Chemistry - Electrochemistry Question with Solution | TestHub

The given cell is a concentration cell. The standard EMF ($E^{\circ }$) for a concentration cell is zero.

The Nernst equation for the cell is:

$$E_{cell}=E^{\circ }-\frac{RT}{nF}\ln Q$$

For a concentration cell, $E^{\circ }=0$.

The cell notation is Pt|H₂(P₁)|H⁺(aq)||H₂(P₂)|Pt.

The half-reactions are:

Anode (oxidation): H₂(P₁) $\to$ 2H⁺(aq) + 2e⁻

Cathode (reduction): 2H⁺(aq) + 2e⁻ $\to$ H₂(P₂)

The overall reaction is H₂(P₁) $\to$ H₂(P₂).

The reaction quotient $Q$ is $\frac{P_2}{P_1}$.

The number of electrons transferred, $n$, is 2.

So, the EMF of the cell is:

$$E_{cell}=0-\frac{RT}{2F}\ln \frac{P_2}{P_1}$$

$$E_{cell}=\frac{RT}{2F}\ln \frac{P_1}{P_2}$$

$$E_{cell}=\frac{RT}{2F}\times 2.303\log \frac{P_1}{P_2}$$

Comparing this with the given options, none of them perfectly match the derived expression. Option B has $\frac{RT}{2f}\log \frac{P_1}{P_2}$, where $f$ is likely intended to be $F$ (Faraday constant). If $f$ is indeed $F$, then option B would be correct. However, given the exact form, "none of these" is the most appropriate answer.

Final Answer: The final answer is $\boxed{D}$