Given below is the plot of the molar conductivity vs sqrt(concentration) for KCl in aqueous solution. If, for the…

Year	2026	2025	2024	Total
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Q jee_main_2025_29_jan_morning Variation of Molar Conductivity with Concentration

The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?

A. A small decrease in molar conductivity is observed at infinite dilution.
B. A small increase in molar conductivity is observed at infinite dilution.
C. Molar conductivity increases sharply with increase in concentration.
D. Molar conductivity decreases sharply with increase in concentration.

Solution

Related Formula

For weak electrolytes, the degree of dissociation $α$ $\alpha$ increases sharply near infinite dilution according to Ostwald's Dilution Law:

α = \sqrt((Kₐ)/(C))

Core Logic

When a weak electrolyte is diluted (concentration $C arrow 0$ $C \rightarrow 0$), its molar conductivity increases steeply. Conversely, when plotted against $\sqrt(C)$ $\sqrt{C}$, as concentration increases, the degree of dissociation drops rapidly, causing a sharp decrease in molar conductivity. This matches the curve given below:

Variation of Molar Conductivity with Concentration diagram for Q28 - JEE Main 2025 Morning

Pattern Recognition

Weak electrolyte plots feature a steep asymptotic exponential-like rise towards the y-axis as $C arrow 0$ $C \rightarrow 0$, meaning a sharp decrease occurs with increasing concentration.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q jee_main_2025_29_jan_morning Nernst Equation

For a

Mg Mg²⁺

(aq)

∥ Ag⁺(aq) Ag

the correct Nernst Equation is :

A. $Ecell = Ecello - (RT)/(2F)ln [Ag⁺][Mg2 +]$
B. $Ecell = Ecell° + RT2 ~F ln [Ag⁺]²[Mg²⁺]$
C. $Ecell = Ecello - (RT)/(2F)ln [Mg2 +][Ag⁺]$
D. $Ecell = Ecello - (RT)/(2F)ln [Ag⁺]²[Mg2 +]$

Solution

Related Formula

Ecell = Ecell° - (RT)/(nF) ln Q

Core Logic

Let us explicitly formulate the complete chemical oxidation-reduction equations : Anode oxidation: $Mg(s) arrow Mg²⁺(aq) + 2e^-$ Cathode reduction: $2Ag⁺(aq) + 2e^- arrow 2Ag(s)$

Net total equation :

Mg(s) + 2Ag⁺(aq) leftharpoons Mg²⁺(aq) + 2Ag(s)

Total transferred moles of electrons $n = 2$ $n = 2$ . Reaction quotient :

Q = [Mg²⁺][Ag⁺]²

Substituting into Nernst form :

Ecell = Ecell° - (RT)/(2F) ln( [Mg²⁺][Ag⁺]² )

Inverting the inside quotient changes the sign of the logarithm term from negative to positive:

Ecell = Ecell° + (RT)/(2F) ln( [Ag⁺]²[Mg²⁺] )

Pattern Recognition

A standard negative logarithmic quotient can always toggle into an addition configuration by inverting the products/reactants variables concentration ratio.

Q82 jee_main_2024_01_february_morning Nernst Equation

The potential for the given half cell at 298K is

(-) \times 10⁻² ~V

.

2H^+(aq) + 2e^- arrow H₂(g)

[H^+] = 1 M, PH₂ = 2 ~atm

Given:

2.303RT/F = 0.06V

,

2 = 0.3

Numerical Answer. Answer: 0.9 to 1

Solution

Related Formula

E = E^° - (2.303RT)/(nF) Q

For the Standard Hydrogen Electrode half-reaction: $2H^+ + 2e^- arrow H₂$ $2H^+ + 2e^- \rightarrow H_2$

EH^+/H₂ = E^°H^+/H₂ - (0.06)/(2) PH₂[H^+]²

Step 1: Substitute the given values

$E^°H^+/H₂ = 0.00 ~V$ $E^\circ_{H^+/H_2} = 0.00 \mathrm{~V}$ (by definition) $[H^+] = 1 ~M$ $[H^+] = 1 \mathrm{~M}$ $PH₂ = 2 ~atm$ $P_{H_2} = 2 \mathrm{~atm}$ $n = 2$ $n = 2$ electrons

E = 0.00 - (0.06)/(2) ( (2)/(1²) )

