What will be the reduction potential of a hydrogen electrode which is filled with HCl solution of pH value 1.0? (at 298 Kelvin)

Options

A. `(-59.15V)`
B. `(+59.15)`
C. `(+59.15mV)`
D. `(-59.15mV)`

Explanation

To find the reduction potential of a hydrogen electrode in a solution of HCl with a pH of 1.0 at 298 K, we can follow these steps:

### Step 1: Understand the relationship between pH and hydrogen ion concentration.
The pH of a solution is related to the concentration of hydrogen ions \([H^+]\) by the formula:
\[ \text{pH} = -\log[H^+] \]
Given that the pH is 1.0, we can find \([H^+]\):
\[ 1 = -\log[H^+] \implies [H^+] = 10^{-1} \, \text{M} = 0.1 \, \text{M} \]

### Step 2: Write the Nernst equation for the hydrogen electrode.
The Nernst equation for the half-cell reaction of the hydrogen electrode is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{1}{[H^+]^2} \right) \]
For the hydrogen electrode, \(E^\circ = 0 \, \text{V}\) and \(n = 2\) (since two electrons are involved in the reduction of \(H^+\) to \(H_2\)).

### Step 3: Substitute the values into the Nernst equation.
Substituting the value of \([H^+]\) into the Nernst equation:
\[ E = 0 - \frac{0.059}{2} \log \left( \frac{1}{(0.1)^2} \right) \]
Calculating \((0.1)^2\):
\[ (0.1)^2 = 0.01 \]
Thus, we have:
\[ E = -\frac{0.059}{2} \log \left( \frac{1}{0.01} \right) \]

### Step 4: Calculate the logarithm.
\[ \log \left( \frac{1}{0.01} \right) = \log(100) = 2 \]
Now substituting this back into the equation:
\[ E = -\frac{0.059}{2} \cdot 2 \]
\[ E = -0.059 \, \text{V} \]

### Step 5: Convert to millivolts.
To express the potential in millivolts:
\[ E = -0.059 \, \text{V} = -59 \, \text{mV} \]

### Final Answer:
The reduction potential of the hydrogen electrode in a HCl solution with a pH of 1.0 at 298 K is \(-59 \, \text{mV}\).

---