The measured voltage of the cell, `Pt(s)|H_2(1.0"atm")|H^+(aq)||Ag^+(1.0M)|Ag(s)` is `1.02 V " at " 25^@C ` . Given `E_("cell")^@` is 0.80 V, calculate the pH of the solution .

Options

A. 1.86
B. 1.69
C. 3.73
D. 7.43

Answer

To solve the problem, we will use the Nernst equation to find the pH of the solution based on the given cell voltage and standard cell potential.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Measured cell voltage, \( E_{\text{cell}} = 1.02 \, \text{V} \)
- Standard cell potential, \( E^{\circ}_{\text{cell}} = 0.80 \, \text{V} \)
- Concentration of \( \text{Ag}^+ = 1.0 \, \text{M} \)
- The half-reaction at the anode: \( \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \)
- The half-reaction at the cathode: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \)

2. **Determine the Number of Electrons Transferred (n):**
- From the half-reactions, we see that 2 electrons are involved in the oxidation of hydrogen and 1 electron in the reduction of silver. Therefore, for the overall reaction, we can consider \( n = 2 \) (as we need to balance the electrons).

3. **Apply the Nernst Equation:**
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]} \right) \]
Since \( [\text{Ag}^+] = 1.0 \, \text{M} \), we can simplify the equation.

4. **Rearranging the Nernst Equation:**
\[ E_{\text{cell}} - E^{\circ}_{\text{cell}} = -\frac{0.0591}{n} \log \left( [\text{H}^+]^2 \right) \]
\[ E_{\text{cell}} - E^{\circ}_{\text{cell}} = -\frac{0.0591}{2} \log \left( [\text{H}^+]^2 \right) \]

5. **Substituting the Known Values:**
\[ 1.02 - 0.80 = -\frac{0.0591}{2} \log \left( [\text{H}^+]^2 \right) \]
\[ 0.22 = -\frac{0.0591}{2} \log \left( [\text{H}^+]^2 \right) \]

6. **Solving for \(\log \left( [\text{H}^+]^2 \right)\):**
Multiply both sides by -2:
\[ -0.44 = 0.0591 \log \left( [\text{H}^+]^2 \right) \]
\[ \log \left( [\text{H}^+]^2 \right) = \frac{-0.44}{0.0591} \]
\[ \log \left( [\text{H}^+]^2 \right) \approx -7.45 \]

7. **Finding \([H^+]\):**
\[ [\text{H}^+]^2 = 10^{-7.45} \]
\[ [\text{H}^+] \approx 10^{-3.725} \]

8. **Calculating pH:**
\[ \text{pH} = -\log [\text{H}^+] \approx 3.73 \]

### Final Answer:
The pH of the solution is approximately **3.73**.