Cell voltages and half-cell potentials are often compeared under standard conditions called the standard state at a specified temperature (298K). The standard state requires all reagents to be pure with all gases at unit fugacity, all ions at unit activity and all electrical connections between half-cells to be made with platinum. Suppose following are half cell reaction, with respective emf (s) are -0.25 V and -0.3V.
Cell reaction I `underset("1 atm")(Pt(H_(2))|HCl)`
Cell reaction II `underset("1 atm")(Pt(H_(2))|H_(2)SO_(4))`
pH value of cell reaction I is

Options

A. 4.23
B. 5.23
C. 5.08
D. 5.32

Answer

To find the pH value of cell reaction I, we will follow these steps:

### Step 1: Understand the Standard Hydrogen Electrode (SHE)
The standard hydrogen electrode (SHE) has a standard reduction potential of 0 V. In our case, the half-cell reaction is given as:
\[ \text{Pt}(H_2) | \text{HCl} \]
The potential for this half-cell reaction is given as -0.25 V.

### Step 2: Use the Nernst Equation
The Nernst equation relates the cell potential (E) to the concentration of hydrogen ions (H⁺) in the solution:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{1}{[H^+]^2} \right) \]
Where:
- \( E \) is the cell potential (-0.25 V in this case)
- \( E^0 \) is the standard electrode potential (0 V for SHE)
- \( n \) is the number of electrons transferred (2 for the reduction of H⁺ to H₂)

### Step 3: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ -0.25 = 0 - \frac{0.059}{2} \log \left( \frac{1}{[H^+]^2} \right) \]

### Step 4: Simplify the Equation
This simplifies to:
\[ -0.25 = -0.0295 \log \left( \frac{1}{[H^+]^2} \right) \]
\[ -0.25 = 0.0295 \log \left( [H^+]^2 \right) \]
Using the property of logarithms, we can rewrite:
\[ -0.25 = 0.059 \log \left( [H^+] \right) \]

### Step 5: Solve for [H⁺]
Now, we can rearrange the equation to solve for the concentration of hydrogen ions:
\[ \log \left( [H^+] \right) = \frac{-0.25}{0.059} \]
Calculating this gives:
\[ \log \left( [H^+] \right) \approx -4.24 \]

### Step 6: Calculate pH
Since pH is defined as:
\[ \text{pH} = -\log \left( [H^+] \right) \]
We can substitute our value:
\[ \text{pH} = -(-4.24) = 4.24 \]

### Final Answer
Thus, the pH value of cell reaction I is approximately:
\[ \text{pH} \approx 4.24 \]