Question

Identify the reducing agent in the following reactions.
$\left( a \right)$ $4N{H_3} + 5{O_2} \to 4NO + 6{H_2}O$
$\left( b \right)$${H_2}O + {F_2} \to HF + HOF$
$\left( c \right)$ $F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}$
$\left( d \right)$ $2{H_2} + {O_2} \to 2{H_2}O$

Answer 1

Hint: A reducing agent (also called a reducer or reductant) is an element or compound that loses (or donates) an electron to oxidized itself and reduces the other element or compound (by accepting the elections).

Complete step by step answer:
In $\left( a \right)$ $\left( {N{H_3}} \right)$ , nitrogen exists in $\left( { - 3} \right)$ oxidation state when react with oxygen $\left( {{O_2}} \right)$ nitrogen get oxidizing itself to $\left( { + 2} \right)$ (by donating the electron) and reduce oxygen $\left( {{O_2}} \right)$ to $\left( {0 \to - 2} \right)$ oxidation states to form $\left( {NO} \right)$ . Therefore, $\left( {N{H_3}} \right)$ act as a reducing agent.
In $\left( b \right)$ $\left( {{H_2}O} \right)$ , Water $\left( {{H_2}O} \right)$ oxidation state of oxygen is $\left( { - 2} \right)$ when react with fluorine $\left( {{F_2}} \right)$ oxygen get oxidizing itself to $\left( 0 \right)$ (by donates the electron) and reduce fluorine $\left( {{F_2}} \right)$ to $\left( {0 \to - 1} \right)$ oxidation states to form$\left( {HF} \right)$. Therefore, $\left( {{H_2}O} \right)$ act as a reducing agent.
In $\left( c \right)$ $\left( {CO} \right)$ , Carbon monoxide $\left( {CO} \right)$ oxidation states of carbon exists in $\left( { + 2} \right)$ oxidation state when it react with Ferrosoferric oxide $\left( {F{e_2}{O_3}} \right)$ (iron oxide) carbon get oxidizing itself to $\left( { + 4} \right)$ (by donates the electron) and reduce Ferrosoferric oxide $\left( {F{e_2}{O_3}} \right)$ to $\left( { + 3 \to 0} \right)$oxidation states to form $\left( {Fe} \right)$ . Therefore, Carbon monoxide $\left( {CO} \right)$ acts as a reducing agent.
In $\left( d \right)$ $\left( {{H_2}} \right)$, Hydrogen exists in $\left( 0 \right)$ oxidation state when react with oxygen $\left( {{O_2}} \right)$ hydrogen get oxidizing itself to $\left( { + 1} \right)$ (by donating the electron) and reduce oxygen $\left( {{O_2}} \right)$ to $\left( {0 \to - 2} \right)$ oxidation states to form $\left( {{H_2}O} \right)$ . Therefore, $\left( {{H_2}} \right)$ act as a reducing agent.
As we discussed above, the reducing agent in $\left( a \right)$ , $\left( b \right)$ , $\left( c \right)$ and $\left( d \right)$ are $\left( {N{H_3}} \right)$ , $\left( {{H_2}O} \right)$ ,$\left( {CO} \right)$ and $\left( {{H_2}} \right)$ respectively.

Note:
If a substance gains oxygen or loses hydrogen during a reaction, it is said to be oxidized and is called a reducing agent. Similarly, if a substance gains hydrogen or loses oxygen during a reaction, it is said to be reduced and is called an oxidizing agent.