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In an electrolytic cell, electrolysis is carried out. Electrical energy is converted into chemical energy. In an electrochemical cell, chemical reaction, i.e., redox reaction occurs and electricity is generated. So chemical energy is converted into electrical energy. Electrolysis is governed by Faraday's laws. The potential difference between the electrodes which is called electromotive force is responsible for the generation of electric energy in the electrochemical cells. Two electrolytic cells, one containing acidified \(\mathrm{FeCl}_{2}\) and another acidified \(\mathrm{FeCl}_{3}\) are connected in series. The ratio of iron deposited at the cathodes in the tow cells will be (a) \(3: 1\) (b) \(2: 1\) (c) \(2: 3\) (d) \(3: 2\)

Short Answer

Expert verified
The ratio of iron deposited is \( 3:2 \) (option d).

Step by step solution

01

Understand the Electrolysis Process

Identify what happens during electrolysis in both cells. In the first cell containing acidified \( \mathrm{FeCl}_{2} \), \( \mathrm{Fe^{2+}} \) ions are reduced at the cathode. In the second cell comprising \( \mathrm{FeCl}_{3} \), \( \mathrm{Fe^{3+}} \) ions are reduced.
02

Write Cathode Reactions

For the first cell, the cathode reaction is \( \mathrm{Fe^{2+} + 2e^- \rightarrow Fe} \). For the second cell, the cathode reaction is \( \mathrm{Fe^{3+} + 3e^- \rightarrow Fe} \).
03

Apply Faraday's Laws of Electrolysis

According to Faraday's second law, the mass of the substance deposited is proportional to the equivalent weight. The equivalent weight for \( \mathrm{Fe^{2+}} \) is \( \frac{M}{2} \) and for \( \mathrm{Fe^{3+}} \) is \( \frac{M}{3} \), where \( M \) is the atomic mass of iron.
04

Determine the Ratio of Iron Deposited

With both cells in series, the same amount of charge passes through each cell. Hence, the mass of iron deposited in the first cell is related to \( \frac{M}{2} \) and in the second cell to \( \frac{M}{3} \). The ratio is given by the inverse of their equivalent weights: \( \frac{3}{2} \).
05

Final Answer

The ratio of iron deposited at the cathodes of the two cells (\( \mathrm{FeCl_2} \) : \( \mathrm{FeCl_3} \)) is \( 3:2 \).

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Key Concepts

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

Electrolytic cells
An electrolytic cell is a device that uses electrical energy to drive a non-spontaneous chemical reaction. It's a wonderful application of electrolysis, turning electrical power into chemical transformation. In an electrolytic cell, you typically have two electrodes: the anode, where oxidation occurs, and the cathode, where reduction takes place. You immerse these electrodes in an electrolyte solution.Imagine you have a battery connected to these electrodes. The battery supplies the necessary energy to push ions in the solution towards the electrodes. This causes a chemical reaction to happen at the electrodes, despite it not wanting to naturally. This is the opposite of what happens in a typical battery, where a chemical reaction produces electrical energy to power devices.In the case of the original exercise, two different electrolytic cells are connected, one with acidified \( \mathrm{FeCl}_2 \) and another with \( \mathrm{FeCl}_3 \). Both require electrical energy to deposit iron at the cathode, illustrating how versatile and useful electrolytic cells are.
Faraday's laws
Faraday's laws are fundamental principles that govern electrolysis. They're named after Michael Faraday, who discovered them in the 19th century. These laws make it easier to calculate how much of a substance will be deposited from an electrolyte solution during electrolysis.

Faraday's First Law

This law states that the amount of chemical change produced by an electric current is proportional to the quantity of electricity passed through the electrolyte. In simpler terms, the more electricity you pass, the more substance you'll get deposited.

Faraday's Second Law

The second law tells us that the amount of different substances liberated by the same quantity of electricity passing through the electrolyte is proportional to their equivalent weights. Equivalent weight is a factor based on the substance's atomic weights and charge, very handy when figuring out how much gets deposited. In the exercise, this law helps determine the ratio of iron deposited, by comparing the equivalent weights of \( \mathrm{Fe}^{2+} \) and \( \mathrm{Fe}^{3+} \) ions.
Redox reactions
Redox reactions are the heart of electrolysis, involving the transfer of electrons leading to reduction and oxidation. "Redox" is shorthand for reduction-oxidation.

