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 (electromotive force) for the given cell represented as `Pt|H2(g,P1)|H+(aq)|H2(g,P2)|Pt`, we can follow these steps:

### Step 1: Identify the Cell Components
The cell consists of:
- Anode: `H2(g, P1)` where hydrogen gas is oxidized.
- Cathode: `H+(aq)` where hydrogen ions are reduced to hydrogen gas at pressure `P2`.
- Inert electrode: Platinum (Pt).

### Step 2: Write the Half-Reactions
At the anode (oxidation):
\[ \text{H}_2(g, P1) \rightarrow 2\text{H}^+ + 2e^- \]

At the cathode (reduction):
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g, P2) \]

### Step 3: Determine the Number of Electrons Transferred (n)
In both half-reactions, 2 electrons are transferred. Thus, \( n = 2 \).

### Step 4: Use the Nernst Equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln \left( \frac{[Products]}{[Reactants]} \right) \]

For this cell, the standard electrode potential \( E^0_{cell} \) for hydrogen is 0 V (both anode and cathode). Therefore:
\[ E^0_{cell} = 0 \]

### Step 5: Substitute into the Nernst Equation
Since \( E^0_{cell} = 0 \):
\[ E_{cell} = 0 - \frac{RT}{nF} \ln \left( \frac{P2}{P1} \right) \]

### Step 6: Simplify the Equation
Substituting \( n = 2 \):
\[ E_{cell} = -\frac{RT}{2F} \ln \left( \frac{P2}{P1} \right) \]

### Step 7: Change the Logarithm
Using the property of logarithms, we can express it as:
\[ E_{cell} = \frac{RT}{2F} \ln \left( \frac{P1}{P2} \right) \]

### Final Answer
Thus, the emf for the given cell is:
\[ E_{cell} = \frac{RT}{2F} \ln \left( \frac{P1}{P2} \right) \]

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