Problem 120
Question
Electrolysis of dilute aqueous \(\mathrm{NaCl}\) solution was carried out by passing 10 mili ampere current. The time required to librate \(0.01\) mole of \(\mathrm{H}_{2}\) gas at the cathode? (a) \(9.65 \times 10^{4} \mathrm{Sec}\) (b) \(19.3 \times 10^{4} \mathrm{Sec}\) (c) \(28.95 \times 10^{4} \mathrm{Sec}\) (d) \(38.6 \times 10^{4} \mathrm{Sec}\)
Step-by-Step Solution
Verified Answer
The time required is \(19.3 \times 10^4 \text{ seconds}\) (option b).
1Step 1: Understanding Electrolysis
In the electrolysis process of dilute aqueous \( \text{NaCl} \), hydrogen gas \( \text{H}_2 \) is liberated at the cathode. The chemical reaction at the cathode for hydrogen evolution can be represented as: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \). Each mole of \( \text{H}_2 \) gas requires 2 moles of electrons.
2Step 2: Calculate Charge Required for Electrolysis
From Faraday's laws of electrolysis, 1 mole of electrons corresponds to a charge of \( 96500 \) C (Faraday's constant). For \( 0.01 \) mole of \( \text{H}_2 \), we need \( 2 \times 0.01 = 0.02 \) moles of electrons. The required charge \( Q \) is calculated as: \( Q = 0.02 \times 96500 = 1930 \text{ C} \).
3Step 3: Calculate Time Required for Electrolysis
To find the time \( t \) in seconds, use the formula \( Q = I \times t \), where \( I \) is the current in amperes. Given \( I = 10 \) mA \( = 0.01 \) A, rearrange to find \( t \): \( t = \frac{Q}{I} = \frac{1930}{0.01} = 193000 \text{ seconds} \).
4Step 4: Determine the Correct Option
Compare 193000 seconds to the options given. It matches option (b) which is \( 19.3 \times 10^4 \) seconds.
Key Concepts
Electrolysis ProcessFaraday's Laws of ElectrolysisHydrogen EvolutionFaraday's Constant
Electrolysis Process
Electrolysis is a chemical process where electrical energy is used to drive a non-spontaneous chemical reaction. In the case of a dilute aqueous sodium chloride (NaCl) solution, electrolysis involves passing an electric current through the solution to cause two main reactions:
This process requires the movement of ions in the solution and is facilitated by the electrodes within the electrolytic cell. The cathode, which is the negative electrode, attracts cations—positively charged ions like hydrogen ions (H⁺). Here, at the cathode, two hydrogen ions accept two electrons each to form hydrogen gas (H₂), a process known as the evolution of hydrogen.
The significance of the Electrolysis Process lies in its application in various industries, such as electroplating, production of chlorine and hydrogen gases, and even water splitting.
- Hydrogen gas is released at the cathode.
- Chlorine gas is released at the anode.
This process requires the movement of ions in the solution and is facilitated by the electrodes within the electrolytic cell. The cathode, which is the negative electrode, attracts cations—positively charged ions like hydrogen ions (H⁺). Here, at the cathode, two hydrogen ions accept two electrons each to form hydrogen gas (H₂), a process known as the evolution of hydrogen.
The significance of the Electrolysis Process lies in its application in various industries, such as electroplating, production of chlorine and hydrogen gases, and even water splitting.
Faraday's Laws of Electrolysis
Faraday's Laws of Electrolysis are fundamental in understanding how electrolysis works.
The first law states that the amount of substance altered at an electrode during electrolysis is proportional to the quantity of electricity that passes through the electrolyte.
The second law elaborates that the amount of different substances liberated or dissolved by the same quantity of electricity is directly proportional to their chemical equivalent weights.
The calculation of the charge required for the electrolysis is underpinned by these laws. Each mole of electrons is equivalent to a charge represented by Faraday's Constant, which is 96500 Coulombs per mole of electrons. Hence, knowing the moles of gas desired, we can use these laws to calculate the necessary charge and thus the achieved by electrolysis.
The first law states that the amount of substance altered at an electrode during electrolysis is proportional to the quantity of electricity that passes through the electrolyte.
The second law elaborates that the amount of different substances liberated or dissolved by the same quantity of electricity is directly proportional to their chemical equivalent weights.
- This means lighter ions like H⁺ (hydrogen) will get reduced more quickly compared to heavier ions like Cl⁻ (chlorine) when the same current is applied.
The calculation of the charge required for the electrolysis is underpinned by these laws. Each mole of electrons is equivalent to a charge represented by Faraday's Constant, which is 96500 Coulombs per mole of electrons. Hence, knowing the moles of gas desired, we can use these laws to calculate the necessary charge and thus the achieved by electrolysis.
Hydrogen Evolution
At the cathode, the process of hydrogen evolution occurs during electrolysis. When electrolysis is carried out in an aqueous solution of NaCl, it is significant that water is primarily reduced at the cathode instead of Na⁺ ions.
The hydrogen ions (H⁺) from the water gain electrons ( 2H⁺ + 2e⁻ → H₂ ) and convert into hydrogen gas, which is then liberated as bubbles.
The efficiency of the hydrogen evolution reaction can be influenced by factors such as electrode material, surface area, and temperature of the solution, which can significantly impact energetics and kinetics of gas production.
The hydrogen ions (H⁺) from the water gain electrons ( 2H⁺ + 2e⁻ → H₂ ) and convert into hydrogen gas, which is then liberated as bubbles.
- This reaction is critical because hydrogen is a valuable industrial gas used in processes like hydrogenation of fats and oils, and as a clean fuel.
- Hydrogen removal is pivotal as it affects the overall yield and efficiency of the electrolysis reaction.
The efficiency of the hydrogen evolution reaction can be influenced by factors such as electrode material, surface area, and temperature of the solution, which can significantly impact energetics and kinetics of gas production.
Faraday's Constant
Faraday's Constant is a cornerstone parameter in electrochemistry, crucial for calculating the charge in electrolysis.
This constant represents the total charge of one mole of electrons and is valued at approximately 96500 Coulombs (C). This value stems from combining Avogadro's number, the charge of a single electron, and expressing it in a useful form for electrochemical reactions.
The usage of Faraday's Constant allows us to connect the macroscopic world (amount of substance) with the microscopic world (number of electrons).
Applications of Faraday's Constant are seen across different scientific and industrial domains, playing a key role in areas from battery chemistry to the synthesis of various elements.
This constant represents the total charge of one mole of electrons and is valued at approximately 96500 Coulombs (C). This value stems from combining Avogadro's number, the charge of a single electron, and expressing it in a useful form for electrochemical reactions.
The usage of Faraday's Constant allows us to connect the macroscopic world (amount of substance) with the microscopic world (number of electrons).
- For instance, knowing how much hydrogen gas (H₂) needs to be produced, you can calculate the needed charge using this constant, enabling precise prediction and operation of the electrolysis system.
Applications of Faraday's Constant are seen across different scientific and industrial domains, playing a key role in areas from battery chemistry to the synthesis of various elements.
Other exercises in this chapter
Problem 117
A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively (a) \(\mathrm{O}_{2}, \mat
View solution Problem 119
A solution containing one mole per litre of each \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ; \mathrm{AgNO}_{3} ; \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)
View solution Problem 121
The standard reduction potentials at \(298 \mathrm{~K}\) for the following half- reactions are given against each \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e} \
View solution Problem 122
For the equation \(\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{NO} .\) The number of electrons in it
View solution