole of copper from $\text{Cu}^{2+}$ requires two Faradays of charge.
R: The reduction of $\text{Cu}^{2+}$ to $\text{Cu}$ involves the gain of two electrons.

Solution: Two electrons per copper ion means $2\ F$ per mole — R explains A.