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 \(\left.E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=0.77\, \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \,\mathrm{~V}\right)\)
- A \(20\)
- B \(21\)
- C \(22\)
- D \(24\)
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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