For the given cell: The standard cell potential of the above reaction is given by:

Year	2026	2025	2024	Total
Questions	13	18	8	39

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

Q82 jee_main_2024_29_jan_morning Faradays Laws of Electrolysis

The mass of zinc produced by the electrolysis of zinc sulphate solution with a steady current of 0.015 A for 15 minutes is

______ \times 10⁻⁴

g. (Atomic mass of zinc

= 65.4

amu)

Numerical Answer. Answer: 45.75 to 46

Solution

Related Formula

W = Z \cdot I \cdot t = (M)/(n \cdot F) \cdot I \cdot t

where, $W$ $W$ = mass deposited $Z$ $Z$ = electrochemical equivalent $I$ $I$ = current in amperes $t$ $t$ = time in seconds $M$ $M$ = molar mass $n$ $n$ = n-factor (electrons exchanged) $F$ $F$ = Faraday's constant ( $96500 C/mol$ $96500 \text{ C/mol}$)

Core Logic

The electrolysis of zinc sulphate ( $ZnSO₄$ $ZnSO_4$) involves the reduction of zinc ions at the cathode:

Zn⁺² + 2e^- arrow Zn

Here, the n-factor ( $n$ $n$) is 2.

Step 1: Calculation

Given values: $I = 0.015 A$ $I = 0.015\text{ A}$ $t = 15 minutes = 15 \times 60 seconds = 900 s$ $M = 65.4 g/mol$ $M = 65.4\text{ g/mol}$ $F \approx 96500 C$ $F \approx 96500\text{ C}$

Plugging the values into Faraday's First Law:

W = (65.4)/(2 \times 96500) \times 0.015 \times 15 \times 60

W = (65.4)/(193000) \times 13.5

W = 3.3886 \times 10⁻⁴ \times 13.5

W = 45.746 \times 10⁻⁴ g

Rounding to two decimal places (or nearest integer depending on convention), we get $45.75 \times 10⁻⁴ g$ $45.75 \times 10^{-4}\text{ g}$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions — Page 7

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

Solution

Related Formula

Core Logic

Step 1: Calculation

Chapter Mix

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