NEET Chemistry - MCQ Practice Questions

Understanding:

We need to find the equilibrium constant $K$ from the standard EMF.

Formula:

The relation between $E_{\text{cell}}^{\circ }$ and $K$ is:

Step 1: Rearrange for $\log K$

Step 2: Find $K$

Using $\frac{RT}{F}=0.0257\text{ V}$:

Answer:

The equilibrium constant is approximately $1.57\times 10^{37}$.

Quick Tip:

When using the $\ln$ form with $\frac{RT}{F}=0.0257\text{ V}$, note that $e^{85.6}=10^{85.6/2.303}=10^{37.16}$, giving $\approx 1.57\times 10^{37}$.

Understanding:

We need to calculate molar conductivity from conductivity and concentration.

Formula:

where $c$ must be in ${\text{mol cm}}^{-3}$ (i.e., ${\text{mol mL}}^{-1}\times 10^{-3}$, or convert using $1\text{ L}=1000{\text{ cm}}^3$).

Step 1: Convert concentration to ${\text{mol cm}}^{-3}$

Step 2: Calculate ${\Lambda }_m$

Answer:

The molar conductivity of the $\text{KCl}$ solution is $141{\text{ S cm}}^2{\text{ mol}}^{-1}$.

Quick Tip:

A very common error is forgetting to convert $c$ from ${\text{mol L}}^{-1}$ to ${\text{mol cm}}^{-3}$. Always divide by $1000$ when $\kappa$ is in ${\text{S cm}}^{-1}$.

Understanding:

We must calculate the standard cell EMF and assess spontaneity.

Formula:

A positive $E_{\text{cell}}^{\circ }$ indicates a spontaneous reaction ($\Delta G^{\circ }<0$).

Step 1: Calculate $E_{\text{cell}}^{\circ }$

Step 2: Assess spontaneity

Since $\Delta G^{\circ }<0$, the reaction is spontaneous.

Answer:

The standard EMF is $+1.10\text{ V}$ and the cell reaction is spontaneous.

Quick Tip:

Always use $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$ (both as reduction potentials). Never reverse the sign of the cathode.

Understanding:

We need to identify the anode reaction when aqueous ${\text{CuSO}}_4$ is electrolysed with copper electrodes.

Step 1: Identify the anode process

At the anode, oxidation occurs. When the electrode itself is made of an active (reactive) metal like copper, the electrode dissolves preferentially over the oxidation of water, because the oxidation potential of copper is more favourable than that of water.

Step 2: Write the anode reaction

Step 3: Identify other electrode reactions

At the cathode, ${\text{Cu}}^{2+}$ ions from solution are reduced and deposited:

This is the principle used in electrorefining of copper.

Answer:

At the anode, copper dissolves into the solution as ${\text{Cu}}^{2+}$ ions.

Quick Tip:

With inert electrodes (like Pt or graphite) and aqueous ${\text{CuSO}}_4$, the anode would evolve ${\text{O}}_2$. With active copper electrodes, the metal itself oxidises — this distinction is a classic NEET trap.

Understanding:

We must apply the Nernst equation to find the electrode potential at non-standard concentration.

Formula:

For the reduction half-reaction ${\text{Zn}}^{2+}+2e^{-}\to \text{Zn}$:

Step 1: Substitute values

Step 2: Round the result

Answer:

The electrode potential for ${\text{Zn}}^{2+}/\text{Zn}$ at $[{\text{Zn}}^{2+}]=0.001\text{ M}$ is $-0.849\text{ V}$.

Quick Tip:

For a reduction half-cell, decreasing the ion concentration makes the potential more negative. The Nernst equation for a reduction is $E=E^{\circ }-\frac{0.0592}{n}\log Q$, where $Q=\frac{1}{[{\text{Zn}}^{2+}]}$ for this half-reaction.

Understanding:

We need to find the charge (in Faradays) needed to deposit $27\text{ g}$ of $\text{Al}$.

Formula:

From Faraday's first law:

where $n_e$ is the number of electrons per atom and $F=96500{\text{ C mol}}^{-1}$.

Rearranged:

Step 1: Find moles of $\text{Al}$

Step 2: Determine electrons required

The cathode half-reaction is:

So $1\text{ mol}$ of $\text{Al}$ requires $3\text{ mol}$ of electrons.

Step 3: Convert to Faradays

Answer:

Depositing $27\text{ g}$ ($1\text{ mol}$) of $\text{Al}$ requires $3$ Faradays of charge.

Quick Tip:

Faradays required $=$ moles of substance $\times$ $n$ (electrons per ion). For ${\text{Al}}^{3+}$, $n=3$, so $1\text{ mol}$ always needs $3\text{ F}$.

Understanding:

We need to derive ${\Lambda }_m^{\circ }({\text{NH}}_4\text{OH})$ using Kohlrausch's law and the given strong electrolytes.

Step 1: Write individual ionic contributions

By Kohlrausch's law:

Step 2: Combine to get ${\Lambda }_m^{\circ }({\text{NH}}_4\text{OH})$

The target is:

Subtracting ${\Lambda }_m^{\circ }(\text{NaCl})$ from the sum of the first two:

Answer:

The correct expression is ${\Lambda }_m^{\circ }({\text{NH}}_4\text{OH})={\Lambda }_m^{\circ }({\text{NH}}_4\text{Cl})+{\Lambda }_m^{\circ }(\text{NaOH})-{\Lambda }_m^{\circ }(\text{NaCl})$.

Quick Tip:

The Hess's law analogy works here — add the equations that give the ions you need and cancel the unwanted ions by subtracting an equation that contains them.

Understanding:

We must calculate $\Delta G^{\circ }$ for the Daniell cell reaction.

Formula:

Step 1: Substitute values

Step 2: Confirm sign

Since $E_{\text{cell}}^{\circ }>0$, $\Delta G^{\circ }<0$, confirming a spontaneous reaction.

Answer:

The standard Gibbs energy change for the reaction is $-212.3{\text{ kJ mol}}^{-1}$.

Quick Tip:

The unit check: ${\text{C mol}}^{-1}\times \text{V}=\text{C}\times {\text{J C}}^{-1}\times {\text{mol}}^{-1}={\text{J mol}}^{-1}$. Divide by $1000$ to get ${\text{kJ mol}}^{-1}$.

Understanding:

We must use the cell constant to find the conductivity of $\text{NaOH}$ solution.

Formula:

The cell constant $G^{\ast }=\kappa \times R$, so:

and then:

Step 1: Calculate the cell constant

Step 2: Calculate ${\kappa }_{\text{NaOH}}$

Answer:

The conductivity of $0.01\text{ M}$ $\text{NaOH}$ solution is $7.05\times 10^{-4}{\text{ S cm}}^{-1}$.

Quick Tip:

The cell constant $G^{\ast }$ (in ${\text{cm}}^{-1}$) is a property of the cell geometry alone — it is the same regardless of the solution. Calibrate it with a known solution ($\text{KCl}$), then apply it to the unknown.