Step 2: Solve the calculation

$E = -0.03 2$ $E = -0.03 \log 2$ Given $2 = 0.3$ $\log 2 = 0.3$ $E = -0.03 \times 0.3$ $E = -0.03 \times 0.3$ $E = -0.009 ~V$ $E = -0.009 \mathrm{~V}$ $E = -0.9 \times 10⁻² ~V$ $E = -0.9 \times 10^{-2} \mathrm{~V}$

Step 3: Match the requested format

The question asks for $(-) \times 10⁻² ~V$ $(-) \dots \times 10^{-2} \mathrm{~V}$. This gives exactly $0.9$ $0.9$. For NAT type with integer expected, $0.9$ $0.9$ can be rounded to $1$ $1$. However, exact calculation yields $0.9$ $0.9$. According to official JEE rounding, $0.9 \approx 1$ $0.9 \approx 1$.

Pattern Recognition

Hydrogen electrode non-standard potential depends strictly on pressure of $H₂$ $H_2$ and concentration of $H^+$ $H^+$. If $[H^+]=1$ $[H^+]=1$, increasing $H₂$ $H_2$ pressure lowers the potential below zero (makes it negative).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q jee_main_2024_29_january_evening Faraday's Laws of Electrolysis

A constant current was passed through a solution of

AuCl₄^-

ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was

1.314g

. The total charge passed through the solution is ________

\times 10⁻²F

. (Given atomic mass of

Au = 197

)

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

Number of equivalents deposited = (W)/(E) = (Q)/(F)

Equivalent Weight (E) = Atomic Massn-factor

Core Logic

In the reduction of gold from the tetrachloroaurate( $III$ $III$) complex anion:

AuCl₄^- + 3e^- arrow Au(s) + 4Cl^- n-factor = 3

Calculate the equivalent weight ( $E$ $E$) of Gold:

E = (197)/(3)

Set up the Faraday equivalence relation to solve for charge ( $Q$ $Q$ in Faradays):

(1.314)/(((197)/(3))) = Q

Step 1: Arithmetic Resolution

Q = (1.314 \times 3)/(197) = (3.942)/(197) = 0.02 F = 2 \times 10⁻² F

Thus, the required integer value is 2.

Pattern Recognition

Always determine the correct change in oxidation state ( $+3$ $+3$ to $0$ $0$) to establish the proper $n$ $n$-factor value for calculations using Faraday's laws.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q81 jee_main_2024_27_jan_morning Faraday's Laws of Electrolysis

The mass of silver (Molar mass of

Ag: 108 g mol⁻¹

) displaced by a quantity of electricity which displaces

5600 mL

of

O₂

at S.T.P. will be

g.

Numerical Answer. Answer: 107 to 108

Solution

Related Formula

By Faraday's Second Law of Electrolysis:

Equivalents of Ag = Equivalents of O₂

Equivalents = MassEquivalent Mass = Moles \times n-factor

Step 1: Calculate equivalents using standard metrics

Let $x$ $x$ grams of Silver be displaced. Using the older STP molar volume baseline ( $22.4 L$ $22.4\text{ L}$ or $22400 mL$ $22400\text{ mL}$):

Moles of O₂ = (5600)/(22400) = 0.25 moles

Since the $n$ $n$-factor of $O₂$ $O_2$ is 4 ( $2O²⁻ arrow O₂ + 4e^-$ ):

Equivalents of O₂ = 0.25 \times 4 = 1

Step 2: Equating equivalents for silver mass

Equivalents of Ag = (x)/(108) \times 1 = 1 x = 108 g

Step 3: Alternative calculation using current STP metric

Using modern STP volume metrics ( $22.7 L$ $22.7\text{ L}$):

(x \times 1)/(108) = (5.6)/(22.7) \times 4 x \approx 106.57 g arrow 107 g

Pattern Recognition

Equivalents equations bypass complex current/time measurements. Always link volume fractions directly to $n$ $n$-factor equivalents.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Some Basic Concepts of Chemistry

Solution & Explanation

Related Formula

Step 1: Setting Up Ratios

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Solution

Related Formula

Step 1: Substitute the given values

Step 2: Solve the calculation

Step 3: Match the requested format

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Arithmetic Resolution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate equivalents using standard metrics

Step 2: Equating equivalents for silver mass

Step 3: Alternative calculation using current STP metric

Pattern Recognition

Chapter Mix

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