Question

Which of the following process represents disproportionation reaction?

• A. $Cu+4HN{O}_3\to Cu(N{O}_3{)}_2+2N{O}_2+2H_{2}O$
• B. $3I_2+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+3H_{2}O$
• C. $C{l}_2+I_2\to 2ICl$
• D. $Zn+2HCl\to ZnC{l}_2+H_2$

Answer 1

The correct option is B $3I_2+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+3H_{2}O$

(a) $\stackrel{0}{C}u+4\stackrel{+5}{H}N{O}_3\to \stackrel{+2}{C}u(N{O}_3{)}_2+23\stackrel{+4}{N}{O}_2+2H_{2}O$
Cu is undergoing oxidation while $HN{O}_3$ is undergoing reduction, hence it is a redox reaction.

(b) Disproportionation reaction are those in which same species oxidized and reduced simultaneously.
$3I_2+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+3H_{2}O$

$3{\stackrel{0}{I}}_2\to \stackrel{+5}{I}{O}_3^{-}$ oxidation
$3{\stackrel{0}{I}}_2\to {\stackrel{-1}{I}}^{-}$ reduction
here $I_2$ is undergoing oxidation as well as reduction.

c) $\stackrel{0}{C}{l}_2+{\stackrel{0}{I}}_2\to 2\stackrel{+}{I}\stackrel{-1}{C}l$
$I_2$ is undergoing oxidation while $C{l}_2$ is undergoing reduction, hence it is a redox reaction.
d) $\stackrel{0}{Z}n+2\stackrel{-1}{H}Cl\to \stackrel{+2}{Z}nC{l}_2+{\stackrel{0}{H}}_2$
$ZnC{l}_2$ is undergoing oxidation while $HCl$ is undergoing reduction, hence it is a redox reaction.