Problem 132
Question
When an electric current is passed through acidulated water, \(112 \mathrm{~mL}\) of hydrogen gas at NTP collects at the cathode in 965 seconds. The current passed, in ampere is (a) \(0.1\) (b) \(0.5\) (c) \(1.0\) (d) \(2.0\)
Step-by-Step Solution
Verified Answer
The current passed is 0.5 A (Answer b).
1Step 1: Understand Key Principles
When electric current is passed through acidulated water, electrolysis takes place. The reaction involves forming hydrogen and oxygen gas. By Faraday's first law of electrolysis, the amount of substance deposited or released at an electrode is proportional to the quantity of electricity passed through the electrolyte.
2Step 2: Recall Faraday's Law
According to Faraday's first law, we use the formula: \( Q = nF \), where \( Q \) is the charge in coulombs, \( n \) is the number of moles of electrons, and \( F \) is Faraday's constant (approximately 96500 C/mol).
3Step 3: Calculate Moles of Hydrogen Gas
At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies \( 22400 \mathrm{~mL} \). Given \( 112 \mathrm{~mL} \) of hydrogen gas is produced, the moles of hydrogen gas formed is \( \frac{112}{22400} \) moles.
4Step 4: Determine the Number of Electrons
Hydrogen gas is formed by the reaction \( 2H^+ + 2e^- \rightarrow H_2 \). Therefore, 1 mole of \( H_2 \) requires 2 moles of electrons. Calculate the moles of electrons: \( 2 \times \frac{112}{22400} \).
5Step 5: Calculate the Charge Passed
Using Faraday's constant, the charge \( Q \) in coulombs is given by \( 2 \times \frac{112}{22400} \times 96500 \).
6Step 6: Calculate Current
Current \( I \) is the charge passed \( Q \) divided by the time \( t \) in seconds. Here, \( t = 965 \) seconds. So, \( I = \frac{Q}{965} \). Substitute \( Q \) calculated in the previous step to find the current.
7Step 7: Solve for Ampere
Substitute the values: \( I = \frac{2 \times \frac{112}{22400} \times 96500}{965} \). Calculate to find \( I \).
Key Concepts
ElectrolysisFaraday's ConstantElectric Current Calculation
Electrolysis
Electrolysis is a fascinating process where an electrical current is used to drive a non-spontaneous chemical reaction.
In simple terms, it occurs when electricity is passed through a substance, often a liquid or solution, causing a chemical change. During electrolysis, ions move towards electrodes, allowing separation or production of elements.
Understanding the concept of electrolysis is crucial for solving problems related to chemical reactions and electricity in chemistry.
In simple terms, it occurs when electricity is passed through a substance, often a liquid or solution, causing a chemical change. During electrolysis, ions move towards electrodes, allowing separation or production of elements.
- The component through which electricity is passed, called an electrolyte, typically contains ions that facilitate this movement.
- The electrode where oxidation occurs (loss of electrons) is the anode.
- The electrode where reduction occurs (gain of electrons) is the cathode. For our exercise, hydrogen gas collects at the cathode.
Understanding the concept of electrolysis is crucial for solving problems related to chemical reactions and electricity in chemistry.
Faraday's Constant
Faraday's constant, often denoted by the symbol \( F \), is a key entity in electrochemistry. It represents the total electric charge carried by one mole of electrons.
The constant is approximately 96500 coulombs per mole of electrons (C/mol). This value is essential when applying Faraday's laws of electrolysis.
Let's break down its use:
The precision Faraday's law offers is invaluable for both scientific study and practical applications.
The constant is approximately 96500 coulombs per mole of electrons (C/mol). This value is essential when applying Faraday's laws of electrolysis.
Let's break down its use:
- Faraday's first law states that the quantity of an element deposited at an electrode is proportional to the charge passed through the substance. This involves calculating \( Q \), which is the charge in coulombs.
- To find \( Q \), you multiply the number of moles of electrons \( n \) by \( F \).
The precision Faraday's law offers is invaluable for both scientific study and practical applications.
Electric Current Calculation
Calculating electric current, particularly in electrolysis problems, involves several steps. Electric current, expressed in amperes (A), is essentially the flow of charge over time.
For our exercise:- First, determine the total charge passed using the formula \( Q = nF \), where \( n \) is the moles of electrons and \( F \) is Faraday's constant.- We previously calculated the charge \( Q \) in coulombs by multiplying the moles of electrons by Faraday's constant.Once you have \( Q \), the current \( I \) can be calculated using the formula:\[I = \frac{Q}{t}\]where \( t \) is the time in seconds over which the charge is passed. In the given exercise, we determined \( t = 965 \) seconds.
Calculating the electric current accurately is essential for predicting the behavior of the electrolytic process and ensures precision in applications such as designing circuits and industrial electrolysis.
For our exercise:- First, determine the total charge passed using the formula \( Q = nF \), where \( n \) is the moles of electrons and \( F \) is Faraday's constant.- We previously calculated the charge \( Q \) in coulombs by multiplying the moles of electrons by Faraday's constant.Once you have \( Q \), the current \( I \) can be calculated using the formula:\[I = \frac{Q}{t}\]where \( t \) is the time in seconds over which the charge is passed. In the given exercise, we determined \( t = 965 \) seconds.
Calculating the electric current accurately is essential for predicting the behavior of the electrolytic process and ensures precision in applications such as designing circuits and industrial electrolysis.
Other exercises in this chapter
Problem 130
The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=\) \(3.0\) at \(25^{\circ} \mathrm{C}\). The potential of hydrogen electrode would be (a) \(-0.1
View solution Problem 131
Three faraday of electricity is passed through aqueous solutions of \(\mathrm{AgNO}_{3}, \mathrm{NiSO}_{4}\) and \(\mathrm{CrCl}_{3}\) kept in three vessels usi
View solution Problem 133
For the electrochemical cell, \(\mathrm{M}\left|\mathrm{M}^{+} \| \mathrm{X}^{-}\right| \mathrm{X}\), \(E^{\circ} \mathrm{M}^{+} / \mathrm{M}=0.44 \mathrm{~V}\)
View solution Problem 135
When a quantity of electricity is passed through \(\mathrm{CuSO}_{4}\) solution, \(0.16 \mathrm{~g}\) of copper gets deposited. If the same quantity of electric
View solution