The `pH` of `LHE` in the following cell is `:`
`Pt, H_(2)(1atm)|H^(o+)(x M)||H^(o+)(0.1M)|H_(2)(0.1atm)Pt` E_(cell)=0.295V`.

Options

A. `6.5`
B. `6.0`
C. `5.5`
D. `4.0`

Explanation

To find the pH of the left-hand electrode in the given electrochemical cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell notation is given as:
\[ \text{Pt, } H_2(1 \text{ atm}) | H^+(x \text{ M}) || H^+(0.1 \text{ M}) | H_2(0.1 \text{ atm}) \text{ Pt} \]
The left-hand side (anode) involves oxidation, while the right-hand side (cathode) involves reduction.

2. **Write the Half-Reactions**:
- At the anode (oxidation):
\[ H_2 \rightarrow 2H^+ + 2e^- \]
- At the cathode (reduction):
\[ 2H^+ + 2e^- \rightarrow H_2 \]

3. **Determine the Equilibrium Constant (K)**:
The equilibrium constant \( K \) for the reaction can be expressed in terms of the concentrations and pressures:
\[ K = \frac{P_{H_2}}{[H^+]^2} \]
Here, \( P_{H_2} \) on the left side is 1 atm, and on the right side is 0.1 atm. The concentration of \( H^+ \) on the right side is 0.1 M, and on the left side is \( x \) M.
Therefore:
\[ K = \frac{1}{(0.1)^2} = 100 \]

4. **Use the Nernst Equation**:
The Nernst equation relates the cell potential \( E \) to the standard potential \( E^\circ \) and the reaction quotient \( Q \):
\[ E_{cell} = E^\circ - \frac{0.059}{n} \log K \]
Given that \( E_{cell} = 0.295 \, V \) and \( n = 2 \):
\[ 0.295 = 0 - \frac{0.059}{2} \log K \]

5. **Solve for \( K \)**:
Rearranging gives:
\[ 0.295 = -0.0295 \log K \]
\[ \log K = -\frac{0.295}{0.0295} \approx -10 \]
Therefore:
\[ K = 10^{-10} \]

6. **Relate \( K \) to Concentration**:
From the equilibrium expression:
\[ K = \frac{1}{x^2} \]
Setting this equal to \( 10^{-10} \):
\[ 10^{-10} = \frac{1}{x^2} \]
Thus:
\[ x^2 = 10^{10} \implies x = 10^5 \, M \]

7. **Calculate pH**:
The pH is calculated using:
\[ pH = -\log[H^+] = -\log(x) = -\log(10^{-5}) = 5 \]

### Final Answer:
The pH of the left-hand side (anode) is approximately **5**.