`100mL` of buffer of `1 M NH_(3)(aq)` and `1 M NH_(4)^(o+)(aq)` are placed in two compartments of a voltaic cell separately. A current of `1.5A` is passed through both cells for `20 m i n`. If only electrolysis of water takes place, then
(a)`ph` of `LHE` half cell will increase
(b)`pH` of `RHE` half cell will increase
(c)`pH` of both half cell will increase
(d)`pH` of both half cell will decrease

Options

A. `ph` of `LHE` half celll will increase
B. `pH` of `RHE` half cell will increase
C. `pH` of both half cell will increase
D. `pH` of both half cell will decrese

Explanation

To solve the question step by step, let's analyze the situation involving the electrolysis of water in a voltaic cell with the given buffer solutions.

### Step 1: Understand the Electrolysis Process
When a current is passed through water, electrolysis occurs, leading to the following half-reactions:
- At the anode (oxidation):
\[ 2 H_2O \rightarrow O_2 + 4 H^+ + 4 e^- \]
- At the cathode (reduction):
\[ 2 H_2O + 2 e^- \rightarrow H_2 + 2 OH^- \]

### Step 2: Analyze the pH Changes
- **At the Anode**: The reaction produces \( H^+ \) ions. The increase in \( H^+ \) concentration will lead to a decrease in pH because pH is inversely related to the concentration of \( H^+ \) ions. Therefore, the pH of the left-hand electrode (LHE) will **decrease**.

- **At the Cathode**: The reaction produces \( OH^- \) ions. The increase in \( OH^- \) concentration will lead to an increase in pH because pH is directly related to the concentration of \( OH^- \) ions. Therefore, the pH of the right-hand electrode (RHE) will **increase**.

### Step 3: Conclusion
Based on the analysis:
- The pH of the left-hand half-cell (anode) will **decrease**.
- The pH of the right-hand half-cell (cathode) will **increase**.

Thus, the correct answer to the question is:
**(b) pH of RHE half cell will increase.**