When a solution of `AgNO_3` (0.5 M) is electrolyzed using platinum anode and cathode. Then the products obtained at two elctrodes are

Options

A. Cu at anode ad Ag at cathode
B. `o_2` at anode and Cu at cathode
C. `o_2` at anode and Ag at cathode
D. `NO_2` at anode and Ag at cathode

Answer

To determine the products obtained at the anode and cathode during the electrolysis of a 0.5 M solution of AgNO₃ using platinum electrodes, we can follow these steps:

### Step 1: Identify the ions present in the solution
When AgNO₃ is dissolved in water, it dissociates into its constituent ions:
- Ag⁺ (silver ions)
- NO₃⁻ (nitrate ions)
- H⁺ (from water)
- OH⁻ (from water)

### Step 2: Determine the reactions at the electrodes
During electrolysis, oxidation occurs at the anode and reduction occurs at the cathode.

**At the Anode:**
- The oxidation of water can occur, leading to the production of oxygen gas (O₂) and hydrogen ions (H⁺):
\[ 2 H₂O \rightarrow O₂ + 4 H⁺ + 4 e⁻ \]
- Alternatively, the oxidation of the nitrate ion (NO₃⁻) could occur, but it is less favorable compared to the oxidation of water.

**At the Cathode:**
- The reduction of silver ions (Ag⁺) occurs, leading to the deposition of silver metal:
\[ Ag⁺ + e⁻ \rightarrow Ag \]

### Step 3: Summarize the products
- At the **anode**, oxygen gas (O₂) is produced.
- At the **cathode**, silver metal (Ag) is deposited.

### Final Answer:
The products obtained at the electrodes are:
- **Anode:** Oxygen gas (O₂)
- **Cathode:** Silver metal (Ag)

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