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. \(1.0 \mathrm{M} \mathrm{CuCl}_{2}\) solution c. \(1.0 \mathrm{MgI}_{2}\) solution

Short answer

In the electrolysis of the given solutions: a. \(1.0 M \mathrm{KF}\) solution: - Anode (oxidation): 2 \(\mathrm{F}^-\) (aq) → \(\mathrm{F}_{2}\) (g) + 4 e\(^-\) - Cathode (reduction): 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq) b. \(1.0 M \mathrm{CuCl}_{2}\) solution: - Anode (oxidation): 2 \(\mathrm{Cl}^-\) (aq) → \(\mathrm{Cl}_{2}\) (g) + 2 e\(^-\) - Cathode (reduction): \(\mathrm{Cu}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Cu}\) (s) c. \(1.0 M \mathrm{MgI}_{2}\) solution: - Anode (oxidation): 2 \(\mathrm{I}^-\) (aq) → \(\mathrm{I}_{2}\) (s) + 2 e\(^-\) - Cathode (reduction): 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq)

Step-by-step solution

Identify the Cathode and Anode Reactions for the \(1.0 M \mathrm{KF}\) Solution

For a \(1.0 M \mathrm{KF}\) solution, we have both \(\mathrm{K}^+\) ions and \(\mathrm{F}^-\) ions. The anode is where oxidation takes place while the cathode is where reduction takes place. At the anode (oxidation), the possible reactions include: -2 \(\mathrm{F}^-\) (aq) → \(\mathrm{F}_{2}\) (g) + 4 e\(^-\) At the cathode (reduction), the possible reactions include: - \(\mathrm{K}^+\) (aq) + e\(^-\) → \(\mathrm{K}\) (s) - 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq) Using standard reduction potentials, we can determine that fluorine has a higher reduction potential than water, so it will be easier for \(\mathrm{F}^-\) ions to be oxidized. Meanwhile, potassium's reduction potential is lower than that of water, so it will be harder to reduce \(\mathrm{K}^+\) ions compared to the reduction of water. So in the \(1.0 M \mathrm{KF}\) solution, the reactions are: Anode (oxidation): 2 \(\mathrm{F}^-\) (aq) → \(\mathrm{F}_{2}\) (g) + 4 e\(^-\) Cathode (reduction): 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq)

Identify the Cathode and Anode Reactions for the \(1.0 M \mathrm{CuCl}_{2}\) Solution

For a \(1.0 M \mathrm{CuCl}_{2}\) solution, we have both \(\mathrm{Cu}^{2+}\) ions and \(\mathrm{Cl}^-\) ions. As before, we must determine the reactions at the anode (oxidation) and cathode (reduction). At the anode (oxidation), the possible reactions include: - 2 \(\mathrm{Cl}^-\) (aq) → \(\mathrm{Cl}_{2}\) (g) + 2 e\(^-\) At the cathode (reduction), the possible reactions include: - \(\mathrm{Cu}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Cu}\) (s) The most likely reactions to occur are: Anode (oxidation): 2 \(\mathrm{Cl}^-\) (aq) → \(\mathrm{Cl}_{2}\) (g) + 2 e\(^-\) Cathode (reduction): \(\mathrm{Cu}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Cu}\) (s)

Identify the Cathode and Anode Reactions for the \(1.0 M \mathrm{MgI}_{2}\) Solution

For a \(1.0 M \mathrm{MgI}_{2}\) solution, we have both \(\mathrm{Mg}^{2+}\) ions and \(\mathrm{I}^-\) ions. Let's determine the reactions at the anode (oxidation) and the cathode (reduction). At the anode (oxidation), the possible reactions include: - 2 \(\mathrm{I}^-\) (aq) → \(\mathrm{I}_{2}\) (s) + 2 e\(^-\) At the cathode (reduction), the possible reactions include: - \(\mathrm{Mg}^{2+}\) (aq) + 2 e\(^-\) → \(\mathrm{Mg}\) (s) - 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq) Using standard reduction potentials, we can determine that magnesium's reduction potential is lower than that of water, so it will be harder to reduce \(\mathrm{Mg}^{2+}\) ions compared to the reduction of water. So in the \(1.0 M \mathrm{MgI}_{2}\) solution, the reactions are: Anode (oxidation): 2 \(\mathrm{I}^-\) (aq) → \(\mathrm{I}_{2}\) (s) + 2 e\(^-\) Cathode (reduction): 2 H\(_{2}\)O (l) + 2 e\(^-\) → H\(_{2}\) (g) + 2 OH\(^-\) (aq)

Key concepts

Cathode Reactions

When a solution undergoes electrolysis, reduction occurs at the cathode. This is where cations gain electrons. For example, in a \(1.0 \ \mathrm{M} \ \mathrm{CuCl}_{2}\) solution, the copper ions (\(\mathrm{Cu}^{2+}\)) are reduced by gaining two electrons to form solid copper (\(\mathrm{Cu}\)): \[\mathrm{Cu}^{2+} (\text{aq}) + 2e^- \rightarrow \mathrm{Cu} (\text{s})\]This reaction is favored because copper ions readily accept electrons. They have a higher reduction potential compared to other cations in the solution.
Similarly, in a \(1.0 \ \mathrm{M} \ \mathrm{MgI}_{2}\) solution, the cathode reduces water rather than magnesium ions, due to the lower reduction potential of \(\mathrm{Mg}^{2+}\) ions compared to water. The cathode reaction is:\[2 \ \mathrm{H}_{2}\mathrm{O} (\text{l}) + 2 e^- \rightarrow \mathrm{H}_2 (\text{g}) + 2 \mathrm{OH}^- (\text{aq})\]

Anode Reactions

The anode is the site of oxidation during electrolysis. Here, anions lose electrons. Let's take a look at what happens in a \(1.0 \ \mathrm{M} \ \mathrm{KF}\) solution. At the anode, fluoride ions (\(\mathrm{F}^-\)) are oxidized to produce fluorine gas:\[2 \ \mathrm{F}^-(\text{aq}) \rightarrow \mathrm{F}_2(\text{g}) + 4e^-\]Fluoride ions have a high tendency to lose electrons, allowing them to be oxidized more readily than other potential reactions in the solution. In the case of \(1.0 \ \mathrm{M} \ \mathrm{CuCl}_{2}\) solution, chloride ions (\(\mathrm{Cl}^-\)) are oxidized:\[2\ \mathrm{Cl}^- (\text{aq}) \rightarrow \mathrm{Cl}_2 (\text{g}) + 2e^-\]This reaction is preferred because chloride ions can easily give up electrons to form chlorine gas.

Standard Reduction Potentials

Standard reduction potentials are crucial for predicting the course of electrochemical reactions. These potentials help determine which substances are more likely to be reduced or oxidized during electrolysis. Standard reduction potential gives us a measure of the tendency of a chemical species to acquire electrons and thereby be reduced.

• A higher standard reduction potential means a species is more likely to gain electrons and be reduced.
• Conversely, a lower standard reduction potential suggests that a species is more easily oxidized.

In the electrolysis of \(1.0 \ \mathrm{M} \ \mathrm{KF}\), the standard reduction potential of water is higher than that of potassium ions (\(\mathrm{K}^+\)). Thus, the reduction of water, forming hydrogen gas, is more favorable.
Understanding these potentials allows us to predict and balance chemical reactions effectively. For instance, in \(1.0 \ \mathrm{M} \ \mathrm{MgI}_{2}\), magnesium has a lower reduction potential than water, which leads to water being reduced at the cathode while iodine ions are oxidized at the anode.