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Chapter 9: Problem 132

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) \(2.0\)

Short Answer

Expert verified
The current passed is 0.5 A (Answer b).

Step by step solution

01

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.
02

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).
03

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.
04

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: \( 2 \times \frac{112}{22400} \).
05

Calculate the Charge Passed

Using Faraday's constant, the charge \( Q \) in coulombs is given by \( 2 \times \frac{112}{22400} \times 96500 \).
06

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.
07

Solve for Ampere

Substitute the values: \( I = \frac{2 \times \frac{112}{22400} \times 96500}{965} \). Calculate to find \( I \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
  • 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.
Electrolysis is incredibly important in various industries, such as for extracting metals from ores or in the electroplating of metals.
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:
  • 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 \( F \).
Using Faraday's constant allows you to precisely calculate the substances involved in electrolysis reactions.
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.

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Most popular questions from this chapter

In which of the following aqueous solutions during electrolysis, \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) are liberated? (a) \(\mathrm{CuCl}_{2}\) (aq) (b) \(\mathrm{KCl}(\mathrm{aq})\) (c) \(\mathrm{MgCl}_{2}(\mathrm{aq})\) (d) \(\mathrm{NaCl}(\mathrm{aq})\)

The standard electrode potential of \(\mathrm{Cu}^{2+} / \mathrm{Cu}=0.34 \mathrm{~V}\). The electrode potential will be zero, when the conc. of \(\mathrm{Cu}^{2+}\) is as \(\mathrm{x} \times 10^{-12} \mathrm{M}\). the value of \(\mathrm{x}\) is ___ .

\(\mathrm{K}_{s}\) of \(\mathrm{BaSO}_{4}\) is \(1 \times 10^{-10 .}\) If the ionic conductances of \(\mathrm{Ba}^{+}\)and \(\mathrm{SO}_{4}^{2-\text { ions are }} 64\) and \(80 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) respectively, then its specific conductance is (a) \(1.44 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (b) \(144 \times 10^{-8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (c) \(1.44 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\) (d) \(144 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

\(2 \mathrm{MnO}_{4}^{-}+\mathrm{aH}^{+}+\mathrm{bH}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{Mn}^{+2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{dO}_{2}\) What are the values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and \(\mathrm{d} ?\) (a) \(4,5,4,5\) (b) \(6,5,8,5\) (c) \(3,2,4,8\) (d) \(2,5,4,6\)

The specific conductance of \(0.1 \mathrm{~N} \mathrm{KCl}\) solution at \(23^{\circ} \mathrm{C}\) is \(0.012 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (a) \(0.142 \mathrm{~cm}^{-1}\) (b) \(0.616 \mathrm{~cm}^{-1}\) (c) \(6.16 \mathrm{~cm}^{-1}\) (d) \(616 \mathrm{~cm}^{-1}\)
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