Question

Which reaction does not represent auto redox or disproportionation?

• A. ${\mathrm{Cl}}_2+{\mathrm{OH}}^{-}\to {\mathrm{Cl}}^{-}+{\mathrm{ClO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}$
• B. $2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$
• C. $2{\mathrm{Cu}}^{+}\to {\mathrm{Cu}}^{2+}+\mathrm{Cu}$
• D. ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to {\mathrm{N}}_2+{\mathrm{Cr}}_2{\mathrm{O}}_3+4{\mathrm{H}}_2\mathrm{O}$

Answer 1

The correct option is D

${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to {\mathrm{N}}_2+{\mathrm{Cr}}_2{\mathrm{O}}_3+4{\mathrm{H}}_2\mathrm{O}$

The explanation of the correct option:

(D) ${\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}\mathbf{}}\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

• The disproportional reaction is a particular type of redox reaction where one compound is converted into two compounds where one gets oxidized to a lower oxidation number and one gets oxidized to a higher oxidation number.
• In the above reaction, Chromium gets reduced and no chromium oxidation occurs.
• In the case of nitrogen, it gets oxidized from $-4to0$
• Hence, one gets oxidized and one gets reduced.

The explanation of the incorrect option:

(A) ${\mathbf{Cl}}_{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{OH}}^{\mathbf{-}}\mathbf{}\mathbf{\to }{\mathbf{Cl}}^{\mathbf{-}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{ClO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

• This is an auto-redox reaction, where only one species gets oxidized and reduced as well.
• Here, chlorine gets oxidized from ${\mathrm{Cl}}_2\to$${\mathrm{ClO}}_3^{-}$
• Chlorine gets reduced from ${\mathrm{Cl}}_2\to {\mathrm{Cl}}^{-}$
• Hence, it is a disproportionate reaction.

(B) $\mathbf{2}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}\mathbf{}}\mathbf{\to }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{O}}_{\mathbf{2}}$

• The above reaction is a disproportionate reaction as oxygen gets reduced and oxidized in the same reaction at the same time.
• Oxygen oxidizes in the form of water molecules.
• Reduces in the form of oxygen molecule.
• Hence, a disproportionate reaction.

(C) $\mathbf{2}{\mathbf{Cu}}^{\mathbf{+}}\mathbf{}\mathbf{\to }{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{Cu}$

• This is a clear-cut direct example of an auto-redox reaction.
• Cuprous ion getting converted to cupric is the oxidation of it.
• While cuprous getting converted to copper metal is a reduction.
• Hence, it is a disproportionate reaction.

Therefore, option (D) ${\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}\mathbf{}}\mathbf{\to }{\mathbf{N}}_{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$doesn't show a disproportionate reaction.