Reduction potential of hydrogen electrode in terms of pH is : (at 1 atm pressure) (T = 298 kelvin )

Options

A. `E=((RT)/(F))xx(PH)`
B. `E= (RT)/(F). (1)/(PH)`
C. `E= (2.303RT)/(F).(PH)`
D. `E=-0.0591PH`

Explanation

To find the reduction potential of the hydrogen electrode in terms of pH at 1 atm pressure and 298 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction at the hydrogen electrode can be represented as:
\[ 2H^+ + 2e^- \leftrightarrow H_2(g) \]

2. **Use the Nernst Equation**:
The Nernst equation for the reduction potential (E) is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where:
- \(E^\circ\) is the standard reduction potential,
- \(R\) is the universal gas constant (8.314 J/(mol·K)),
- \(T\) is the temperature in Kelvin (298 K),
- \(n\) is the number of moles of electrons transferred (2 for the hydrogen electrode),
- \(F\) is Faraday's constant (96500 C/mol),
- \(Q\) is the reaction quotient.

3. **Determine the Standard Potential**:
For the hydrogen electrode, the standard reduction potential \(E^\circ\) is defined as 0 V.

4. **Calculate the Reaction Quotient (Q)**:
For the hydrogen electrode, the reaction quotient \(Q\) is given by:
\[ Q = \frac{P_{H_2}}{[H^+]^2} \]
At 1 atm pressure, \(P_{H_2} = 1\), so:
\[ Q = \frac{1}{[H^+]^2} \]

5. **Substitute into the Nernst Equation**:
Substituting \(Q\) into the Nernst equation gives:
\[ E = 0 - \frac{RT}{nF} \ln \left(\frac{1}{[H^+]^2}\right) \]
This simplifies to:
\[ E = -\frac{RT}{nF} \ln \left(\frac{1}{[H^+]^2}\right) = \frac{RT}{nF} \ln [H^+]^2 \]

6. **Convert Natural Log to Logarithm Base 10**:
Using the conversion \( \ln x = 2.303 \log x \), we can rewrite the equation:
\[ E = \frac{RT}{nF} \cdot 2.303 \log [H^+] \]

7. **Substitute Values**:
Now substituting the values:
- \(R = 8.314 \, \text{J/(mol·K)}\)
- \(T = 298 \, \text{K}\)
- \(n = 2\)
- \(F = 96500 \, \text{C/mol}\)
\[ E = \frac{(8.314)(298)}{(2)(96500)} \cdot 2.303 \log [H^+] \]

8. **Calculate the Coefficient**:
Performing the calculation:
\[ E = -0.059 \log [H^+] \]

9. **Relate to pH**:
Since \(pH = -\log [H^+]\), we can rewrite the equation as:
\[ E = -0.059 \cdot (-pH) = 0.059 \cdot pH \]

### Final Answer:
Thus, the reduction potential of the hydrogen electrode in terms of pH at 1 atm pressure and 298 K is:
\[ E = -0.059 \cdot pH \]