Problem 81 - Chemistry Principles and Practice

Question

How many coulombs of charge are needed to accomplish each of the following conversions by electrolysis? (a) Produce \(0.50 \mathrm{~mol} \mathrm{Al}\) by electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\). (b) Reduce all of the \(\mathrm{Cu}^{2+}\) in \(100 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) (c) Make \(10.0 \mathrm{~g} \mathrm{Cl}_{2}\) by electrolysis of molten \(\mathrm{NaCl}\). (d) Deposit \(0.32 \mathrm{~g}\) silver from an aqueous \(\mathrm{AgNO}_{3}\) solution.

Step-by-Step Solution

Answer

(a) 144750 C, (b) 3860 C, (c) 27213 C, (d) 286.73 C.

Step 1: Calculate Charge for Al Electrolysis

To produce 0.50 mol of Al, we need to know the half-reaction for Al from \(\text{Al}_2\text{O}_3\):\[\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\]For 1 mol of Al, 3 mol of electrons are required. Therefore, for 0.50 mol of Al:\[\text{Electrons needed} = 0.50 \times 3 = 1.5 \text{ mol electrons}\]Using Faraday's constant (\( 96500 \text{ C/mol electrons}\)), we find the charge:\[\text{Charge} = 1.5 \times 96500 = 144750 \text{ C (Coulombs)}\]

Step 2: Calculate Charge for Cu Electrolysis

We need to calculate the moles of \(\text{Cu}^{2+}\) in 100 mL of \(0.20 \text{ M} \) \(\text{Cu(NO}_3\text{)}_2\):\[\text{Moles of Cu}^{2+} = 0.1 \times 0.20 = 0.02 \text{ mol}\]For the reduction \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\), each mol of \(\text{Cu}^{2+}\) needs 2 mol of electrons.\[\text{Electrons needed} = 0.02 \times 2 = 0.04 \text{ mol electrons}\]\[\text{Charge} = 0.04 \times 96500 = 3860 \text{ C (Coulombs)}\]

Step 3: Calculate Charge for Cl2 Production

To make \(10.0 \text{ g} \) \(\text{Cl}_2\), find the moles of \(\text{Cl}_2\):\[\frac{10.0}{70.90} = 0.141 \text{ mol of Cl}_2\]The reaction is \(2\text{Cl}^{-} \rightarrow \text{Cl}_2 + 2e^-\). Each mole of \(\text{Cl}_2\) requires 2 moles of electrons.\[\text{Electrons needed} = 0.141 \times 2 = 0.282 \text{ mol electrons}\]\[\text{Charge} = 0.282 \times 96500 = 27213 \text{ C (Coulombs)}\]

Step 4: Calculate Charge for Ag Deposition

For depositing \(0.32 \text{ g} \) \(\text{Ag}\), find the moles of \(\text{Ag}\):\[\frac{0.32}{107.87} = 0.00297 \text{ mol of Ag}\]The reaction is \(\text{Ag}^{+} + e^- \rightarrow \text{Ag}\). Each mole of \(\text{Ag}\) needs 1 mole of electrons.\[\text{Electrons needed} = 0.00297 \times 1 = 0.00297 \text{ mol electrons}\]\[\text{Charge} = 0.00297 \times 96500 = 286.73 \text{ C (Coulombs)}\]

Key Concepts

Coulomb's Law

Coulomb's Law helps us understand how electrical charges interact. It is fundamental in electrochemistry and describes the force between two point charges. The force is proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this is expressed as:\[ F = k \frac{{|q_1 q_2|}}{{r^2}} \]where:

• \( F \) is the force between the charges,
• \( k \) is Coulomb's constant,
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• \( r \) is the distance between the centers of the two charges.

This law helps predict how charged particles will behave in an electrochemical cell. A strong understanding of this interaction aids in calculating forces during electrolysis, allowing you to predict how substances will behave when they are broken down into ions. It's crucial in ensuring the correct voltages and currents in experiments and industrial processes.

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is vital for calculating the quantity of charge needed to cause a chemical change through electrolysis. It states that the amount of substance produced at an electrode is directly proportional to the quantity of electricity that passes through the electrolyte. The formula representing this relationship is:\[ m = \frac{{Q}}{{F}} \times \frac{{M}}{{n}} \]where:

• \( m \) is the mass of the substance,
• \( Q \) is the total electric charge passed through the substance,
• \( F \) is Faraday's constant (approximately 96500 C/mol),
• \( M \) is the molar mass of the substance,
• \( n \) is the number of moles of electrons exchanged per mole of substance.

In the context of the original problem, this means determining the number of moles of electrons needed for reactions such as producing aluminum from aluminum oxide and then multiplying by Faraday's constant to find the charge in coulombs needed for the reaction.

Chemical Reactions

Chemical reactions underlie the process of electrolysis. During electrolysis, electrical energy drives chemical changes that wouldn't otherwise occur naturally. In this context, reactions occur at the electrodes where substances gain or lose electrons, undergoing reduction or oxidation. This process is often referred to as redox because it includes both reduction (gain of electrons) and oxidation (loss of electrons).Some key reactions that occur in electrolysis include:

• Reduction: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \) where copper ions gain electrons to form copper metal.
• Oxidation: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \) where chloride ions lose electrons to produce chlorine gas.

Understanding these reactions helps in predicting how electrolysis can be used to produce metals or gases from ionic compounds, and is essential for correctly calculating the needed electric charge in a given electrolysis experiment.

Electrochemistry

Electrochemistry is the study of chemical processes that cause electrons to move. It plays a crucial role in understanding electrolysis where electricity prompts chemical change. It involves both the study of redox reactions and the transformation of chemical energy into electrical energy, and vice versa. In electrolysis:

• Electrical energy is used to drive non-spontaneous chemical reactions.
• Ionic compounds are decomposed into their elements.
• Electrodes are used to carry current into and out of the solution.

Electrochemical reactions are governed by the Nernst equation and other electrochemical principles, explaining how voltage relates to concentration and temperature. Knowing electrochemistry helps guide experiments in cells and batteries, allowing precise calculations of electricity needed, as illustrated in determining charge requirements for metal deposition or gas production as seen in previous exercises.