Question

Using data from Table 11.1, place the following in order of increasing strength as oxidizing agents (all under standard conditions).

${\mathbf{Cd}}^{\mathbf{2}\mathbf{+}}\mathbf{,}{{\mathbf{IO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{,}{\mathbf{K}}^{\mathbf{+}}\mathbf{,}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}{{\mathbf{AuCI}}_{\mathbf{4}}}^{\mathbf{-}}\mathbf{,}{\mathbf{andI}}_{\mathbf{2}}$

Short answer

The strengthincreasing order is ${\mathrm{K}}^{+}<{\mathrm{H}}_2\mathrm{O}<{\mathrm{Cd}}^{2+}<{{\mathrm{AuCI}}_4}^{-}<{{\mathrm{IO}}_3}^{-}<{\mathrm{I}}_2$.

Step-by-step solution

Definition of standard reduction potential

An increase in the value of standard reduction potential means anincrease in acidic strength. So, to make increasing the order of strength of the given oxidizing agent, first , observe the value of standard reduction potential.

Explanation of increasing order of strength

The more the value of ${\mathrm{E}}^0$(standard reduction potential), more the strength of oxidation, thus the order ofspecies in the increasing order of strength is,

$\begin{array}{l}{\mathrm{K}}^{+}+{\mathrm{e}}^{-}\to \mathrm{K},{\mathrm{E}}^0=-2.92\mathrm{V}\\ 2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-},{\mathrm{E}}^0=-0.83\mathrm{V}\\ {\mathrm{Cd}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cd},{\mathrm{E}}^0=-0.40\mathrm{V}\\ {\mathrm{I}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{I}}^{-},{\mathrm{E}}^0=0.54\mathrm{V}\\ {\mathrm{AuCI}}_4+3{\mathrm{e}}^{-}\to \mathrm{Au}+4{\mathrm{CI}}^{-},{\mathrm{E}}^0=0.99\mathrm{V}\\ {{\mathrm{IO}}_3}^{-}+6{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{I}}_2+3{\mathrm{H}}_2\mathrm{O},{\mathrm{E}}^0=1.20\mathrm{V}\end{array}$

From the strength of oxidation, thus the order of species in the increasing order of strength is determined.