A cell contain two hydrogen electrodes. The negatove electrode is in contact with a solution of `pH = 5.5`. The emf of the cell is `0.118 V` at `25^(@)C`. Calculate the `pH` of solution positive electrode. (assume pressure of `H_(2)` in the both electrondes `= 1` bar )

Options

A. `3 . 5`
B. `7 . 5`
C. `4.5`
D. `6.5`

Explanation

To solve the problem of calculating the pH of the solution at the positive electrode, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- pH of the negative electrode solution = 5.5
- EMF of the cell (E) = 0.118 V
- Temperature = 25°C
- Pressure of H₂ gas = 1 bar

2. **Calculate the Concentration of H⁺ Ions at the Negative Electrode**:
- pH = 5.5
- \[ [H^+] = 10^{-pH} = 10^{-5.5} \, \text{mol/L} \]

3. **Write the Half-Reactions**:
- At the anode (negative electrode):
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]
- At the cathode (positive electrode):
\[ 2H^+(aq) + 2e^- \rightarrow H_2(g) \]

4. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
Where:
- \( E^\circ \) for the hydrogen electrode = 0 V
- \( n = 2 \) (number of electrons transferred)
- \( Q \) is the reaction quotient.

5. **Calculate the Reaction Quotient (Q)**:
- For the reaction at the anode:
\[ Q = \frac{[H^+]^2}{[H_2]} \]
Since the pressure of H₂ is 1 bar, we can substitute:
\[ Q = \frac{(10^{-5.5})^2}{1} = 10^{-11} \]

6. **Substitute into the Nernst Equation**:
\[ 0.118 = 0 - \frac{0.0591}{2} \log(10^{-11}) \]
\[ 0.118 = -0.02955 \cdot (-11) \]
\[ 0.118 = 0.32505 \]

7. **Calculate the Concentration of H⁺ at the Cathode**:
Rearranging the Nernst equation to find the unknown concentration \( x \):
\[ 0.118 = -0.02955 \log \left( \frac{10^{-5.5}}{x^2} \right) \]
\[ 0.118 = 0.02955 \log \left( \frac{x^2}{10^{-11}} \right) \]
\[ 0.118 = 0.02955 \left( 2 \log x + 11 \right) \]

8. **Solve for x**:
Rearranging gives:
\[ 0.118 = 0.0591 \log x + 0.32505 \]
\[ 0.0591 \log x = 0.118 - 0.32505 \]
\[ \log x = \frac{-0.20705}{0.0591} \]
\[ x \approx 10^{-3.5} \]

9. **Calculate pH of the Solution at the Positive Electrode**:
\[ pH = -\log [H^+] = -\log(10^{-3.5}) = 3.5 \]

### Final Answer:
The pH of the solution at the positive electrode is **3.5**.

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