Question

In the chemical reaction between stoichiometric quantities of ${\mathrm{KMnO}}_4$ and $\mathrm{KI}$ in a weakly basic solution, what is the number of moles of ${\mathrm{I}}_2$ released for 4 moles of ${\mathrm{KMnO}}_4$ consumed?

Answer 1

• In a weakly basic solution, ${\mathrm{KMnO}}_4$ and $\mathrm{KI}$ react to produce manganese dioxide and iodine.
• ${\mathrm{KMnO}}_4+\mathrm{KI}\to {\mathrm{MnO}}_2+{\mathrm{I}}_2$ (weakly basic solution)
• The oxidation state of $\mathrm{Mn}$ in ${\mathrm{KMnO}}_4$ is +7. The oxidation state of $\mathrm{Mn}$ in ${\mathrm{MnO}}_2$ is +4.
• ${\mathrm{KMnO}}_4$,in neutral medium form ${\mathrm{MnO}}_2$(+7 top +4) that is it gains 3 electrons hence n-factor of $\mathrm{KMnO}4$is +3, n-factor of ${\mathrm{I}}_2$ is +2. $(2{\mathrm{I}}^{-}\to {\mathrm{I}}_2)$
• $3\times molesofKMn{O}_4=2\times molesof{I}_2$
• $3\times 4=2\times molesof{I}_2$
• $\frac{12}{2}=molesof{I}_2$
• $Molesof{I}_2=6.$

So, the correct answer is 6.