The emf of a cell corresponding to the reaction
`Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1` atm is `0.28` volt at `25^(@)C`. Calculate the `pH` of the solution at the hydrogen electrode.
`E_(Zn^(2+)//Zn)^(@) =- 0.76` volt and `E_(H^(+)//H_(2))^(@) = 0`

Options

A. 7.05
B. 8.62
C. 8.75
D. 9.57

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the standard reduction potentials
We have the following standard reduction potentials:
- \( E^\circ (Zn^{2+}/Zn) = -0.76 \, \text{V} \)
- \( E^\circ (H^+/H_2) = 0 \, \text{V} \)

### Step 2: Calculate the standard cell potential \( E^\circ_{cell} \)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
In our case, the cathode is the hydrogen electrode and the anode is zinc:
\[ E^\circ_{cell} = E^\circ (H^+/H_2) - E^\circ (Zn^{2+}/Zn) = 0 - (-0.76) = 0.76 \, \text{V} \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[Zn^{2+}]}{[H^+]^2} \right) \]
where \( n \) is the number of moles of electrons transferred (which is 2 in this case).

### Step 4: Substitute the known values into the Nernst equation
We know:
- \( E_{cell} = 0.28 \, \text{V} \)
- \( E^\circ_{cell} = 0.76 \, \text{V} \)
- \( [Zn^{2+}] = 0.1 \, \text{M} \)

Substituting these values into the Nernst equation:
\[ 0.28 = 0.76 - \frac{0.0591}{2} \log \left( \frac{0.1}{[H^+]^2} \right) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0.28 - 0.76 = - \frac{0.0591}{2} \log \left( \frac{0.1}{[H^+]^2} \right) \]
\[ -0.48 = - \frac{0.0591}{2} \log \left( \frac{0.1}{[H^+]^2} \right) \]
Multiplying both sides by -1:
\[ 0.48 = \frac{0.0591}{2} \log \left( \frac{0.1}{[H^+]^2} \right) \]

### Step 6: Solve for the logarithmic term
Multiply both sides by \( \frac{2}{0.0591} \):
\[ \frac{0.48 \times 2}{0.0591} = \log \left( \frac{0.1}{[H^+]^2} \right) \]
Calculating the left side:
\[ \frac{0.96}{0.0591} \approx 16.24 \]
Thus,
\[ \log \left( \frac{0.1}{[H^+]^2} \right) = 16.24 \]

### Step 7: Exponentiate to find \([H^+]\)
Taking the antilogarithm:
\[ \frac{0.1}{[H^+]^2} = 10^{16.24} \]
This implies:
\[ [H^+]^2 = \frac{0.1}{10^{16.24}} \]
Calculating \( 10^{16.24} \):
\[ [H^+]^2 \approx 0.1 \times 10^{-16.24} \approx 0.1 \times 5.75 \times 10^{-17} \approx 5.75 \times 10^{-18} \]
Taking the square root:
\[ [H^+] \approx \sqrt{5.75 \times 10^{-18}} \approx 2.39 \times 10^{-9} \, \text{M} \]

### Step 8: Calculate the pH
The pH is calculated using the formula:
\[ pH = -\log [H^+] \]
Substituting the value of \([H^+]\):
\[ pH = -\log (2.39 \times 10^{-9}) \approx 8.62 \]

### Final Answer
The pH of the solution at the hydrogen electrode is approximately **8.62**.