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COMEDK · Chemistry · 18. Electrochemistry

What is the reduction potential of a half-cell consisting of a Pt electrode dipped in \(2.2 \mathrm{M} \mathrm{Fe}^{2+}\) and \(0.04 \mathrm{M} \mathrm{Fe}^{3+}\) solution where the reaction taking place is conversion of \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ?
\(\mathrm{E}^0\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771 \mathrm{~V}\)

  1. A 0.598 V
  2. B 0.668 V
  3. C 0.719 V
  4. D 0.723 V
Verified Solution

Answer & Solution

Correct Answer

(B) 0.668 V

Step-by-step Solution

Detailed explanation

The reduction half-reaction is given by \(\mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+}\). Using the Nernst equation for the electrode potential at \(298 \mathrm{K}\): \(E = E^{0} - \dfrac{0.0591}{n} \log \dfrac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}\) Given…