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What is Faraday's contribution to quantitative electrolysis?

Short Answer

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Faraday's contribution includes his two laws of electrolysis, which relate the mass of substances altered at electrodes to the electric charge used.

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01

Understanding Faraday's Laws of Electrolysis

Faraday's contribution to quantitative electrolysis is encapsulated in his two fundamental laws of electrolysis. The first law states that the amount of substance deposited at an electrode during electrolysis is directly proportional to the total electric charge passed through the electrolyte. Mathematically, this is expressed as \( m = kQ \), where \( m \) is the mass of the substance, \( Q \) is the charge, and \( k \) is a proportionality constant.
02

Introducing the Second Law of Electrolysis

Faraday's second law of electrolysis establishes the relationship between the masses of different substances liberated by the same quantity of electricity. It states that the mass of a substance deposited or liberated is proportional to its equivalent weight when the same electric charge is used, expressed as \( m = \frac{Q}{F} \times \text{equivalent weight} \), where \( F \) is the Faraday constant.
03

Applying Faraday's Laws

These laws can be applied to determine the quantities of various substances involved in electrolysis reactions. They provide the basis for calculating how much of a material can be separated from its ions in an electrolytic process, such as determining how much copper can be plated from a copper sulfate solution by passing a specific amount of electric charge.

Key Concepts

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

Quantitative Electrolysis
Quantitative electrolysis is a fascinating concept that stems from the principles laid out by Michael Faraday. It involves the quantitative measurement of substances transformed or deposited during the electrolytic process.
The essence of quantitative electrolysis lies in understanding how electric charge affects the amount of material formed at electrodes.
Faraday's first law of electrolysis provides a foundation by expressing that the mass of a substance formed is directly proportional to the total electric charge passed through the electrolyte. This implies that if you double the charge, the mass of the substance will also double. It's a direct, mathematical relationship that is crucial for calculations in electrochemical processes.
So, when you see a chemical reaction taking place in an electrolytic cell, you can quantify the results precisely by knowing the charge that has flowed.
Electric Charge
The concept of electric charge is central to understanding Faraday’s laws and electrolysis in general. It is the charge that drives the chemical reactions in the electrolytic process.
Electric charge, denoted by the symbol \( Q \), is measured in coulombs and is essentially the currency of electrochemistry.
During electrolysis, charges are carried by ions through an electrolyte - that is, a solution capable of conducting electricity. The electric charge causes ions to move towards electrodes, where they are either reduced or oxidized, leading to the deposition or liberation of substances.
Understanding the flow of electric charge helps in predicting and controlling what happens in an electrolytic cell, making it a vital component of quantitative electrolysis.
Electrolytic Process
The electrolytic process is a method of driving a non-spontaneous chemical reaction using electric energy. It's a technique that transforms ions in a solution into a solid material.
Essentially, an electrolysis setup consists of an electrolytic cell with two electrodes submerged in an electrolyte solution.
When electric current passes through, it prompts ions to move towards respective electrodes. At the cathode, reduction occurs because electrons are gained, while at the anode, oxidation happens as electrons are lost.
This procedure can be controlled by the amount of electric charge provided, which subsequently determines the mass of substances deposited or dissolved, illustrating the practical application of Faraday's laws.
Mass of Substance
In electrolysis, the mass of a substance separated at an electrode is a key factor for understanding the entire process. Faraday’s laws give precise methodologies to calculate this mass based on the electric charge involved.
According to Faraday's first law, the equation \( m = kQ \) shows that the mass \( m \) of the substance is directly proportional to the electric charge \( Q \).
Similarly, Faraday's second law relates the mass of different substances liberated by the same quantity of electricity to their equivalent weights. This means that when a consistent charge is applied, heavier or lighter deposits can be predicted based on their equivalent weights.
By quantifying the mass, industries can plan efficient electrolytic processes for purposes like metal plating, refining, and in the production of chemicals, highlighting the importance of precise electrochemical calculations.

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Most popular questions from this chapter

The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable: The net transformation is \(\mathrm{Zn}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnO}(s)\) (a) Write the half-reactions at the zinc-air electrodes, and calculate the standard emf of the battery at \(25^{\circ} \mathrm{C}\). (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from \(1 \mathrm{~kg}\) of the metal) of the zinc electrode? (d) If a current of \(2.1 \times 10^{5} \mathrm{~A}\) is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the battery every second? Assume that the temperature is \(25^{\circ} \mathrm{C}\) and the partial pressure of oxygen is 0.21 atm.

Balance the following redox equations by the half-reaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution) (e) \(\mathrm{Mn}^{2+}+\mathrm{BiO}_{3}^{-} \longrightarrow \mathrm{Bi}^{3+}+\mathrm{MnO}_{4}^{-}\) (in acidic solution)

For a number of years, it was not clear whether mercury(I) ions existed in solution as \(\mathrm{Hg}^{+}\) or as \(\mathrm{Hg}_{2}^{2+}\). To distinguish between these two possibilities, we could set up the following system: $$ \operatorname{Hg}(l) \mid \text { soln } \mathrm{A} \| \operatorname{soln} \mathrm{B} \mid \operatorname{Hg}(l)$$ where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained \(2.63 \mathrm{~g}\) mercury(I) nitrate per liter. If the measured emf of such a cell is \(0.0289 \mathrm{~V}\) at \(18^{\circ} \mathrm{C},\) what can you deduce about the nature of the mercury(I) ions?

Define the term overvoltage. How does overvoltage affect electrolytic processes?

Based on the following standard reduction potentials: $$\begin{aligned}\mathrm{Fe}^{2+}(a q)+2 e^{-} & \longrightarrow \mathrm{Fe}(s) & & E_{1}^{\circ}=-0.44 \mathrm{~V} \\ \mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E_{2}^{\circ} &=0.77 \mathrm{~V} \end{aligned}$$ calculate the standard reduction potential for the halfreaction:$$\mathrm{Fe}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Fe}(s) \quad E_{3}^{\circ}=?$$

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