Question

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution

Short answer

In the electrolysis of: a. \(1.0 M \mathrm{NiBr}_{2}\) solution: - Cathode (reduction) reaction: \(\mathrm{Ni}^{2+} + 2e^- \rightarrow \mathrm{Ni}\) - Anode (oxidation) reaction: \(2\mathrm{Br}^- \rightarrow \mathrm{Br}_{2} + 2e^-\) b. \(1.0 M \mathrm{AlF}_{3}\) solution: - Cathode (reduction) reaction: \(\mathrm{Al}^{3+} + 3e^- \rightarrow \mathrm{Al}\) - Anode (oxidation) reaction: \(2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{O}_{2} + 4\mathrm{H}^+ + 4e^-\) (water undergoes oxidation instead of \(\mathrm{F}^-\)) c. \(1.0 M \mathrm{MnI}_{2}\) solution: - Cathode (reduction) reaction: \(\mathrm{Mn}^{2+} + 2e^- \rightarrow \mathrm{Mn}\) - Anode (oxidation) reaction: \(2\mathrm{I}^- \rightarrow \mathrm{I}_{2} + 2e^-\)

Step-by-step solution

Reviewing reduction potentials

For this exercise, we must rely on reduction potentials of different species for figuring out half-reactions at the electrodes. For metals, reduction half-reaction can be written as: \( M^{n+} + ne^- \rightarrow M \). For halogen elements, the oxidation half-reactions are: \( 2X^- \rightarrow X_2 + 2e^- \) where X is a halogen (F, Cl, Br, I). Now let's analyze each case.

Case a: \(\mathrm{NiBr}_{2}\) solution electrolysis

Given a \(\mathrm{NiBr}_{2}\) solution, we will have \(\mathrm{Ni}^{2+}\) and \(\mathrm{Br}^-\) ions in the solution. The cathode reaction will be the reduction of \(\mathrm{Ni}^{2+}\) and the anode reaction will be the oxidation of \(\mathrm{Br}^-\). Cathode (reduction) reaction: \[ \mathrm{Ni}^{2+} + 2e^- \rightarrow \mathrm{Ni} \] Anode (oxidation) reaction: \[ 2\mathrm{Br}^- \rightarrow \mathrm{Br}_{2} + 2e^- \]

Case b: \(\mathrm{AlF}_{3}\) solution electrolysis

Given a \(\mathrm{AlF}_{3}\) solution, we will have \(\mathrm{Al}^{3+}\) and \(\mathrm{F}^-\) ions in the solution. The cathode reaction will be the reduction of \(\mathrm{Al}^{3+}\) and the anode reaction will be the oxidation of \(\mathrm{F}^-\). Cathode (reduction) reaction: \[ \mathrm{Al}^{3+} + 3e^- \rightarrow \mathrm{Al} \] Anode (oxidation) reaction: \[ 2\mathrm{F}^- \rightarrow \mathrm{F}_{2} + 2e^- \] However, the oxidation of \(\mathrm{F^-}\) to \(\mathrm{F}_{2}\) has a very high standard potential. Since the electrolysis takes place under standard conditions, it's more likely that the water will undergo oxidation instead of the fluoride ions. Anode (oxidation) reaction (water instead of \(\mathrm{F}^-\)): \[ 2\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{O}_{2} + 4\mathrm{H}^+ + 4e^- \]

Case c: \(\mathrm{MnI}_{2}\) solution electrolysis

Given a \(\mathrm{MnI}_{2}\) solution, we will have \(\mathrm{Mn}^{2+}\) and \(\mathrm{I}^-\) ions in the solution. The cathode reaction will be the reduction of \(\mathrm{Mn}^{2+}\) and the anode reaction will be the oxidation of \(\mathrm{I}^-\) Cathode (reduction) reaction: \[ \mathrm{Mn}^{2+} + 2e^- \rightarrow \mathrm{Mn} \] Anode (oxidation) reaction: \[ 2\mathrm{I}^- \rightarrow \mathrm{I}_{2} + 2e^- \] In summary, during the electrolysis of each given solution, there will be a reduction half-reaction involving metal cations at the cathode and an oxidation half-reaction involving halogen anions or water at the anode.

