The products of electrolysis of `CuSO_4(aq)` between two Pt electrodes

Options

A. Cu at cathode and `O_2` at anode
B. Cu at cathode and `SO_2` at anode
C. `H_2` at cathode and `O_2` at anodo
D. `H_2` at cathode end `O_2` at anode

Explanation

To determine the products of electrolysis of an aqueous solution of copper sulfate (CuSO₄) between two platinum electrodes, we can follow these steps:

### Step 1: Identify the Electrolyte and Electrodes
- The electrolyte is copper sulfate (CuSO₄) in aqueous solution.
- The electrodes are platinum, which are inert electrodes.

### Step 2: Determine the Ions Present
- In the aqueous solution of CuSO₄, the dissociated ions are:
- Cations: Cu²⁺ (copper ions) and H⁺ (hydrogen ions from water)
- Anions: SO₄²⁻ (sulfate ions) and OH⁻ (hydroxide ions from water)

### Step 3: Identify the Reactions at the Electrodes
- At the **anode** (positive electrode), oxidation occurs.
- At the **cathode** (negative electrode), reduction occurs.

### Step 4: Analyze the Reactions at the Cathode
- The cations competing for reduction are Cu²⁺ and H⁺.
- The standard reduction potential for Cu²⁺ is +0.34 V, while for H⁺ (from water) it is 0 V.
- Since Cu²⁺ has a higher reduction potential, it will be reduced:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
- This means copper metal will be deposited at the cathode.

### Step 5: Analyze the Reactions at the Anode
- The anions competing for oxidation are SO₄²⁻ and OH⁻.
- The oxidation potential for OH⁻ is -1.22 V, while for SO₄²⁻ it is -2.05 V.
- Since OH⁻ has a higher oxidation potential, it will be oxidized:
\[ 4\text{OH}^- \rightarrow \text{O}_2 (g) + 2\text{H}_2\text{O} + 4e^- \]
- This means oxygen gas will be released at the anode.

### Step 6: Summarize the Products
- At the cathode, copper (Cu) is deposited.
- At the anode, oxygen gas (O₂) is released.

### Conclusion
The products of electrolysis of CuSO₄(aq) between two platinum electrodes are:
- Copper metal (Cu) at the cathode.
- Oxygen gas (O₂) at the anode.