During the electrolysis of the aqueous solution of copper sulphate using `Pt` electrode, the reaction taking place at anode electrode is

Options

A. `Cu^(2+)+2e^(-) rarr Cu`
B. `Cu rarr Cu^(2+)+2e^(-)`
C. `2H_(2)O rarr 4H^(o+)+O_(2)+4e^(-)`
D. `H_(2)O+e^(-) rarr overset(c-)(O)H+1//2H_(2)`

Explanation

To determine the reaction taking place at the anode during the electrolysis of an aqueous solution of copper sulfate using a platinum electrode, we can follow these steps:

### Step 1: Understand the Electrolysis Process
Electrolysis involves the breakdown of a compound in solution by passing an electric current through it. In this case, we are dealing with an aqueous solution of copper sulfate (CuSO₄).

### Step 2: Identify the Electrodes
In electrolysis, there are two electrodes:
- **Anode**: The electrode where oxidation occurs.
- **Cathode**: The electrode where reduction occurs.

### Step 3: Determine the Reactions at the Electrodes
- At the **cathode**, copper ions (Cu²⁺) in the solution will gain electrons (reduction) and be deposited as solid copper (Cu).

Reaction at cathode:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]

- At the **anode**, oxidation occurs. In an aqueous solution, water can be oxidized to produce oxygen gas (O₂), protons (H⁺), and electrons (e⁻).

### Step 4: Write the Anode Reaction
The oxidation of water can be represented by the following half-reaction:
\[ 2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4e^- \]

### Step 5: Conclusion
Thus, during the electrolysis of an aqueous solution of copper sulfate using a platinum electrode, the reaction taking place at the anode is the oxidation of water to produce oxygen gas.

### Final Answer
The reaction taking place at the anode is:
**Oxidation of water to oxygen gas.**

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