To find the electromotive force (emf) 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 Anode and Cathode
In the cell representation, the left side represents the anode and the right side represents the cathode.
- **Anode Reaction:** At the anode, hydrogen gas (H2) at pressure P1 is oxidized to hydrogen ions (H+), releasing electrons:
\[
\text{H}_2(g, P_1) \rightarrow 2\text{H}^+(aq) + 2e^-
\]
- **Cathode Reaction:** At the cathode, hydrogen ions (H+) are reduced to hydrogen gas (H2) at pressure P2:
\[
2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g, P_2)
\]
### Step 2: Write the Overall Cell Reaction
Combining the two half-reactions, we can simplify the overall reaction:
\[
\text{H}_2(g, P_1) \rightarrow \text{H}_2(g, P_2)
\]
This can be expressed as:
\[
\frac{1}{2}\text{H}_2(g, P_1) \rightarrow \frac{1}{2}\text{H}_2(g, P_2)
\]
### Step 3: Apply the Nernst Equation
The Nernst equation is given by:
\[
E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q
\]
Where:
- \(E^\circ_{\text{cell}}\) is the standard cell potential (for hydrogen, it is 0 V).
- \(R\) is the universal gas constant (8.314 J/mol·K).
- \(T\) is the temperature in Kelvin.
- \(n\) is the number of moles of electrons transferred (which is 2 in this case).
- \(F\) is Faraday's constant (96485 C/mol).
- \(Q\) is the reaction quotient.
### Step 4: Determine the Reaction Quotient (Q)
For the reaction:
\[
\frac{1}{2}\text{H}_2(g, P_1) \rightarrow \frac{1}{2}\text{H}_2(g, P_2)
\]
The reaction quotient \(Q\) is:
\[
Q = \frac{P_2^{1/2}}{P_1^{1/2}} = \sqrt{\frac{P_2}{P_1}}
\]
### Step 5: Substitute into the Nernst Equation
Substituting \(E^\circ_{\text{cell}} = 0\), \(n = 2\), and \(Q\) into the Nernst equation:
\[
E_{\text{cell}} = 0 - \frac{RT}{2F} \ln \left(\sqrt{\frac{P_2}{P_1}}\right)
\]
This simplifies to:
\[
E_{\text{cell}} = -\frac{RT}{2F} \ln \left(\frac{P_2}{P_1}\right)
\]
### Step 6: Final Expression
To remove the negative sign, we can express it as:
\[
E_{\text{cell}} = \frac{RT}{2F} \ln \left(\frac{P_1}{P_2}\right)
\]
### Conclusion
Thus, the emf for the given cell is:
\[
E_{\text{cell}} = \frac{RT}{2F} \ln \left(\frac{P_1}{P_2}\right)
\]
