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 CuSO₄ (aqueous) between two platinum electrodes, we can follow these steps:

### Step 1: Understand the Electrolysis Process
Electrolysis involves the dissociation of the electrolyte (in this case, CuSO₄) into its constituent ions when an electric current is passed through the solution.

### Step 2: Dissociation of CuSO₄
When CuSO₄ is dissolved in water, it dissociates into:
- Cu²⁺ ions (copper ions)
- SO₄²⁻ ions (sulfate ions)

Additionally, water also dissociates into:
- H⁺ ions (hydrogen ions)
- OH⁻ ions (hydroxide ions)

### Step 3: Identify the Electrodes
In this scenario, we have:
- Cathode: where reduction occurs (gain of electrons)
- Anode: where oxidation occurs (loss of electrons)

### Step 4: Determine the Reaction at the Cathode
At the cathode, we have two positive ions: Cu²⁺ and H⁺. The ion with the higher reduction potential will be reduced first. The reduction potential of Cu²⁺ is higher than that of H⁺, so Cu²⁺ will be reduced to copper metal (Cu).

The reduction reaction can be represented as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

To balance the reaction:
\[ 2 \text{Cu}^{2+} + 4e^- \rightarrow 2 \text{Cu} \]

### Step 5: Determine the Reaction at the Anode
At the anode, we have two negative ions: SO₄²⁻ and OH⁻. The ion with the higher oxidation potential will be oxidized first. The oxidation potential of OH⁻ is higher than that of SO₄²⁻, so OH⁻ will be oxidized to produce oxygen gas (O₂).

The oxidation reaction can be represented as:
\[ 4 \text{OH}^- \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 + 4e^- \]

### Step 6: Conclusion
From the above reactions, we can conclude:
- At the cathode, copper (Cu) is deposited.
- At the anode, oxygen gas (O₂) is liberated.

Thus, the products of electrolysis of CuSO₄ (aq) between two Pt electrodes are:
**Cu at cathode and O₂ at anode.**