The cell `Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag` has emf, `E_(298KK) =0`. The electrode potential for the reaction `AgI +e^(-) rarr Ag + I^(Theta)` is `-0.151` volt. Calculate the `pH` value:-

Options

A. `3.37`
B. `5.26`
C. `2.56`
D. `4.62`

Explanation

To solve the problem step by step, we will analyze the given electrochemical cell and apply the Nernst equation to find the pH value.

### Step 1: Identify the half-reactions
The cell is represented as:
\[ \text{Pt}(H_2)(1 \text{ atm}) | H^+ (pH = ?) | AgI(s), Ag \]

- **Anode Reaction**: The oxidation reaction at the anode involves hydrogen gas (H2) converting to protons (H+):
\[ \text{H}_2 \rightarrow 2 \text{H}^+ + 2e^- \]

- **Cathode Reaction**: The reduction reaction at the cathode involves silver iodide (AgI) being reduced to silver (Ag):
\[ \text{AgI} + e^- \rightarrow \text{Ag} + \text{I}^- \]

### Step 2: Determine the standard electrode potentials
- The standard electrode potential for the cathode reaction (AgI + e^- → Ag + I^-) is given as:
\[ E^\circ_{\text{cathode}} = -0.151 \text{ V} \]

- The standard electrode potential for the anode reaction (standard hydrogen electrode) is:
\[ E^\circ_{\text{anode}} = 0 \text{ V} \]

### Step 3: Calculate the cell EMF
The EMF of the cell is given as:
\[ E_{\text{cell}} = E_{\text{cathode}} + E_{\text{anode}} \]

Since the EMF is zero:
\[ 0 = -0.151 + E_{\text{anode}} \]
Thus, we can find \( E_{\text{anode}} \):
\[ E_{\text{anode}} = 0.151 \text{ V} \]

### Step 4: Apply the Nernst equation
The Nernst equation at room temperature (298 K) is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]

For the anode reaction:
- \( n = 2 \) (since 2 electrons are transferred)
- The concentration of H2 is 1 atm (standard condition).

The Nernst equation for the anode becomes:
\[ E_{\text{anode}} = 0 - \frac{0.0591}{2} \log \frac{[H^+]^2}{1} \]

### Step 5: Substitute known values
Substituting the values we have:
\[ 0.151 = -\frac{0.0591}{2} \log [H^+]^2 \]

This simplifies to:
\[ 0.151 = -0.02955 \log [H^+]^2 \]

### Step 6: Solve for \([H^+]\)
Rearranging gives:
\[ \log [H^+]^2 = -\frac{0.151}{0.02955} \]
\[ \log [H^+]^2 = -5.11 \]

Taking the antilog:
\[ [H^+]^2 = 10^{-5.11} \]
\[ [H^+] = 10^{-2.555} \]

### Step 7: Calculate pH
Since pH is defined as:
\[ \text{pH} = -\log [H^+] \]
We find:
\[ \text{pH} = -\log(10^{-2.555}) \]
\[ \text{pH} \approx 2.555 \]

### Final Answer
The calculated pH value is approximately:
\[ \text{pH} \approx 2.5 \]