On the basis of the following $\mathrm{E}^{\circ}$ values, the strongest oxidizing agent is…
On the basis of the following $\mathrm{E}^{\circ}$ values, the strongest oxidizing agent is
$\begin{array}{ll}
{\left[\mathrm{Fe}(\mathrm{CN})_\theta\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_\theta\right]^{3-}+\mathrm{e}^{-1} ;} & \mathrm{E}^0=-0.35 \mathrm{~V} \\
\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-1} ; & \mathrm{E}^0=-0.77 \mathrm{~V}
\end{array}$
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
$\mathrm{Fe}^{2+}$
$\mathrm{Fe}^{3+}$
Solution
For the strongest oxidizing agent, the oxidizing potential should be least. Here, the oxidizing potential of $\mathrm{Fe}^{+2}$ is less than that of $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$. Therefore, $\mathrm{Fe}^{+2}$ is stronger oxidizing agent than $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$. Also, the stronger oxidizing agent should easily reduce itself. Here, $\mathrm{Fe}^{+3}$ is easily reduced than $\mathrm{Fe}^{+2}$. Therefore, among all the four, $\mathrm{Fe}^{+3}$ is the stronger oxidizing agent.