Question

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{ll} {\text{}} & \quad{ \mathscr{E}^{\circ} } \\ \hline {\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}} & {2.01 \mathrm{V}} \\ {\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}} & {1.23 \mathrm{V}} \\ {2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}} & {-0.83 \mathrm{V}} \\\ {\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}} & {-2.71 \mathrm{V}}\end{array}$$

Short answer

During the electrolysis of an aqueous solution of \(Na_2SO_4\), the reaction at the anode (oxidation) is \(S_2O_{8}^{2-} \longrightarrow 2SO_{4}^{2-} + 2e^-\), and the reaction at the cathode (reduction) is \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\).

Step-by-step solution

Identify possible half-reactions.

From the given table, we have four possible half-reactions: 1. \(S_2O_{8}^{2-} + 2e^- \longrightarrow 2SO_{4}^{2-}\) , E°=2.01 V 2. \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\) , E°=1.23 V 3. \(2H_2O + 2e^- \longrightarrow H_2 + 2OH^-\) , E°=-0.83 V 4. \(Na^+ + e^- \longrightarrow Na\) , E°=-2.71 V

Determine the reaction occurring at the anode.

The anode is the electrode where oxidation occurs. Therefore, we need to consider the most positive oxidation potential. In this case, the half-reaction \(S_2O_{8}^{2-} + 2e^- \longrightarrow 2SO_{4}^{2-}\) has the highest E° value (2.01 V). Thus, at the anode, the reaction is as follows: \(S_2O_{8}^{2-} \longrightarrow 2SO_{4}^{2-} + 2e^-\)

Determine the reaction occurring at the cathode.

The cathode is the electrode where reduction occurs. Therefore, we need to consider the most positive reduction potential. From the table, the half-reaction \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\) has the highest E° value (1.23 V). Thus, at the cathode, the reaction is as follows: \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\)

Write the reactions at the anode and cathode.

The reactions at the anode and cathode are: Anode (oxidation): \(S_2O_{8}^{2-} \longrightarrow 2SO_{4}^{2-} + 2e^-\) Cathode (reduction): \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\) These are the reactions occurring during the electrolysis of an aqueous solution of Na2SO4 under standard conditions.

Key concepts

Aqueous Solution

An aqueous solution is a solution where water is the solvent. This means that the solute is dissolved in water, creating a homogenous mixture. When considering electrolysis in an aqueous solution, it is important to understand that water itself can play an active role in the reactions. This can impact what reactions occur at the electrodes. For instance, in the electrolysis of sodium sulfate (\( \mathrm{Na}_2 \mathrm{SO}_4 \)), the compound dissolves in water to form sodium ions (\( \mathrm{Na}^+ \)) and sulfate ions (\( \mathrm{SO}_4^{2-} \)).

The presence of water alongside these ions opens up the possibility for additional reactions involving water molecules. This is why it's crucial to evaluate the standard electrode potentials to predict which half-reactions will occur. An important takeaway is that in an aqueous solution, both the dissolved ions and water can be involved in chemical reactions, potentially sharing or exchanging electrons with the electrodes.

Anode Reaction

During electrolysis, the anode is characterized as the electrode where oxidation occurs. Oxidation involves the loss of electrons. To determine the reactions that occur at the anode in an aqueous solution of sodium sulfate, we review the standard electrode potentials.

For this scenario, we need to look for the most positive oxidation potential, which generally indicates a favorable reaction. According to the presented data, the possible oxidation half-reaction at the anode with the highest electrode potential is
\( \mathrm{S}_2\mathrm{O}_8^{2-} \longrightarrow 2\mathrm{SO}_4^{2-} + 2\mathrm{e}^- \) with an E° of 2.01 V.

This is the oxidation reaction that occurs at the anode. It is important to remember that electrons are being lost from the ions, moving through the external circuit to the cathode.

Cathode Reaction

On the cathode side, reduction—the gain of electrons—takes place. For an aqueous sodium sulfate solution undergoing electrolysis, determining the cathode reaction requires examining the most positive reduction potential among the possible half-reactions.

In this case, the reaction with the highest possibility is \( \mathrm{O}_2 + 4\mathrm{H}^+ + 4\mathrm{e}^- \longrightarrow 2\mathrm{H}_2\mathrm{O} \) that comes with an E° of 1.23 V.

This reaction is favored for the cathode under standard conditions. The oxygen and protons in the environment become reduced, gaining electrons to eventually produce water. This process is critical for balancing the redox reactions that happen in the electrochemical cell. Understanding which reactions occur at each electrode can help in analyzing the efficiency and products of the electrolysis process.

Standard Electrode Potentials

Standard electrode potentials (\( E^o \)) are critical in electrochemical reactions because they help to predict the direction of flow of electrons. They are measured in volts and indicate how easily a substance can gain or lose electrons compared to the standard hydrogen electrode, which has a potential of 0 V.

For any half-reaction, a more positive \( E^o \) means a greater tendency to occur as a reduction half-reaction. Conversely, a less positive or negative \( E^o \) is favored for oxidation. In the context of aqueous sodium sulfate electrolysis, standard electrode potentials direct us in choosing the correct half-reactions at the anode and cathode respectively.

• The anode reaction corresponds to the highest potential for oxidation, \( 2.01 \, \mathrm{V} \) for \( \mathrm{S}_2\mathrm{O}_8^{2-} \longrightarrow 2\mathrm{SO}_4^{2-} \).
• The cathode reaction aligns with the maximum reduction potential, \( 1.23 \, \mathrm{V} \) for \( \mathrm{O}_2 + 4\mathrm{H}^+ + 4\mathrm{e}^- \longrightarrow 2\mathrm{H}_2\mathrm{O} \).

These calculations are pivotal in electrochemistry to forecast the course of electrode reactions and harnessing the reactions effectively, whether for industrial electrolysis or laboratory experiments.