Question

Predict the products of electrolysis in each of the following: An aqueous solution of CuCl\(_2\)

Answer 1

In aqueous CuCl\(_2\):

• Cations: Cu\(^{2+}\), H\(^+\)
• Anions: Cl\(^-\), OH\(^-\)

At the cathode (reduction):
The reduction reaction at the cathode involves the Cu\(^{2+}\) ion, which has a higher reduction potential than water. The copper ions gain two electrons and are reduced to form copper metal: \[ Cu^{2+} + 2e^- \rightarrow Cu(s) \]
At the anode (oxidation):
At the anode, chloride ions (Cl\(^-\)) are preferentially oxidized to chlorine gas. This is due to the easier oxidation of chloride ions compared to water. The oxidation reaction is: \[ 2Cl^- \rightarrow Cl_2(g) + 2e^- \]
Products:

• Cathode: Copper metal (Cu)
• Anode: Chlorine gas (Cl\(_2\))

Answer 2

Platinum electrodes are inert.
At the cathode (reduction):
At the cathode, the reduction of hydrogen ions (H\(^+\)) occurs. Two hydrogen ions gain two electrons to form hydrogen gas: \[ 2H^+ + 2e^- \rightarrow H_2(g) \]
At the anode (oxidation):
At the anode, water undergoes oxidation, releasing oxygen gas, hydrogen ions (H\(^+\)), and electrons: \[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
Products:

• Cathode: Hydrogen gas (H\(_2\))
• Anode: Oxygen gas (O\(_2\))

Answer 3

Reduction reaction:
The reduction of silver ions (\(Ag^+\)) involves the gain of one electron to form solid silver: \[ Ag^+ + e^- \rightarrow Ag \] One mole of \(Ag^+\) requires 1 mole of electrons for the reduction.
1 Faraday = charge of 1 mole of electrons
Since 1 mole of electrons is required for the reduction of 1 mole of \(Ag^+\), the charge required for this reaction is equal to 1 Faraday. \[ \therefore \text{Charge required = 1 Faraday} \]

Answer 4

Faraday's Second Law of Electrolysis:
Faraday’s second law of electrolysis states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the chemical equivalent of the substance and the quantity of charge passed through the electrolyte.

In mathematical form, the law can be expressed as:
\[ \frac{m_1}{m_2} = \frac{E_1}{E_2} \] where:
- \( m_1 \) and \( m_2 \) are the masses of the substances deposited at the electrodes,
- \( E_1 \) and \( E_2 \) are the chemical equivalents of the substances,
- \( m_1/m_2 \) is the ratio of the masses deposited,
- \( E_1/E_2 \) is the ratio of their chemical equivalents.

Explanation:
This law implies that when the same quantity of electricity passes through different electrolytes, the masses of the substances deposited at the electrodes are proportional to their equivalent weights. It highlights the relationship between the quantity of electric charge passed through the electrolyte and the amount of material deposited or dissolved at the electrodes.

Answer 5

Feasibility of Reactions at the Anode:
To determine which reaction is feasible at the anode during the electrolysis of aqueous sodium chloride solution, we need to compare the standard electrode potentials (E°) of the two reactions occurring at the anode.

The two reactions that can occur at the anode are:
1. \[ \text{Cl}^- \rightarrow \frac{1}{2} \text{Cl}_2 + e^- \quad E^\circ = 1.36\,V \]
2. \[ 2 \text{H}_2 \text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4 e^- \quad E^\circ = 1.23\,V \]

The reaction that occurs at the anode is the one with the **lower reduction potential** (since oxidation is the reverse of reduction). In other words, we need to look for the oxidation reaction with the smaller value of \( E^\circ \), which favors oxidation.

Step-by-step analysis:
- For the reaction involving chloride ions (Cl⁻), the standard electrode potential is \( +1.36\,V \), which means the reduction of chloride ions to chlorine gas is energetically favorable.
- For the reaction involving water, the standard electrode potential is \( +1.23\,V \), meaning the reduction of water to oxygen is also energetically favorable, but slightly less than the reduction of chloride ions.

Since oxidation occurs in the reverse direction of reduction, we consider the reverse of these reactions: - Oxidation of Cl⁻ to form Cl₂: \( 1.36\,V \) - Oxidation of H₂O to form O₂: \( 1.23\,V \)

Conclusion:
The reaction involving chloride ions (\( \text{Cl}^- \)) to form chlorine gas (\( \text{Cl}_2 \)) is more feasible because it has a higher reduction potential. This means it is easier to oxidize chloride ions at the anode compared to oxidizing water molecules.
Thus, the reaction \( \text{Cl}^- \rightarrow \frac{1}{2} \text{Cl}_2 + e^- \) is the one that occurs at the anode during the electrolysis of aqueous sodium chloride solution.