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

Concept: For identical chemical electrodes forming a concentration cell, \(E^\circ_{\text{cell}} = 0\text{ V}\). The overall potential is generated by concentration gradients according to the Nernst Equation: \[ E_{\text{cell}} = -\frac{0.0591}{n} \log\left(\frac{[\text{H}^+]_{\text{anode}}}{[\text{H}^+]_{\text{cathode}}}\right) \]

Step 1: Convert pH indices to active ion molarity.
Using the standard relationship \(\text{pH} = -\log[\text{H}^+]\): \[ [\text{H}^+]_{\text{anode}} = 10^{-3}\text{ M}, \quad [\text{H}^+]_{\text{cathode}} = 10^{-1}\text{ M} \]

Step 2: Calculate potential with single-electron transfer (\(n=1\)).
\[ E_{\text{cell}} = -\frac{0.0591}{1} \log\left(\frac{10^{-3}}{10^{-1}}\right) = -0.0591 \log(10^{-2}) \] \[ E_{\text{cell}} = -0.0591 \times (-2) = 0.1182\text{ V} \]