Problem 79
Question
The products formed in the electrolysis of aqueous \(\mathrm{CuSO}_{4}\) are \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) Write equations for the anode and cathode reactions.
Step-by-Step Solution
Verified Answer
Cathode: \( \mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}(s) \); Anode: \( 2\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4\mathrm{H}^+ + 4\mathrm{e}^- \).
1Step 1: Identify the Electrolyte Solutions and Electrolyte Components
Electrolysis involves breaking down the components of an electrolyte solution. Here, we have an aqueous solution of copper sulfate, \( \mathrm{CuSO}_4 \). In the solution, this dissociates into \( \mathrm{Cu}^{2+} \) ions and \( \mathrm{SO}_4^{2-} \) ions. Additionally, the water (\( \mathrm{H}_2\mathrm{O} \)) in the solution can also dissociate into \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions.
2Step 2: Determine the Cathode Reaction
At the cathode, reduction occurs. The \( \mathrm{Cu}^{2+} \) ions in the solution are reduced to solid copper. This can be written as: \[ \mathrm{Cu}^{2+}(aq) + 2\mathrm{e}^- \rightarrow \mathrm{Cu}(s) \] Here, copper ions gain two electrons to become copper metal.
3Step 3: Determine the Anode Reaction
At the anode, oxidation occurs. In this scenario, water is more readily oxidized than sulfate ions. The oxidation of water at the anode forms oxygen gas, and can be described by the equation: \[ 2\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \] Thus, water molecules lose electrons and form oxygen gas.
Key Concepts
Anode ReactionCathode ReactionElectrolyte Solution
Anode Reaction
The anode is where oxidation takes place during electrolysis. In the case of the electrolysis of aqueous copper sulfate \( \mathrm{CuSO}_4 \), it is crucial to identify which species undergoes oxidation. In our scenario:
- The potential candidates for oxidation are the sulfate ion \( \mathrm{SO}_4^{2-} \) and water \( \mathrm{H}_2\mathrm{O} \).
- However, water is more easily oxidized than the sulfate ions.
- Each water molecule loses electrons.
- Oxygen gas \( \mathrm{O}_2 \) is generated.
- Four hydrogen ions \( \mathrm{H}^+ \) are produced.
Cathode Reaction
At the cathode, the process of reduction occurs. This means that ions gain electrons, resulting in the formation of a solid element. In the electrolysis of \( \mathrm{CuSO}_4 \) solution, the reduction process taking place involves the \( \mathrm{Cu}^{2+} \) ions:
- Copper(II) ions, \( \mathrm{Cu}^{2+} \), are present in the solution.
- These ions pick up electrons from the cathode.
- Two electrons are gained by each \( \mathrm{Cu}^{2+} \) ion.
- This electron gain converts the ions into copper metal \( \mathrm{Cu} \).
Electrolyte Solution
An electrolyte solution is a key component in the process of electrolysis. It consists of a solvent and an electrolyte, which is a substance that, when dissolved, results in an electrically conductive solution. For our particular example with aqueous \( \mathrm{CuSO}_4 \):
- The solvent is water \( \mathrm{H}_2\mathrm{O} \).
- The solute producing the electrolyte is copper sulfate \( \mathrm{CuSO}_4 \).
- The resulting solution can conduct electricity due to the presence of free ions.
- \( \mathrm{CuSO}_4 \) dissociates into copper ions \( \mathrm{Cu}^{2+} \) and sulfate ions \( \mathrm{SO}_4^{2-} \).
- The water itself dissociates slightly into \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions.
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