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Chapter 19: Problem 43

The half-reaction at an electrode is: \(\mathrm{Mg}^{2+}(\) molten \()+2 e^{-} \longrightarrow \mathrm{Mg}(s)\) Calculate the number of grams of magnesium that can be produced by supplying \(1.00 \mathrm{~F}\) to the electrode.

Short Answer

Expert verified
12.15 grams of magnesium can be produced.

Step by step solution

01

Understanding Faraday's Law

Faraday's Law of Electrolysis tells us that 1 faraday ( 1 F ) corresponds to the charge of 1 mole of electrons. This is approximately 96500 C (Coulombs), the charge of 1 mole of electrons.
02

Determine Moles of Electrons

Since 1.00 F is used, this means that 1 mole of electrons are provided for the half-reaction.
03

Relate Electrons to Moles of Magnesium

The half-reaction shows that 2 e^{-} are needed to produce 1 mole of Mg . Therefore, 1 mole of electrons will produce 0.5 moles of Mg.
04

Calculate Molar Mass of Magnesium

The molar mass of magnesium ( Mg ) is approximately 24.3 g/mol.
05

Calculate Mass of Magnesium Produced

To find the mass of 0.5 moles of Mg , multiply the moles by the molar mass: 0.5 moles imes 24.3 g/mol = 12.15 g .

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Key Concepts

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

Understanding Half-Reaction
In electrochemistry, a half-reaction is a representation of just one part of the overall redox (reduction-oxidation) reaction. It focuses on either the reduction or the oxidation that takes place at the electrodes during electrolysis. In the provided exercise, the half-reaction is: \( \mathrm{Mg}^{2+} + 2 e^{-} \rightarrow \mathrm{Mg} \).
This equation shows the reduction process, where magnesium ions \( (\mathrm{Mg}^{2+}) \) gain electrons \((2\, e^{-})\) to form solid magnesium \((\mathrm{Mg})\). Understanding the half-reaction is crucial because it helps determine how many electrons are involved and the relationships between ions and the solid produced.
Basics of Electrolysis
Electrolysis is the process of using electricity to cause a non-spontaneous chemical reaction. During electrolysis, an electric current is passed through a substance to drive a chemical change. In the context of the half-reaction \( \mathrm{Mg}^{2+} + 2 e^{-} \rightarrow \mathrm{Mg} \), this process involves reducing magnesium ions to form magnesium metal.
The device where electrolysis takes place is called an electrolytic cell. In this cell, the source of electrical energy usually compels ions to move, enabling previously unattainable compound separation or formulation.
Calculating Moles of Electrons
Moles of electrons are an essential concept when dealing with Faraday's Law of Electrolysis. Faraday's law indicates that the amount of substance transformed at an electrode is directly proportional to the number of moles of electrons exchanged.
In this exercise, 1.00 Faraday (\( \mathrm{F} \)) of electrical charge supplies exactly 1 mole of electrons. Since the half-reaction \( \mathrm{Mg}^{2+} + 2 e^{-} \rightarrow \mathrm{Mg} \) requires 2 electrons (\(2 e^{-}\)) to convert magnesium ions into magnesium metal, 1.00 Faraday enables the production of 0.5 moles of magnesium.
Understanding Molar Mass of Magnesium
Molar mass is the mass of one mole of a given substance, expressed in grams per mole (\( \mathrm{g/mol} \)). For magnesium, the molar mass is approximately \( 24.3 \) g/mol. This means that one mole of magnesium atoms weighs roughly 24.3 grams.
In the context of our problem, using 1.00 Faraday of charge produces 0.5 moles of magnesium. To calculate the mass of magnesium produced, we need to multiply the number of moles by the molar mass: \[ 0.5\, \text{moles} \times 24.3\, \text{g/mol} = 12.15\, \text{g} \].Thus, 12.15 grams of magnesium can be produced, showcasing the importance of understanding the molar mass to find the mass of substances generated in electrolysis.

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Most popular questions from this chapter

A spoon was silver-plated electrolytically in an \(\mathrm{AgNO}_{3}\) solution. (a) Sketch a diagram for the process. (b) If \(0.884 \mathrm{~g}\) of Ag was deposited on the spoon at a constant current of \(18.5 \mathrm{~mA}\), how long (in min) did the electrolysis take?

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