Given below are two statements: 1 M aqueous solution of each of Cu(NO3)2, AgNO3, Hg2(NO3)2; Mg(NO3)2 are electrolysed…

Q: Given below are two statements: 1 M aqueous solution of each of Cu(NO3)2, AgNO3, Hg2(NO3)2; Mg(NO3)2 are electrolysed using inert electrodes, Given: EAg^+/Agtheta = 0.80V , EHg2^2+/Hgtheta = 0.79V, EC

- At the cathode, the metal ion with the highest standard reduction potential (E0) gets reduced and deposited first. Arranging the given potentials: Thus, deposition follows the order Ag ightarrow Hg ightarrow Cu as voltage is steadily increased, confirming Statement I. - For Mg2+, its reduction potential is highly negative (-2.37 V), much lower than that of water (-0.83 V). Consequently, water undergoes reduction at the cathode instead of magnesium: This results in the evolution of **Hydrogen g

Year	2026	2025	2024	Total
Questions	13	19	8	40

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Ease of discharge at Cathode \propto Standard Reduction Potential (E⁰)

Core Logic

At the cathode, the metal ion with the highest standard reduction potential ( $E⁰$ $E^0$) gets reduced and deposited first. Arranging the given potentials:

E⁰Ag^+/Ag (0.80V) > E⁰Hg₂²⁺/Hg (0.79V) > E⁰Cu²⁺/Cu (0.24V)

Thus, deposition follows the order $Ag arrow Hg arrow Cu$ as voltage is steadily increased, confirming Statement I.

For $Mg²⁺$ $\text{Mg}^{2+}$, its reduction potential is highly negative ( $-2.37 V$ $-2.37\text{ V}$), much lower than that of water ( $-0.83 V$ $-0.83\text{ V}$). Consequently, water undergoes reduction at the cathode instead of magnesium:

2H₂O + 2e^- arrow H₂(g) + 2OH^-

This results in the evolution of Hydrogen gas at the cathode, not oxygen gas. Oxygen gas is evolved at the anode via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044]

Step 1: Conclusion Match

Since Statement I is correct and Statement II is incorrect, we select option (2).

Pattern Recognition

Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and $Al$ $\text{Al}$), $H₂$ $\text{H}_2$ gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas ( $O₂$ $\text{O}_2$) is an anodic product generated by water oxidation.

Chapter Mix

Class 12 Chemistry: Electrochemistry

More Electrochemistry Previous-Year Questions — Page 8

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

Q80 jee_main_2024_30_january_evening Standard Electrode Potential

Reduction potential of ions are given below:

ClO₄^- E° = 1.19V

IO₄^- E° = 1.65V

BrO₄^- E° = 1.74V

The correct order of their oxidising power is:

A. $ClO₄^- > IO₄^- > BrO₄^-$
B. $BrO₄^- > IO₄^- > ClO₄^-$
C. $BrO₄^- > ClO₄^- > IO₄^-$
D. $IO₄^- > BrO₄^- > ClO₄^-$

Solution

Core Logic

The Standard Reduction Potential ( $E^°$ $E^\circ$) measures a species' tendency to undergo reduction (gain electrons). A higher, more positive $E^°$ $E^\circ$ value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent.

Comparing the given $E^°$ $E^\circ$ values: $BrO₄^-$ $\mathrm{BrO}_4^-$: $1.74V$ $1.74\mathrm{V}$ $IO₄^-$ $\mathrm{IO}_4^-$: $1.65V$ $1.65\mathrm{V}$ $ClO₄^-$ $\mathrm{ClO}_4^-$: $1.19V$ $1.19\mathrm{V}$

The order of oxidizing power follows the magnitude of the reduction potential: $BrO₄^- > IO₄^- > ClO₄^-$

Pattern Recognition

Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: The p Block Elements

Q65 jee_main_2024_31_jan_morning Batteries

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:

A. $B, C and E only$ $\text{B, C and E only}$
B. $A, B, C, D and E$ $\text{A, B, C, D and E}$
C. $A, B, C and D only$ $\text{A, B, C and D only}$
D. $B, D and E only$ $\text{B, D and E only}$

Solution

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.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q67 jee_main_2024_31_jan_morning Electrolytic Conductance

Identify the factor from the following that does not affect electrolytic conductance of a solution.

A. $The nature of the electrolyte added.$
B. $The nature of the electrode used.$
C. $Concentration of the electrolyte.$
D. $The nature of solvent used.$ $\text{The nature of solvent used.}$

Solution

Core Logic

Conductivity of an electrolytic cell is affected by the concentration of the electrolyte, the nature of the electrolyte, and the nature of the solvent. It does not depend on the nature of the electrode used.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Q90 jee_main_2024_31_jan_morning Faraday's Laws of Electrolysis

One Faraday of electricity liberates

x \times 10⁻¹

gram atom of copper from copper sulphate, x is

Numerical Answer. Answer: 5 to 5

Solution

Core Logic

The reduction reaction for copper is:

Cu²⁺ + 2e^- arrow Cu

This shows that 2 moles of electrons (2 Faraday) are required to deposit 1 mole (or 1 gram atom) of Cu.

Therefore, 1 Faraday of electricity will deposit:

(1)/(2) = 0.5 moles of Cu

Step 1: Finding x

0.5 mole = 0.5 gram atom = 5 \times 10⁻¹ gram atom

Hence, $x = 5$ $x = 5$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Solution & Explanation

Related Formula

Core Logic

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Pattern Recognition

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