Question

On the basis of following $E^{\circ }$ values, the strongest oxidising agent is:
$\left[Fe\right(CN{)}_4{]}^{4^{-}}\to \left[Fe\right(CN{)}_6{]}^{3-}+e^{-}$
$\begin{array}{cc}& E^{\circ }=-0.35\text{V}\\ F{e}^{2+}\to F{e}^{3+}+e^{-}& E^{\circ }=-0.77\text{V}\end{array}$

• A. $F{e}^{2+}$
• B. $F{e}^{3+}$
• C. $\left[Fe\right(CN{)}_6{]}^{3-}$
• D. $\left[Fe\right(CN{)}_6{]}^{4-}$

Answer 1

The correct option is B $F{e}^{3+}$

The species that has the most positive value of the reduction potential would get reduced the easiest, and hence would be the strongest oxidising agent. Here, if we reverse the second reaction, we get $E_{F{e}^{3+}/F{e}^{2+}}=0.77\text{V}$
This means that $F{e}^{3+}$ can gain an electron easily to form $F{e}^{2+}$ making it the strongest oxidising agent.