Question

Electrolysis of aqueous \(\mathrm{CuSO}_{4}\) with inert electrodes gives (a) Cu at cathode, anode gets dissolved (b) Cu at cathode, \(\mathrm{O}_{2}\) at anode (c) \(\mathrm{O}_{2}\) at anode, \(\mathrm{H}_{2}\) at cathode (d) \(\mathrm{O}_{2}\) at anode, cathode gets dissolved

Short answer

Question: In an electrolysis of aqueous CuSO₄ with inert electrodes, the products formed at cathode and anode are: (a) H₂ at cathode, O₂ at anode (b) Cu at cathode, O₂ at anode (c) Cu at cathode, SO₃ at anode (d) H₂ at cathode, SO₃ at anode Answer: (b) Cu at cathode, O₂ at anode

Step-by-step solution

Identify the reactions happening at the cathode and anode

When an aqueous \(\mathrm{CuSO}_{4}\) solution undergoes electrolysis, the possible reactions that can occur are: Cathode (reduction occurs): 1. \(\mathrm{Cu^{2+}} + 2\mathrm{e^{-}} \longrightarrow \mathrm{Cu}\) 2. \(\mathrm{2H_{2}O + 2\mathrm{e^{-}} \longrightarrow \mathrm{H}_{2}+ 2\mathrm{OH^{-}}\) Anode (oxidation occurs): 1. \(\mathrm{2H_{2}O \longrightarrow O_{2} + 4\mathrm{H^{+}} + 4\mathrm{e^{-}}\) 2. \(\mathrm{SO^{2-}_{4} \longrightarrow SO_{3}+ \mathrm{e^{-}}\)

Determine the electrode potentials of the reactions

To predict which reaction is more likely to happen in each case, let's use the standard electrode potentials of the possible reactions: Cathode (reduction): 1. Copper reduction: \(E^{\circ}=+0.337\,\mathrm{V}\) 2. Water reduction (formation of \(\mathrm{H}_{2}\)): \(E^{\circ}=-0.83\,\mathrm{V}\) Anode (oxidation): 1. Water oxidation (formation of \(\mathrm{O}_{2}\)): \(E^{\circ}=+1.23\,\mathrm{V}\) 2. Sulfate oxidation (formation of \(\mathrm{SO}_{3}\)): This reaction does not occur easily, as sulfate ions are very stable and have a very high positive electrode potential.

Evaluate the overall reaction

By comparing the electrode potentials, we can determine which reactions are more likely to happen: Cathode: 1. Copper reduction is more likely as it has a higher electrode potential (positive and higher in value compared to water reduction). Anode: 1. Water oxidation, since sulfate oxidation does not occur easily and is not favored in electrolysis.

Choose the correct option

Based on the electrode potentials and our analysis of the likely reactions in the electrolysis of the aqueous \(\mathrm{CuSO}_{4}\), we can determine that the formed products are Cu at the cathode and \(\mathrm{O}_{2}\) at the anode. Therefore, the correct option is: (b) Cu at cathode, \(\mathrm{O}_{2}\) at anode

Key concepts

Electrode Reactions

During electrolysis, different reactions happen at the two electrodes: the cathode and the anode. At the cathode, we usually see reduction reactions. Here, ions gain electrons to form elements in their reduced state. In our exercise, the copper ions (\(\mathrm{Cu^{2+}}\)) receive electrons to become neutral copper atoms (\(\mathrm{Cu}\)). This process is represented by the equation:

• \(\mathrm{Cu^{2+}} + 2\mathrm{e^{-}} \longrightarrow \mathrm{Cu}\)

At the anode, oxidation reactions occur. Molecules or atoms lose electrons here. For our aqueous\(\mathrm{CuSO_4}\), water molecules are oxidized to form oxygen gas, hydrogen ions, and electrons:

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

Understanding these reactions helps predict the substances produced during electrolysis, like copper at the cathode and oxygen at the anode.

Electrochemical Series

The electrochemical series is a chart of standard electrode potentials, arranged from the most negative to the most positive. This series helps us predict which substances are more likely to gain or lose electrons during electrolysis. A substance with a more positive electrode potential will likely undergo reduction, while one with a more negative potential is prone to oxidation. For instance, the reduction potential for copper is +0.337 V, making it favorable for copper ions to be reduced over other possible reactions at the cathode, such as water reduction, which has a lower potential of -0.83 V.

At the anode, our exercise shows the oxidation of water to oxygen is favored over sulfate oxidation because the potential for water oxidation (+1.23 V) is more viable than the challenging oxidation of stable sulfate ions. Thus, the electrochemical series is crucial in determining the likely reactions during electrolysis.

Aqueous Solutions

When dealing with electrolysis in aqueous solutions, water itself can participate in electrode reactions. This is important because the presence of water may introduce competition with solute ions, such as\(\mathrm{Cu^{2+}}\)and\(\mathrm{SO_4^{2-}}\)in our solution.

In such cases, the reactions are not as straightforward as in molten salts. Water can be reduced at the cathode or oxidized at the anode, depending on the electrode potentials of all competing species. This adds complexity, as seen with the potential formation of hydrogen gas at the cathode or oxygen gas at the anode, due to the involvement of water. Understanding the role of water in aqueous solutions is critical for predicting the products of electrolysis aside from the solute contributions.

Reduction and Oxidation Reactions

Reduction and oxidation reactions, often abbreviated as redox reactions, involve the transfer of electrons between chemical species. Each part of the redox process occurs at one of the electrodes: reduction at the cathode and oxidation at the anode.

In our exercise, at the cathode, reduction allows copper ions to gain electrons and form solid copper. The equation for copper reduction is central to this understanding:

• \(\mathrm{Cu^{2+}} + 2\mathrm{e^{-}} \longrightarrow \mathrm{Cu}\)

At the anode, oxidation involves water losing electrons to form oxygen gas. This reaction not only releases\(\mathrm{O_2}\)but also influences the pH of the solution by forming\(\mathrm{H^+}\)ions:

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

Understanding these redox reactions is essential for predicting and explaining product formation during electrolysis.