Question

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. 1.0 \(M \mathrm{KF}\) solution b. 1.0\(M \mathrm{CuCl}_{2}\) solution c. 1.0 \( M \mathrm{MgI}_{2}\) solution

Short answer

For the electrolysis of: a. 1.0 M KF solution: Cathode: \(2H_{2}O + 2e^{-} \rightarrow H_{2} (g) + 2OH^{-}\) ; Anode: \(2H_{2}O \rightarrow O_{2} (g) + 4H^{+} + 4e^{-}\) b. 1.0 M CuCl₂ solution: Cathode: \(Cu^{2+} (aq) + 2e^{-} \rightarrow Cu (s)\) ; Anode: \(2Cl^{-} (aq) \rightarrow Cl_{2} (g) + 2e^{-}\) c. 1.0 M MgI₂ solution: Cathode: \(Mg^{2+} (aq) + 2e^{-} \rightarrow Mg (s)\) ; Anode: \(2I^{-} (aq) \rightarrow I_{2} (s) + 2e^{-}\)

Step-by-step solution

Identify the possible reduction and oxidation half-reactions

We first look at the possible reduction half-reaction happening at the cathode: \(F^{-} (aq) + e^{-} \rightarrow F (s)\) The standard reduction potential is -2.87V. This process would require more energy to reduce the F- to F so it is not feasible. Which means that water will get reduced at the cathode to form hydrogen gas and hydroxide ions: \(2H_{2}O + 2e^{-} \rightarrow H_{2} (g) + 2OH^{-}\) Now let's look at the oxidation half-reaction happening at the anode: \(K(s) \rightarrow K^{+} (aq) + e^{-}\) The potential is -2.93V, meaning it is also not feasible. Instead, water oxidation will occur at the anode: \(2H_{2}O \rightarrow O_{2} (g) + 4H^{+} + 4e^{-}\)

Write the overall reaction

Combine the two half-reactions at the cathode and the anode: \(2H_{2}O + 2F^{-} (aq) \rightarrow H_{2} (g) + O_{2} (g) + 2OH^{-} (aq) + 2F^{-} (aq)\) b. Electrolysis of 1.0 M CuCl₂ solution

Identify the possible reduction and oxidation half-reactions

Possible reduction half-reaction at the cathode: \(Cu^{2+} (aq) + 2e^{-} \rightarrow Cu (s)\) Possible oxidation half-reactions at the anode: \(2Cl^{-} (aq) \rightarrow Cl_{2} (g) + 2e^{-}\)

Write the overall reaction

Combine the two half-reactions at the cathode and the anode: \(Cu^{2+} (aq) + 2Cl^{-} (aq) \rightarrow Cu (s) + Cl_{2} (g)\) c. Electrolysis of 1.0 M MgI₂ solution

Identify the possible reduction and oxidation half-reactions

Possible reduction half-reaction at the cathode: \(Mg^{2+} (aq) + 2e^{-} \rightarrow Mg (s)\) Possible oxidation half-reactions at the anode: \(2I^{-} (aq) \rightarrow I_{2} (s) + 2e^{-}\)

Write the overall reaction

Combine the two half-reactions at the cathode and the anode: \(Mg^{2+} (aq) + 2I^{-} (aq) \rightarrow Mg (s) + I_{2} (s)\)

Key concepts

Cathode Reaction

In electrolysis, the cathode is the negative electrode where reduction reactions take place. Reducing agents lose electrons here, as opposites attract—positive ions from the solution are drawn to the negatively charged cathode. These ions gain electrons, leading to reduction.

Consider a classic example in electrolysis of a 1.0 M CuCl₂ solution. Here, copper ions, \(Cu^{2+}\), approach the cathode. They gain two electrons each to form solid copper metal:

• \(Cu^{2+} (aq) + 2e^{-} \rightarrow Cu (s)\)

This process is favorable because the standard reduction potential is positive, making copper deposition efficient.

When dealing with a KF solution, water reduces preferentially over fluoride ions due to a lower potential requirement:

• \(2H_{2}O + 2e^{-} \rightarrow H_{2} (g) + 2OH^{-}\)

In summary, the cathode hosts reduction reactions, converting ions in solution to their elemental forms or simpler molecules, dependent on electrode potential.

Anode Reaction

The anode in electrolysis is the positive electrode, where oxidation takes place. Here, ions or atoms lose electrons. This loss of electrons results in an oxidation reaction.

In a 1.0 M CuCl₂ solution, chloride ions approach the anode. They lose electrons to form chlorine gas:

• \(2Cl^{-} (aq) \rightarrow Cl_{2} (g) + 2e^{-}\)

During the electrolysis of 1.0 M MgI₂ solution, iodide ions oxidize upon reaching the anode, resulting in iodine gas formation:

• \(2I^{-} (aq) \rightarrow I_{2} (s) + 2e^{-}\)

In solutions with water, such as KF, water oxidizes to form oxygen gas:

• \(2H_{2}O \rightarrow O_{2} (g) + 4H^{+} + 4e^{-}\)

Thus, the anode's oxidation process determines the chemical transformation during electrolysis, relying on potential differences to release electrons.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are chemical processes involving the transfer of electrons between two substances. In electrolysis, redox is split into two half-reactions: reduction at the cathode and oxidation at the anode.

These reactions are driven by the application of an external voltage. In the electrolytic cell, this external voltage forces the non-spontaneous reaction of electron flow: from the anode, where oxidation occurs, to the cathode for reduction. Each reaction involves a specific potential that determines the feasibility of the reaction; electrons flow "downhill" energetically speaking.

In electrolysis of a 1.0 M CuCl₂ solution, the anode reaction involves the oxidation of chloride ions, while the cathode sees the reduction of copper ions. These balanced reactions are:

• Oxidation: \(2Cl^{-} (aq) \rightarrow Cl_{2} (g) + 2e^{-}\)
• Reduction: \(Cu^{2+} (aq) + 2e^{-} \rightarrow Cu (s)\)

Combining these gives a redox reaction where electrons are transferred from chloride ions to copper ions, leading to elemental chlorine and copper.

Understanding redox reactions helps illuminate the choreography of ions during electrolysis, where electron movement is fundamentally transforming compounds.