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 step by step, we will use the Nernst equation and the information provided in the question.

### Step 1: Understand the Cell Reaction
The cell is represented as:
\[ \text{Pt(s)} | \text{H}_2(1.0 \text{ atm}) | \text{H}^+(aq) || \text{Ag}^+(1.0 \text{ M}) | \text{Ag(s)} \]

At the anode, hydrogen gas is oxidized:
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]

At the cathode, silver ions are reduced:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

The overall balanced reaction is:
\[ \text{H}_2 + 2\text{Ag}^+ \rightarrow 2\text{H}^+ + 2\text{Ag} \]

### Step 2: Write the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

Where:
- \( E_{\text{cell}} \) is the measured cell potential (1.02 V)
- \( E_{\text{cell}}^0 \) is the standard cell potential (0.80 V)
- \( n \) is the number of moles of electrons transferred (2 in this case)

### Step 3: Substitute Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ 1.02 = 0.80 - \frac{0.0591}{2} \log \left( \frac{[\text{H}^+]^2}{[\text{Ag}^+]^2} \right) \]

Since the concentration of \(\text{Ag}^+\) is given as 1.0 M, we can simplify:
\[ 1.02 = 0.80 - \frac{0.0591}{2} \log \left( \frac{[\text{H}^+]^2}{(1.0)^2} \right) \]

### Step 4: Simplify the Equation
This simplifies to:
\[ 1.02 = 0.80 - 0.02955 \log \left( [\text{H}^+]^2 \right) \]

Using properties of logarithms:
\[ \log \left( [\text{H}^+]^2 \right) = 2 \log \left( [\text{H}^+] \right) \]

Thus, we can rewrite the equation as:
\[ 1.02 = 0.80 - 0.0591 \log \left( [\text{H}^+] \right) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 1.02 - 0.80 = -0.0591 \log \left( [\text{H}^+] \right) \]
\[ 0.22 = -0.0591 \log \left( [\text{H}^+] \right) \]

### Step 6: Solve for \([\text{H}^+]\)
Dividing both sides by -0.0591:
\[ \log \left( [\text{H}^+] \right) = \frac{0.22}{-0.0591} \]
\[ \log \left( [\text{H}^+] \right) \approx -3.73 \]

### Step 7: Calculate the pH
Since \( \text{pH} = -\log \left( [\text{H}^+] \right) \):
\[ \text{pH} \approx 3.73 \]

Thus, the pH of the solution is approximately **3.73**.