On the basis of the following $\mathrm{E}^{\circ}$ values, the strongest oxidizing agent is : $\left…

On the basis of the following $\mathrm{E}^{\circ}$ values, the strongest oxidizing agent is :
$\left.\mathrm{Fe}(\mathrm{CN})_{6}ight]^{4-} ightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}ight]^{3-}+\mathrm{e}^{-} ; \mathrm{E}^{\circ}=-0.35 \mathrm{~V}$
$\mathrm{Fe}^{2+} ightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \quad \mathrm{E}^{\circ}=-0.77 \mathrm{~V}$
  1. $\left[\mathrm{Fe}(\mathrm{CN})_{6}ight]^{4}$
  2. $\mathrm{Fe}^{2+}$
  3. $\mathrm{Fe}^{3+}$
  4. $\left[\mathrm{Fe}(\mathrm{CN})_{6}ight]^{3-}$

Solution

From the given data we find $\mathrm{Fe}^{3+}$ is strongest oxidising agent. More the positive value of $\mathrm{E}^{\circ}$, more is the tendency to get oxidized. Thus correct option is $(\mathrm{c})$.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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