JEE Chemistry - MCQ Practice Questions

Step 1: Identify the cathode and anode

In a galvanic cell, the electrode with the higher (more positive) standard reduction potential undergoes reduction at the cathode, and the electrode with the lower standard reduction potential undergoes oxidation at the anode.

Step 2: Apply the standard cell potential formula

Step 3: Substitute the values

In a galvanic cell, the anode is the electrode at which oxidation occurs, and the cathode is the electrode at which reduction occurs.

At the anode, a species loses electrons:

At the cathode, a species gains electrons:

A useful mnemonic is $\text{AN OX}$ and $\text{RED CAT}$: the anode undergoes oxidation, and the cathode undergoes reduction.

Therefore, the electrode reaction that occurs at the anode is:

Since $[{\text{H}}^{+}]=[{\text{OH}}^{-}]$, the solution is neutral.

Step 1: Verify the ion product of water.

At $25^{\circ }\text{C}$, the water dissociation constant is

Step 2: Calculate pH and pOH.

Step 3: Confirm neutrality.

Because $[{\text{H}}^{+}]=[{\text{OH}}^{-}]=1\times 10^{-7}\text{M}$, and $\text{pH}=\text{pOH}=7$, the solution is neither acidic nor basic.

Step 1: Identify the half-reactions

In the Zn–Cu galvanic cell, copper acts as the cathode (reduction) and zinc acts as the anode (oxidation):

Step 2: Apply the cell potential formula

Step 3: Substitute and calculate

Color of reactants has no effect on reaction rate. Temperature, catalyst, and concentration all significantly affect reaction rate.

H₂O has maximum boiling point due to strong hydrogen bonding compared to other hydrides in group 16

Cu²⁺ requires 2 electrons per atom. 1 Faraday = 1 mol of electrons; moles of Cu = $\frac{1}{2}$ = 0.5 mol

Diamond has a 3D tetrahedral covalent network structure with C-C bonds in all directions, while graphite has layered structure with weak van der Waals forces between layers

Kc depends only on temperature. Pressure, concentration changes, and catalysts do not alter Kc value—they shift equilibrium position but not the constant itself.

Cp - Cv = R = 8.314 J/(mol·K). Therefore, Cv = 29 - 8.314 ≈ 20.7 ≈ 20.8 J/(mol·K). This indicates a diatomic gas (Cp = $\frac{7}{2}$ R, Cv = $\frac{5}{2}$ R).

For first-order reaction, each half-life reduces concentration by 50%. After 1st half-life (10 min): 50% remains. After 2nd half-life (20 min): 25% remains = $\frac{1}{4}$th. Total time = 2 × 10 = 20 minutes.