What will be the emf for the given cell ?
`Pt|H_(2)(g,P_(1))|H^(+)(aq)|H_(2)(g,P_(2))|Pt`

Options

A. `(RT)/(F)ln(P_(1))/(P_(2))`
B. `(RT)/(2F)ln(P_(1))/(P_(2))`
C. `(RT)/(F)ln(P_(2))/(P_(1))`
D. None of these

Answer

To find the EMF for the given cell \( \text{Pt} | \text{H}_2(g, P_1) | \text{H}^+(aq) | \text{H}_2(g, P_2) | \text{Pt} \), we will follow these steps:

### Step 1: Identify the Anode and Cathode
In the given cell notation, the left side represents the anode and the right side represents the cathode. Therefore:
- Anode: \( \text{H}_2(g, P_1) \)
- Cathode: \( \text{H}_2(g, P_2) \)

### Step 2: Write the Half-Reactions
At the anode, oxidation occurs:
\[ \text{H}_2(g) \rightarrow 2 \text{H}^+ + 2e^- \]
At the cathode, reduction occurs:
\[ 2 \text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \]

### Step 3: Use the Nernst Equation
The Nernst equation for the cell is given by:
\[ E = E^0 - \frac{RT}{nF} \ln \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
In this case, since we are dealing with gases, the concentrations can be replaced by partial pressures.

### Step 4: Determine Standard EMF (\(E^0\))
For the hydrogen electrode, the standard electrode potential \(E^0\) is 0 V. Therefore:
\[ E^0 = 0 \text{ V} \]

### Step 5: Identify \(n\), \(R\), and \(F\)
- \(n = 2\) (number of electrons transferred)
- \(R = 8.314 \, \text{J/(mol K)}\) (universal gas constant)
- \(F = 96500 \, \text{C/mol}\) (Faraday's constant)

### Step 6: Substitute into the Nernst Equation
Substituting the values into the Nernst equation:
\[ E = 0 - \frac{RT}{2F} \ln \left( \frac{P_2}{P_1} \right) \]
This simplifies to:
\[ E = -\frac{RT}{2F} \ln \left( \frac{P_2}{P_1} \right) \]

### Step 7: Rearranging the Equation
We can rearrange the equation to:
\[ E = \frac{RT}{2F} \ln \left( \frac{P_1}{P_2} \right) \]

### Final Answer
Thus, the EMF for the given cell is:
\[ E = \frac{RT}{2F} \ln \left( \frac{P_1}{P_2} \right) \]