Key concepts

Reduction Potentials

Reduction potentials are like the preferences of ions and molecules for gaining electrons. Imagine it as how much they "want" to be reduced. These potentials are measured in volts and help us predict which species will be reduced or oxidized during electrolysis.

• Every atom or ion has a specific reduction potential. This value tells us how eager it is to gain electrons and be reduced.
• Reduction potentials are usually found in tables for standard conditions.

Positive reduction potentials mean the species strongly wants electrons. For example, in electrolysis, we look at these potentials to decide what will happen at the electrodes.
In our cases, metals like nickel, aluminum, and manganese are reduced at the cathode because they have the potential to gain electrons and form solid metals.

Cathode Reaction

During electrolysis, the cathode is the site of the reduction reaction. Here, ions gain electrons to form neutral atoms or molecules. Remember this by thinking "cathode" and "cations" both relate to gaining something, like electrons.

• Metals undergo reduction at the cathode because they "gain" electrons, turning metal ions into metal atoms.
• The formula for these reactions generally looks like: \( M^{n+} + ne^- \rightarrow M \).

At the cathode for \( NiBr_2 \), \( AlF_3 \), and \( MnI_2 \), the reactions reduce \( Ni^{2+} \), \( Al^{3+} \), and \( Mn^{2+} \) to their respective metals. This is why electroplating and metal extraction often rely on controlled cathode reactions.

Anode Reaction

At the anode, oxidation takes place. Oxidation means losing electrons. Here, substances like halides or water give away electrons to become other forms.

• The general rule at the anode is "oxidation occurs."
• The typical equations for halogens would be: \( 2X^- \rightarrow X_2 + 2e^- \), where \( X \) is a halogen.

For \( NiBr_2 \) and \( MnI_2 \), the anodes oxidize \( Br^- \) and \( I^- \) to form \( Br_2 \) and \( I_2 \), respectively.
In \( AlF_3 \), the strong stability of \( F^- \) makes it more favorable for water to undergo oxidation, producing oxygen gas instead. This demonstrates the variation in anode reactions depending on the species involved.

Oxidation

Oxidation is the process of losing electrons. It can easily be remembered as "OIL" from "OIL RIG," which stands for "Oxidation Is Loss (of electrons)."

• In electrolysis, oxidation happens at the anode, where negative ions (anions) lose electrons.
• Oxidation reactions can produce gases like chlorine, bromine, or oxygen, depending on the ions present.

For example, in the electrolysis of \( NiBr_2 \), the oxidation of \( Br^- \) ions results in the production of \( Br_2 \) gas. It's oxygen that gets produced during the oxidation of water if fluoride ions are involved, showcasing how versatile oxidation processes can be.

Reduction

Reduction means gaining electrons, and this can be remembered by "RIG" from "OIL RIG," meaning "Reduction Is Gain (of electrons)."

• During electrolysis, reduction is the process happening at the cathode, where metal ions gain electrons to form solid metals.
• The reduction process is favored when the reduction potential is more positive.

In solutions like \( NiBr_2 \), \( AlF_3 \), and \( MnI_2 \), we observe the reduction of metal ions, \( Ni^{2+} \), \( Al^{3+} \), and \( Mn^{2+} \), into solid metal forms at the cathode. This is a crucial step in recovering metals from their salts.

Standard Conditions

Standard conditions in chemistry mean you're at 25°C (298 K), 1 atm pressure, and, generally, 1 M concentrations for solutions. These conditions help in comparing the spontaneity of reactions.

• Electrolysis often assumes standard conditions to predict reactions accurately using standard reduction potentials.
• However, deviations such as high concentrations or different temperatures can shift reaction possibilities.

During the electrolysis examples, standard conditions allow us to predict using typical reduction potentials. Yet, adjustments might be required when conditions change, like in a real lab environment. This helps in understanding that standard conditions are a crucial baseline for analyzing reactions.