Question

What is the cell potential (\(E_{\text{cell}}\)) for a concentration cell consisting of two hydrogen electrodes at $298\text{ K}$, where the anode compartment is at \(\text{pH} = 3\) and the cathode compartment is at \(\text{pH} = 1\) under standard pressure conditions?

• A. \(0.0591\text{ V} \)
• B. \(0.1182\text{ V} \)
• C. \(-0.1182\text{ V} \)
• D. \(0.0000\text{ V} \)

Hint:

For symmetrical hydrogen ion concentration cells at \(298\text{ K}\), you can use this simple shortcut: \(E_{\text{cell}} = 0.0591 \times (\text{pH}_{\text{anode}} - \text{pH}_{\text{cathode}})\).

Answer 1

The Correct Option is B

Step 1 : Understanding the Question:
The topic of this question is Electrochemistry, specifically Concentration Cells. A concentration cell is an electrolytic cell that consists of two half-cells with identical electrodes but different concentrations of the same electrolyte. Because the electrodes are the same, the standard cell potential ($E^\circ_{cell}$) is zero. The electrical potential is generated entirely by the difference in concentration (or pH) between the two compartments. We need to calculate the cell potential using the Nernst equation for a hydrogen electrode system.
Step 2 : Key Formulas and approach:
The approach involves three main steps:
1. Converting pH to $[H^+]$ concentration using: $[H^+] = 10^{-\text{pH}}$.
2. Identifying the number of electrons transferred ($n = 1$ for $H^+ \rightarrow \frac{1}{2}H_2$).
3. Applying the Nernst equation at $298\text{ K}$:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \frac{[H^+]_{anode}}{[H^+]_{cathode}} \]
Since $E^\circ_{cell} = 0$ for concentration cells, the formula simplifies significantly.
Step 3 : Detailed Explanation:

First, we calculate the concentrations of hydrogen ions in both compartments. For the anode, $\text{pH} = 3$, so $[H^+]_{anode} = 10^{-3}\text{ M}$. For the cathode, $\text{pH} = 1$, so $[H^+]_{cathode} = 10^{-1}\text{ M}$.

The reaction at each electrode is $H^+ + e^- \rightarrow \frac{1}{2}H_2$. Therefore, the number of electrons exchanged ($n$) is 1.

We now substitute these values into the simplified Nernst equation: $E_{cell} = -0.0591 \log \frac{10^{-3}}{10^{-1}}$.

Simplifying the fraction inside the log: $\frac{10^{-3}}{10^{-1}} = 10^{-3 - (-1)} = 10^{-2}$.

The equation becomes: $E_{cell} = -0.0591 \times \log(10^{-2})$.

Using the property of logarithms ($\log a^b = b \log a$), we get: $E_{cell} = -0.0591 \times (-2) \times \log(10)$.

Since $\log(10) = 1$, the final calculation is: $E_{cell} = +0.1182\text{ V}$.

The positive value indicates that the cell reaction is spontaneous in the direction calculated, with the higher concentration of $H^+$ at the cathode pulling electrons from the lower concentration anode.

Step 4 : Final Answer:
The cell potential for the given hydrogen concentration cell is 0.1182 V. Thus, the correct option is (B).