Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 5.
Q1499 (2024)
A constant current was passed through a solution of $\mathrm{AuCl}_4^-$ ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was $1.314\mathrm{g}$. The total charge passed through the solution is ________ $\times 10^{-2}\mathrm{F}$. (Given atomic mass of $\mathrm{Au} = 197$)
### Related Formula
$$\text{Number of equivalents deposited} = \frac{W}{E} = \frac{Q}{F}$$
$$\text{Equivalent Weight } (E) = \frac{\text{Atomic Mass}}{n\text{-factor}}$$
### Core Logic
In the reduction of gold from the tetrachloroaurate($III$) complex anion:
$$\mathrm{AuCl}_4^- + 3e^- \rightarrow \mathrm{Au}(s) + 4\mathrm{Cl}^- \implies n\text{-factor} = 3$$
Calculate the equivalent weight ($E$) of Gold:
$$E = \frac{197}{3}$$
Set up the Faraday equivalence relation to solve for charge ($Q$ in Faradays):
$$\frac{1.314}{\left(\frac{197}{3}\right)} = Q$$
### Step 1: Arithmetic Resolution
$$Q = \frac{1.314 \times 3}{197} = \frac{3.942}{197} = 0.02\text{ F} = 2 \times 10^{-2}\text{ F}$$
Thus, the required integer value is **2**.
### Pattern Recognition
Always determine the correct change in oxidation state ($+3$ to $0$) to establish the proper $n$-factor value for calculations using Faraday's laws.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q81 (2024)
The mass of silver (Molar mass of $\text{Ag}: 108\text{ g mol}^{-1}$) displaced by a quantity of electricity which displaces $5600\text{ mL}$ of $O_2$ at S.T.P. will be $\text{\quad\quad}$ g.
### Related Formula
By Faraday's Second Law of Electrolysis:
$$\text{Equivalents of Ag} = \text{Equivalents of O}_2$$
$$\text{Equivalents} = \frac{\text{Mass}}{\text{Equivalent Mass}} = \text{Moles} \times n\text{-factor}$$
### Step 1: Calculate equivalents using standard metrics
Let $x$ grams of Silver be displaced.
Using the older STP molar volume baseline ($22.4\text{ L}$ or $22400\text{ mL}$):
$$\text{Moles of O}_2 = \frac{5600}{22400} = 0.25\text{ moles}$$
Since the $n$-factor of $O_2$ is 4 ($2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$):
$$\text{Equivalents of O}_2 = 0.25 \times 4 = 1$$
### Step 2: Equating equivalents for silver mass
$$\text{Equivalents of Ag} = \frac{x}{108} \times 1 = 1 \implies x = 108\text{ g}$$
### Step 3: Alternative calculation using current STP metric
Using modern STP volume metrics ($22.7\text{ L}$):
$$\frac{x \times 1}{108} = \frac{5.6}{22.7} \times 4 \implies x \approx 106.57\text{ g} \rightarrow 107\text{ g}$$
### Pattern Recognition
Equivalents equations bypass complex current/time measurements. Always link volume fractions directly to $n$-factor equivalents.
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Some Basic Concepts of Chemistry
Q82 (2024)
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)
### Related Formula
$$W = Z \cdot I \cdot t = \frac{M}{n \cdot F} \cdot I \cdot t$$
where,
$W$ = mass deposited
$Z$ = electrochemical equivalent
$I$ = current in amperes
$t$ = time in seconds
$M$ = molar mass
$n$ = n-factor (electrons exchanged)
$F$ = Faraday's constant ($96500 \text{ C/mol}$)
### Core Logic
The electrolysis of zinc sulphate ($ZnSO_4$) involves the reduction of zinc ions at the cathode:
$$Zn^{+2} + 2e^- \rightarrow Zn$$
Here, the n-factor ($n$) is 2.
### Step 1: Calculation
Given values:
$I = 0.015\text{ A}$
$t = 15\text{ minutes} = 15 \times 60\text{ seconds} = 900\text{ s}$
$M = 65.4\text{ g/mol}$
$F \approx 96500\text{ C}$
Plugging the values into Faraday's First Law:
$$W = \frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60$$
$$W = \frac{65.4}{193000} \times 13.5$$
$$W = 3.3886 \times 10^{-4} \times 13.5$$
$$W = 45.746 \times 10^{-4}\text{ g}$$
Rounding to two decimal places (or nearest integer depending on convention), we get $45.75 \times 10^{-4}\text{ g}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q80 (2024)
Reduction potential of ions are given below:
$\mathrm{ClO}_4^- \quad E^{\circ} = 1.19\mathrm{V}$
$\mathrm{IO}_4^- \quad E^{\circ} = 1.65\mathrm{V}$
$\mathrm{BrO}_4^- \quad E^{\circ} = 1.74\mathrm{V}$
The correct order of their oxidising power is:
- $\mathrm{ClO}_4^- > \mathrm{IO}_4^- > \mathrm{BrO}_4^-$
- $\mathrm{BrO}_4^- > \mathrm{IO}_4^- > \mathrm{ClO}_4^-$
- $\mathrm{BrO}_4^- > \mathrm{ClO}_4^- > \mathrm{IO}_4^-$
- $\mathrm{IO}_4^- > \mathrm{BrO}_4^- > \mathrm{ClO}_4^-$
### Core Logic
The Standard Reduction Potential ($E^\circ$) measures a species' tendency to undergo reduction (gain electrons).
A higher, more positive $E^\circ$ value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent.
Comparing the given $E^\circ$ values:
$\mathrm{BrO}_4^-$: $1.74\mathrm{V}$
$\mathrm{IO}_4^-$: $1.65\mathrm{V}$
$\mathrm{ClO}_4^-$: $1.19\mathrm{V}$
The order of oxidizing power follows the magnitude of the reduction potential:
$\mathrm{BrO}_4^- > \mathrm{IO}_4^- > \mathrm{ClO}_4^-$
### Pattern Recognition
Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 12 Chemistry: The p Block Elements
Q65 (2024)
The metals that are employed in the battery industries are
A. Fe
B. Mn
C. Ni
D. Cr
E. Cd
Choose the correct answer from the options given below:
- $\text{B, C and E only}$
- $\text{A, B, C, D and E}$
- $\text{A, B, C and D only}$
- $\text{B, D and E only}$
### Core Logic
Mn, Ni, and Cd metals are predominantly used in battery industries.
- Mn is used in dry cells (Leclanche cell).
- Ni and Cd are used in Nickel-Cadmium (Ni-Cd) rechargeable batteries.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