Oxidation and Reduction

- **Oxidation:** Loss of electrons. Occurs at the anode in an electrolytic cell.- **Reduction:** Gain of electrons. Happens at the cathode. In the initial exercise, two different ions undergo reduction:- \( \mathrm{Fe}^{2+} \) ions reduce by gaining two electrons to form metallic iron.- \( \mathrm{Fe}^{3+} \) ions reduce by gaining three electrons.This difference in electron requirements influences how much iron is deposited, as described by Faraday's laws. Understanding these reactions allows you to predict and control chemical transformations in electrolytic and electrochemical cells.
Iron deposition
Iron deposition through electrolysis is a fascinating process where iron ions in a solution become solid iron on an electrode. This is a clear, real-world application of all the principles we've explored.In the exercise, two different iron ions are involved: \( \mathrm{Fe}^{2+} \) and \( \mathrm{Fe}^{3+} \). Each requires a different number of electrons to reduce into solid iron:- \( \mathrm{Fe}^{2+} \) captures two electrons.- \( \mathrm{Fe}^{3+} \) needs three electrons.Faraday's laws come into play here, determining how much iron is deposited when the same amount of electricity is distributed in series through both cells. Since the equivalent weight of iron for \( \mathrm{Fe}^{3+} \) is lower than for \( \mathrm{Fe}^{2+} \), less iron is deposited in the \( \mathrm{FeCl}_3 \) cell compared to the \( \mathrm{FeCl}_2 \) cell.By understanding this process, industries can efficiently deposit metals on other materials, which is essential in manufacturing and engineering. This not only illustrates a fundamental chemical process but also shows the practical application of electrochemistry in industrial settings.

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Most popular questions from this chapter

\(2 \mathrm{MnO}_{4}^{-}+\mathrm{aH}^{+}+\mathrm{bH}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{Mn}^{+2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{dO}_{2}\) 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) \(2,5,4,6\)

Four elements \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) can form diatomic molecules and monoatomic anions with \(-1\) charge. Consider the following reactions about these. \(2 \mathrm{~B}^{-}+\mathrm{C}_{2} \longrightarrow 2 \mathrm{C}^{-}+\mathrm{B}_{2}\) \(\mathrm{B}_{2}+2 \mathrm{D}^{-} \longrightarrow 2 \mathrm{~B}^{-}+\mathrm{D}_{2}\) \(2 \mathrm{~A}^{-}+\mathrm{C}_{2}\) no reaction Select correct statement about these. (1) \(\mathrm{A}_{2}\) is strongest oxidizing agent while \(\mathrm{D}\) is strongest reducing agent (2) \(\mathrm{D}_{2}\) is strongest oxidizing agent while \(\mathrm{A}\) is strongest reducing agent (3) \(\mathrm{C}_{2}\) will oxidize \(\mathrm{B}^{-}\)and also \(\mathrm{D}^{-}\)to form \(\mathrm{B}_{2}\) and \(\mathrm{D}_{2}\) (4) \(\mathrm{E}^{\circ} \mathrm{A}_{2} / \mathrm{A}^{-}\)is the lowest (a) 2 and 3 (b) 1 and 3 (c) 2 and 4 (d) 1,2 and 3

\(\mathrm{K}_{s}\) 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) \(144 \times 10^{8} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

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) \(\mathrm{NaCl}(\mathrm{aq})\)

In the ionic equation \(\mathrm{xBrO}_{3}^{-}+\mathrm{yCr}^{3+}+\mathrm{zH}_{2} \mathrm{O} \longrightarrow \mathrm{Br}_{2}+\mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+}\) (a) \(\mathrm{x}=6, \mathrm{y}=8, \mathrm{z}=10\) (b) \(x=10, y=6, z=22\) (c) \(x=6, y=8, z=10\) (d) \(x=6, y=10, z=22\)

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