
A certain charge liberates 0.8 gm of ${O_2}$. The same charge will liberate how many gm of silver
A) 108 gm
B) 10.8 gm
C) 0.8 gm
D) 108/0.8 gm
Answer
163.2k+ views
Hint: We have to use Faraday’s first law to solve this question. According to the first law, the amount of chemical change a current causes at an electrode-electrolyte contact is proportional to the amount of electricity consumed.
Complete step by step solution:The mass of the substance released or deposited on an electrode during electrolysis is directly proportional to the amount of electric charge carried through the electrolyte, according to Faraday's first law.
If m = mass of a substance liberated or deposited at an electrode and q = charge. Then according to Faraday's law of electrolysis,
$m \propto q \Rightarrow m = zq$
Where z = electrochemical equivalent of the substance. The z is calculated by the given equation.
$z = \dfrac{{Atomic~mass}}{{Valency}} \times \dfrac{1}{{96500}}gm{C^{ - 1}}$
The charge is same for both elements. The equations for the two cases will be,${m_1} = {z_1}q$ and ${m_2} = {z_2}q$. Now comparing the two equations and we will get,
$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{z_1}}}{{{z_2}}}$
\[{m_1}\] = mass of oxygen = 0.8 gm
\[{z_1}\] = electrochemical equivalent of oxygen = $\dfrac{{16}}{2} \times \dfrac{1}{{96500}} = \dfrac{8}{{96500}}gm{C^{ - 1}}$
\[{z_2}\] = electrochemical equivalent of silver = $\dfrac{{108}}{1} \times \dfrac{1}{{96500}} = \dfrac{{108}}{{96500}}gm{C^{ - 1}}$
\[{m_2}\] = mass of silver
$\dfrac{m_1}{m_2}=\dfrac{z_1}{z_2}=\dfrac{0.8}{m_2}=\dfrac{\dfrac{8}{96500}}{\dfrac{108}{96500}}$
$\dfrac{{0.8}}{{{m_2}}} = \dfrac{8}{{108}}$
${m_2} = \dfrac{{0.8 \times 108}}{8} = 10.8g$
Hence, the correct option is Option (B).
Additional Information:According to Faraday's second law of electrolysis, the masses of ions deposited at the electrodes are inversely proportional to their chemical equivalents when the same amount of electricity is applied to various electrolytes.
The Faraday constant represents the amount of electric charge carried by one mole, or Avogadro's number of electrons. It is a crucial constant in physics, electronics, and chemistry. The measurement is given in coulombs per mole (C/mol).
Note: Electrolysis is a technique for eliminating iron oxide. By applying a tiny electrical charge to the rusted metal from a battery or battery charger to stimulate ion exchange while the device is submerged in an electrolyte solution.
Complete step by step solution:The mass of the substance released or deposited on an electrode during electrolysis is directly proportional to the amount of electric charge carried through the electrolyte, according to Faraday's first law.
If m = mass of a substance liberated or deposited at an electrode and q = charge. Then according to Faraday's law of electrolysis,
$m \propto q \Rightarrow m = zq$
Where z = electrochemical equivalent of the substance. The z is calculated by the given equation.
$z = \dfrac{{Atomic~mass}}{{Valency}} \times \dfrac{1}{{96500}}gm{C^{ - 1}}$
The charge is same for both elements. The equations for the two cases will be,${m_1} = {z_1}q$ and ${m_2} = {z_2}q$. Now comparing the two equations and we will get,
$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{z_1}}}{{{z_2}}}$
\[{m_1}\] = mass of oxygen = 0.8 gm
\[{z_1}\] = electrochemical equivalent of oxygen = $\dfrac{{16}}{2} \times \dfrac{1}{{96500}} = \dfrac{8}{{96500}}gm{C^{ - 1}}$
\[{z_2}\] = electrochemical equivalent of silver = $\dfrac{{108}}{1} \times \dfrac{1}{{96500}} = \dfrac{{108}}{{96500}}gm{C^{ - 1}}$
\[{m_2}\] = mass of silver
$\dfrac{m_1}{m_2}=\dfrac{z_1}{z_2}=\dfrac{0.8}{m_2}=\dfrac{\dfrac{8}{96500}}{\dfrac{108}{96500}}$
$\dfrac{{0.8}}{{{m_2}}} = \dfrac{8}{{108}}$
${m_2} = \dfrac{{0.8 \times 108}}{8} = 10.8g$
Hence, the correct option is Option (B).
Additional Information:According to Faraday's second law of electrolysis, the masses of ions deposited at the electrodes are inversely proportional to their chemical equivalents when the same amount of electricity is applied to various electrolytes.
The Faraday constant represents the amount of electric charge carried by one mole, or Avogadro's number of electrons. It is a crucial constant in physics, electronics, and chemistry. The measurement is given in coulombs per mole (C/mol).
Note: Electrolysis is a technique for eliminating iron oxide. By applying a tiny electrical charge to the rusted metal from a battery or battery charger to stimulate ion exchange while the device is submerged in an electrolyte solution.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
