Q83 / 200 · NEET 2008 (Screening) · Chemistry
On the basis of the following \(E^{\circ}\) values, the strongest oxidizing agent is :-
\(\begin{aligned}
& {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}+e^{-1}, \mathrm{E}^{\circ}=-0.35 \mathrm{~V}} \\
& \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+e^{-1} ; \quad \mathrm{E}^{\circ}=-0.77 \mathrm{~V}
\end{aligned}\)
- A$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
- B$\mathrm{Fe}^{2+}$
- Correct answerC$\mathrm{Fe}^{3+}$
- D$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
Solution
Key Idea :
Oxidised form + ne $e^{-} \rightarrow$ Reduced form
The substances which have lower reduction substances which have higher reduction potential are stronger oxidising agent.
$\begin{array}{l}
{\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}+e^{-}} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} ; E^{\circ}=0.35 \mathrm{~V} \\
\mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} ; E^{\circ}=0.77 \mathrm{~V}
\end{array}$
The reduction potential of $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is higher, hence $\mathrm{Fe}^{3+}$ is strongest oxidising agent.
Hence, the correct answer is option: 'C'.