Question

In an aqueous solution of Fe\(^{2+}\) ions with concentration of 10\(^{-4}\) M at 298 K and atmospheric pressure, the reduction potential of Fe in volt is ................... (round off to 2 decimal places).
Given: Standard reduction potential, \(E^{\circ}_{\text{Fe}^{2+}/\text{Fe}} = -0.44\) V
Faraday's constant, F = 96500 C per mole of electrons
Universal gas constant, R = 8.314 J mol\(^{-1}\)K\(^{-1}\)

Hint:

For electrochemical calculations at room temperature (298 K or 25°C), memorizing the term \(0.0592/n\) in the base-10 log version of the Nernst equation can save significant time compared to calculating \(RT/F\) from scratch.

Answer 1

Step 1: Understanding the Concept:
The reduction potential of an electrode under non-standard conditions (i.e., concentration not equal to 1 M) is calculated using the Nernst equation. The equation relates the measured cell potential to the standard potential and the concentrations of the species involved.
Step 2: Key Formula or Approach:
The half-cell reaction is: \( \text{Fe}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Fe}(\text{s}) \). The Nernst equation for this reaction is: \[ E = E^\circ - \frac{RT}{nF} \ln(Q) \] The reaction quotient \(Q\) is \( \frac{a_{\text{Fe(s)}}}{a_{\text{Fe}^{2+}(\text{aq})}} \). The activity of the solid iron, \(a_{\text{Fe(s)}}\), is 1, and the activity of the ion, \(a_{\text{Fe}^{2+}(\text{aq})}\), is approximated by its molar concentration, \( [\text{Fe}^{2+}] \). So, \( Q = \frac{1}{[\text{Fe}^{2+}]} \). The equation becomes: \[ E = E^\circ - \frac{RT}{nF} \ln\left(\frac{1}{[\text{Fe}^{2+}]}\right) = E^\circ + \frac{RT}{nF} \ln\left([\text{Fe}^{2+}]\right) \] At T = 298 K, \( \frac{2.303RT}{F} \approx 0.0592 \) V, simplifying the equation to: \[ E \approx E^\circ + \frac{0.0592}{n} \log_{10}\left([\text{Fe}^{2+}]\right) \] Step 3: Detailed Calculation:
Given values:
- \(E^\circ = -0.44\) V
- \(n = 2\) (two electrons are transferred)
- \( [\text{Fe}^{2+}] = 10^{-4} \) M
- \(T = 298\) K
Using the simplified Nernst equation at 298 K: \[ E = -0.44 + \frac{0.0592}{2} \log_{10}(10^{-4}) \] \[ E = -0.44 + (0.0296) \times (-4) \] \[ E = -0.44 - 0.1184 \] \[ E = -0.5584 \text{ V} \] Rounding to 2 decimal places: \[ E = -0.56 \text{ V} \] Step 4: Final Answer:
The reduction potential of Fe is -0.56 V.