The reduction potential of a half-cell consisting of a Pt electrode immersed in 1.5 M Fe2+ and 0.015 M Fe3+ solution at 25°C is \((E^o_{Fe^{3+}/Fe^{2+}} = 0.770V)\):
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 0.652 V.
To calculate the reduction potential of the half-cell involving the \(Fe^{2+}/Fe^{3+}\) redox couple at given concentrations, we'll use the Nernst equation. Here are the given data and the calculation steps:
Given,
Standard reduction potential \( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.770 \text{ V} \)
Concentration of \(Fe^{2+}\): \( [\text{Fe}^{2+}] = 1.5 \text{ M} \)
Concentration of \(Fe^{3+}\): \( [\text{Fe}^{3+}] = 0.015 \text{ M} \)
Temperature \( T = 25^\circ \text{C} = 298 \text{ K} \)
Number of electrons transferred, \( n = 1 \)
The Nernst equation for the reaction:
\(\text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq)\) is given by:
\( E = E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} - \frac{0.0592}{n} \log \frac{[\text{Fe}^{2+}]}{[\text{Fe}^{3+}]}\)
\(⇒ E = 0.770 - \frac{0.0591}{1} \log\frac{1.5}{0.015}\)
\(⇒ E =0.770 - 0.0591 \times \log(100)\)
\(⇒ E = 0.770 - 0.0591 \times 2\)
\(⇒ E = 0.770 - 0.1182\)
\(⇒ E = 0.6518\)
\(⇒ E \approx 0.652\, \ V\)
Thus, the reduction potential of the half-cell consisting of a Pt electrode immersed in 1.5 M \(Fe^{2+}\) and 0.015 M \(Fe^{3+}\) solution at 25°C is 0.652 V