Question

\( E_{\text{cell}} \) of the following cell is
\( \text{Pt(s)} \mid \text{H}_2\text{(g), 1 bar} \mid \text{H}^+ \text{ (1 M)} \mid\mid \text{H}^+ \text{ (0.1 M)} \mid \text{H}_2\text{(g), 1 bar} \mid \text{Pt(s)} \)

• A. $\frac{-2.303 RT}{F}$
• B. $\frac{2.303 RT}{F}$
• C. $\frac{-2.303 RT}{2F}$
• D. $\frac{2.303 RT}{2F}$
• E. $\frac{RT}{2F}$

Hint:

In a concentration cell, the cell potential is positive only if the concentration at the cathode is higher than at the anode.

Answer 1

The Correct Option is A

Concept: This is a concentration cell. The standard cell potential ($E^\circ_{\text{cell}}$) is zero because the same electrode is used. The cell potential ($E_{\text{cell}}$) is found using the Nernst equation: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log Q$.

Step 1: {Determine the cell reaction and $n$.} The reactions are: Anode: $\text{H}_2(g) \rightarrow 2\text{H}^+(1 \text{ M}) + 2e^-$ Cathode: $2\text{H}^+(0.1 \text{ M}) + 2e^- \rightarrow \text{H}_2(g)$ Overall: $2\text{H}^+(0.1 \text{ M}) \rightarrow 2\text{H}^+(1 \text{ M})$. Here, $n = 2$ (or $n=1$ for $\text{H}^+$ transfer).

Step 2: {Set up the Nernst equation.} $$E_{\text{cell}} = 0 - \frac{2.303 RT}{F} \log \left( \frac{[\text{H}^+]_{\text{anode}}}{[\text{H}^+]_{\text{cathode}}} \right)$$ $$E_{\text{cell}} = - \frac{2.303 RT}{F} \log \left( \frac{1}{0.1} \right)$$

Step 3: {Solve for $E_{\text{cell}}$.} $$\log \left( \frac{1}{0.1} \right) = \log(10) = 1$$ $$E_{\text{cell}} = - \frac{2.303 RT}{F} \times 1$$ $$E_{\text{cell}} = \frac{-2.303 RT}{F}$$