Understanding Electrolysis of Aqueous CuSO₄ with Platinum Electrodes

Let's analyze the electrolysis of an aqueous solution of copper(II) sulfate ($\text{CuSO}_4$) using platinum electrodes. Platinum electrodes are considered inert, meaning they do not participate chemically in the electrode reactions. In an aqueous solution of $\text{CuSO}_4$, the following species are present:

• Ions from $\text{CuSO}_4$: $\text{Cu}^{2+}$ and $\text{SO}_4^{2-}$
• From water: $\text{H}_2\text{O}$ (which can ionize slightly into $\text{H}^{+}$ and $\text{OH}^{-}$ ions)

Electrolysis involves reduction at the cathode (negative electrode) and oxidation at the anode (positive electrode).

Reactions at the Cathode (Reduction)

At the cathode, species that can be reduced compete. The possible species are $\text{Cu}^{2+}$ and $\text{H}_2\text{O}$ (or $\text{H}^{+}$ from water). We need to consider their standard reduction potentials:

• Reduction of $\text{Cu}^{2+}$ ions: $\text{Cu}^{2+}\text{(aq)} + 2e^{-} \rightarrow \text{Cu}\text{(s)}$; $E^\circ = +0.34 \text{ V}$
• Reduction of water: $2\text{H}_2\text{O}\text{(l)} + 2e^{-} \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^{-}\text{(aq)}$; $E^\circ = -0.83 \text{ V}$ (at pH 7)

A species with a more positive (or less negative) standard reduction potential is preferentially reduced. Comparing $+0.34 \text{ V}$ for $\text{Cu}^{2+}$ and $-0.83 \text{ V}$ for water, $\text{Cu}^{2+}$ has a significantly higher reduction potential. Therefore, $\text{Cu}^{2+}$ ions will be reduced at the cathode, and copper metal will be deposited.

Statement A: Copper will deposit at cathode. This is correct.

Reactions at the Anode (Oxidation)

At the anode, species that can be oxidized compete. The possible species are $\text{SO}_4^{2-}$ and $\text{H}_2\text{O}$ (or $\text{OH}^{-}$ from water). We need to consider their oxidation potentials (or compare their reduction potentials in reverse).

• Oxidation of sulfate ions ($\text{SO}_4^{2-}$): $2\text{SO}_4^{2-}\text{(aq)} \rightarrow \text{S}_2\text{O}_8^{2-}\text{(aq)} + 2e^{-}$ (This reaction is difficult)
• Oxidation of water: $2\text{H}_2\text{O}\text{(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^{+}\text{(aq)} + 4e^{-}$; $E^\circ = -1.23 \text{ V}$ (at pH 7, oxidation potential is $+1.23 \text{ V}$)

Comparing the ease of oxidation, water is more easily oxidized than sulfate ions. Water will be oxidized at the anode, producing oxygen gas and hydrogen ions.

Statement C: Oxygen will be released at anode. This is correct.

Analyzing Other Statements

Statement B: Copper will deposit at anode. This is incorrect. Copper ions are cations and move towards the negative electrode (cathode), where reduction occurs. The anode is the positive electrode where oxidation occurs.

Statement D: Copper will dissolve at anode. This would happen if the anode were made of copper (an active electrode) and the electrolyte contained ions that allow copper to oxidize. However, the electrode is platinum, which is inert and does not dissolve.

Summary of Events at Electrodes

Based on the analysis, statements A (Copper will deposit at cathode) and C (Oxygen will be released at anode) describe what will happen during the electrolysis of aqueous $\text{CuSO}_4$ using platinum electrodes.

Revision Table: Key Electrolysis Concepts

Additional Information on Electrolysis Reactions

The reactions that occur at the electrodes during electrolysis of an aqueous solution depend on the nature of the ions present, the nature of the electrode material (inert or active), and the concentration of ions.

When using inert electrodes like platinum or graphite, the competition at the cathode is usually between the metal cation (if present) and water (or $\text{H}^{+}$ ions), and at the anode between the anion (if it can be oxidized, like halides) and water (or $\text{OH}^{-}$ ions).

The reduction potential values help predict which species will be reduced at the cathode. Species with higher reduction potentials are reduced first. Similarly, oxidation potentials help predict which species will be oxidized at the anode; species with higher oxidation potentials are oxidized first (which corresponds to lower reduction potentials).

In the case of aqueous $\text{CuSO}_4$ with platinum electrodes:

• At cathode: $\text{Cu}^{2+}$ is reduced instead of water because $\text{Cu}^{2+}/\text{Cu}$ has a higher reduction potential ($+0.34 \text{ V}$) than $\text{H}_2\text{O}/\text{H}_2$ ($-0.83 \text{ V}$).
• At anode: Water is oxidized instead of $\text{SO}_4^{2-}$ because the oxidation of water to $\text{O}_2$ happens at a lower potential (around $+1.23 \text{ V}$) compared to the oxidation of $\text{SO}_4^{2-}$ to $\text{S}_2\text{O}_8^{2-}$ (which requires much higher potential, around $+2.01 \text{ V}$).

This confirms that copper deposition at the cathode and oxygen release at the anode are the expected outcomes.