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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

Consider the cell at \(25^{\circ} \mathrm{C}\) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}),(1 \mathrm{M}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})\) The fraction of total iron present as \(\mathrm{Fe}^{3+}\) ion at the cell potential of \(1.500\, \mathrm{~V}\) is \(\mathrm{X} \times 10^{-2}\). The value of \(x\) is \(.....\) (Nearest integer). \(\left(\right.\) Given \(\left.E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=0.77\, \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \,\mathrm{~V}\right)\)

  1. A \(20\)
  2. B \(21\)
  3. C \(22\)
  4. D \(24\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(24\)

Step-by-step Solution

Detailed explanation

\(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}\) \(2 \mathrm{Fe}^{3+} \longrightarrow \mathrm{e}^{-}+\mathrm{e}^{2+}\) ___________________________________________________________…
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