Problem 106 - Chemistry The Molecular Science

Question

In an electrolytic cell, a 10.0 -A direct current passes through an aqueous copper(II) nitrate solution and \(9.50 \mathrm{~g}\) metallic copper plates out. (a) Calculate how long it took for this mass of copper to be deposited at the cathode. Assume \(100 \%\) efficiency. (b) A gas is produced at the anode and collected. Identify the gas and calculate its volume. The gas was collected at \(25^{\circ} \mathrm{C}\) and \(0.945 \mathrm{~atm} .\)

Step-by-Step Solution

Answer

(a) 2884 seconds; (b) Oxygen, 1.941 L.

Step 1: Determine the Number of Moles of Copper Plated

First, calculate the number of moles of copper that were deposited using its molar mass. The molar mass of copper (Cu) is approximately 63.55 g/mol.Number of moles of copper, \( n = \frac{9.50 \text{ g}}{63.55 \text{ g/mol}} \approx 0.1495 \text{ moles} \).

Step 2: Relate Moles of Copper to Moles of Electrons

Copper deposits as Cu from Cu²⁺ ions, which means 2 moles of electrons are required per mole of copper. Thus, total moles of electrons, \( n_e = 2 \times 0.1495 \text{ moles} = 0.299 \text{ moles} \).

Step 3: Calculate Time using Faraday’s Law

Faraday's law of electrolysis states that \( moles = \frac{It}{nF} \), where:- \( I = 10 \text{ A} \) (current),- \( t \) = time in seconds,- \( n = 2 \) (from copper's equation),- \( F = 96485 \text{ C/mol} \) (Faraday's constant).Rearrange to solve for time, \( t = \frac{n_e \times F}{I} = \frac{0.299 \text{ moles} \times 96485 \text{ C/mol}}{10 \text{ A}} \approx 2884 \text{ seconds} \).

Step 4: Identify the Gas at the Anode

In an aqueous copper nitrate solution, the gas produced at the anode is typically oxygen (O₂) due to the oxidation of water.

Step 5: Use the Ideal Gas Law to Find Volume of Gas

The ideal gas law is \( PV = nRT \).- Temperature \( T \) must be in Kelvin: \( 25^{ ext{o}}C = 298 ext{ K} \).- Universal gas constant \( R = 0.0821 \text{ L atm/mol K} \).- Pressure, \( P = 0.945 \text{ atm} \).First, calculate moles of O₂ produced using \( rac{1}{4} \) of the total moles of electrons (since each O₂ requires 4 electrons):\( n_{O_2} = \frac{0.299}{4} \approx 0.07475 \text{ moles} \).Now apply the ideal gas law:\( V = \frac{nRT}{P} = \frac{0.07475 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K}}{0.945 \text{ atm}} \approx 1.941 \text{ L} \).

Key Concepts

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a key concept that describes how electric current interacts with chemical substances. It explains that the amount of substance produced at each electrode is proportional to the amount of electric charge passed through the cell. For electrolytic processes, this means:

• The mass of an element deposited is proportional to the quantity of electricity.
• The equation used is: \( \, moles = \frac{It}{nF} \, \), where \( \, I \, \) is current in amperes, \( \, t \, \) is time in seconds, \( \, n \, \) is the number of electrons exchanged per mole of substance, and \( \, F \, \), Faraday's constant, is approximately \( \, 96485 \text{ C/mol} \, \).

Consequently, this law helps to determine the time needed or the mass of deposited substance when the current and conditions are known. In our exercise, it was used to find out how long it took for copper to deposit.

Copper Deposition

Copper deposition occurs when copper ions gain electrons and become solid copper atoms in an electrolytic cell. This process is represented by the chemical equation:

• Copper ions \((\text{Cu}^{2+})\) receive 2 electrons \((2e^-)\) to become solid copper \((\text{Cu})\).
• This means \(2\, \text{moles of electrons are needed for each mole of copper deposited.}\)

The amount of deposited copper can be calculated using its molar mass. In electrolytic cells, as observed in the exercise, the copper ions are reduced and plate out on the cathode as pure copper. This mirrors how electrochemical reactions transform ions in a solution into solid metal.

Ideal Gas Law

The Ideal Gas Law is an essential tool for understanding gases in various conditions and is represented as:

• \( PV = nRT \).

The variables represent:

• \( P \): Pressure of the gas
• \( V \): Volume of the gas
• \( n \): Number of moles of the gas
• \( R \): Universal gas constant, approximately \(0.0821 \text{ L atm/mol K}\)
• \( T \): Temperature in Kelvin

In the exercise, the Ideal Gas Law was used to discover the volume of oxygen gas produced at the anode. By converting temperature to Kelvin and using the provided pressure, the volume of the gas could be accurately determined.

Electrolysis Efficiency

Electrolysis efficiency represents how well the electrolytic cell converts electric energy into chemical energy, resulting in material deposition. In theoretical exercises such as the one provided, it's usually assumed to be 100% efficient:

• 100% efficiency means every electron passing through contributes to the reaction and product formation.
• In real-world applications, efficiency may vary due to factors like energy losses, side reactions, or material impurities.

For the exercise calculation, this 100% efficiency assumption makes things simpler by indicating that all supplied current directly contributes to copper deposition.

Anode and Cathode Reactions

Reactions at the anode and cathode are critical in electrolysis. At each electrode different chemical changes occur:

• Cathode: Reduction occurs. For copper, \(\, \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\,\) is the typical reaction.
• Anode: Oxidation is the primary process. In the case of the exercise, oxygen gas \((\text{O}_2)\) was produced via water oxidation, especially as \(\text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-}\).

The components and ions in a solution, along with physical settings, determine the exact reactions occurring. The reactions at both electrodes are governed by the movement of electrons, confirming the potential for various reactions given the electrolyte types.