Given
\(\mathrm{Fe^{3+}(aq)} + e^- \rightarrow \mathrm{Fe^{2+}(aq)};\ E^0 = +0.77\ \mathrm{V}\)
\(\mathrm{Al^{3+}(aq)} + 3e^- \rightarrow \mathrm{Al(s)};\ E^0 = -1.66\ \mathrm{V}\)
\(\mathrm{Br_2(aq)} + 2e^- \rightarrow 2\mathrm{Br^-};\ E^0 = +1.09\ \mathrm{V}\)
Considering the electrode potentials, which of the following represents the correct order of reducing power?
\(\mathrm{Fe^{3+}(aq)} + e^- \rightarrow \mathrm{Fe^{2+}(aq)};\ E^0 = +0.77\ \mathrm{V}\)
\(\mathrm{Al^{3+}(aq)} + 3e^- \rightarrow \mathrm{Al(s)};\ E^0 = -1.66\ \mathrm{V}\)
\(\mathrm{Br_2(aq)} + 2e^- \rightarrow 2\mathrm{Br^-};\ E^0 = +1.09\ \mathrm{V}\)
Considering the electrode potentials, which of the following represents the correct order of reducing power?
Show Hint
- \(\mathrm{Fe^{2+}} \lt \mathrm{Al} \lt \mathrm{Br^-}\)
- \(\mathrm{Br^-} \lt \mathrm{Fe^{2+}} \lt \mathrm{Al}\)
- \(\mathrm{Al} \lt \mathrm{Br^-} \lt \mathrm{Fe^{2+}}\)
- \(\mathrm{Al} \lt \mathrm{Fe^{2+}} \lt \mathrm{Br^-}\)
The Correct Option is B
Solution and Explanation
\(E^0\): \(\mathrm{Br_2/Br^-} = +1.09\) V, \(\mathrm{Fe^{3+}/Fe^{2+}} = +0.77\) V, \(\mathrm{Al^{3+}/Al} = -1.66\) V.
Order of \(E^0\): \(\mathrm{Al^{3+}/Al} \lt \mathrm{Fe^{3+}/Fe^{2+}} \lt \mathrm{Br_2/Br^-}\), so the reducing power order is \(\mathrm{Br^-} \lt \mathrm{Fe^{2+}} \lt \mathrm{Al}\).
Answer: (2) \(\mathrm{Br^-} \lt \mathrm{Fe^{2+}} \lt \mathrm{Al}\)
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