Question

In which of the following reactions, the underlined substance has been reduced?

• A. $CO+CuO\to C{O}_2+Cu$
• B. $CuO+2HCl\to CuC{l}_2+H_{2}O$
• C. $4H_{2}O+3Fe\to 4H_2+F{e}_3{O}_4$
• D. $C+4HN{O}_3\to C{O}_2+2H_{2}O+4N{O}_2$

Answer 1

Answer: (C)

$4H_{2}O+3Fe\to 4H_2+F{e}_3{O}_4$

Oxidation is loss of electron and reduction is the gain of electron. Thereby assigning the oxidation number on central atom in each molecule for each conversion will determine oxidation or reduction process. Considering oxidation number of $H=+1,O=-2$ we get oxidation state as:

(A) Oxidation number of $C$ and $Cu$ in $CO+CuO\to C{O}_2+Cu$

$\stackrel{+2}{C}O+\stackrel{+2}{Cu}O\to \stackrel{+4}{C}{O}_2+\stackrel{0}{Cu}$

Oxidation number of $C$ changes from +2 to +4 thus gets oxidized

Oxidation number of $Cu$ changes from +2 to 0 thus gets reduced

(B) Oxidation number of $Cu$ and $Cl$ in $CuO+2HCl\to CuC{l}_2+H_{2}O$

$\stackrel{+2}{Cu}O+2H\stackrel{-1}{Cl}\to \stackrel{+2}{Cu}\stackrel{-1}{C{l}_2}+H_{2}O$

Oxidation number of $Cu$ and $Cl$ does not change, so no oxidation or reduction.

(C) Oxidation number of $H$ and $Fe$ in $4H_{2}O+3Fe\to 4H_2+F{e}_3{O}_4$

$\stackrel{+1}{H_2}O+3\stackrel{0}{Fe}\to 4\stackrel{0}{H_2}+\stackrel{+8/3}{F{e}_3}{O}_4$

Oxidation number of $H$ and $Fe$ changes where $H_{2}O$ gets reduced and $Fe$ is oxidized.

(D)Oxidation number of $H$ and $Fe$ in $C+4HN{O}_3\to C{O}_2+2H_{2}O+4N{O}_2$

$\stackrel{0}{C}+4H\stackrel{+5}{N}{O}_3\to \stackrel{+4}{C}{O}_2+4\stackrel{+4}{N}{O}_2$

Oxidation number of $C$ and $N$ changes where $HN{O}_3$ gets reduced and $C$ is oxidized.