100 mL of a buffer of 1 M `NH_(3)(aq)` and `1M NH_(4)^(+) (aq)` are placed in two compartments of a voltaic cell separately. A current of 1.5 A is passed through both cell for 20 min. If electrolysis of water only takes place, then pH of the :

Options

A. LHE half – cell will increase
B. RHE half – cell will increase
C. Both half – cell will increase
D. Both half – cells will decrease

Explanation

To solve the problem, we need to analyze the electrolysis of water occurring in the voltaic cell with the given buffer solutions of NH3 and NH4+. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a voltaic cell with two compartments, each containing a buffer solution of 1 M NH3 (ammonia) and 1 M NH4+ (ammonium ion). A current of 1.5 A is passed through the cell for 20 minutes.

### Step 2: Identify the Reactions at the Electrodes
In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. Since the question states that only the electrolysis of water takes place, we can write the half-reactions:

- **At the Anode (Oxidation)**:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
This reaction produces oxygen gas and releases hydrogen ions (H+), which will decrease the pH of the solution.

- **At the Cathode (Reduction)**:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
This reaction produces hydrogen gas and hydroxide ions (OH-), which will increase the pH of the solution.

### Step 3: Analyze the Effect on pH
- **At the Anode**: The production of H+ ions will lead to an increase in acidity, thus decreasing the pH.
- **At the Cathode**: The production of OH- ions will lead to a decrease in acidity (or increase in basicity), thus increasing the pH.

### Step 4: Conclusion
- The pH of the left-hand side (anode) will **decrease**.
- The pH of the right-hand side (cathode) will **increase**.

### Final Answer
The pH of the left-hand side (anode) will decrease, and the pH of the right-hand side (cathode) will increase.