Radii of iron atom and its ions follow `Fe > Fe^(2+) > Fe^(3+)` - explain.

Adding powdered lead and iron to a solution that is 1M in both Pb^(2+) and Fe^(2+) ions, would results to a reaction in which (E^@ Fe^(2+)//Fe = -0.44V, E^@Pb^(2+)//Pb = -0.13V)

The oxidation state of Fe in [Fe(CN)_6]^(-3) ion is

Standared electrode potential of two half-reactions are given below : Fe^(2+) iff Fe " "E^(@) = -0.44 V , Fe^(3+) iff Fe^(2+) " "E^(@) = + 0.77V If Fe^(2+) , Fe^(3+) and Fe are kept together :

What is the correct order of radii of Fe^(2+), Fe^(3+) and neutral Fe atom. Why?

Which of the following order is correct for the size of Fe^(3+) , Fe and Fe^(2+) ?

Explain the decrease in the trend of the radius: Fe gt Fe^(2+) gt Fe^(3+)

A quantity of 25.0 mL of solution containing both Fe^(2+) and Fe^(3+) ions is titrated with 25.0 mL of 0.0200 M KMnO_(4) (in dilute H_(2)SO_(4) ). As a result, all of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same KMnO_(4) solution for oxidation to Fe^(3+) . MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O Zinc aded in the second titration wil