Question

Identify the oxidizing agent and the reducing agent in the following reaction: $H_{2}S+HNO_{3} \longrightarrow NO+S+H_{2}O$

Answer 1

"$2{ HN }^{ V }{ O }_{ 3 }\quad +\quad 3{ H }_{ 2 }{ S }^{ -II }\quad \rightarrow \quad 2{ N }^{ II }O\quad +\quad 4{ H }_{ 2 }O\quad +\quad 3{ S }^{ 0 }$

This is an oxidation-reduction (redox) reaction:

$3{ S }^{ -II }\quad -\quad 6{ e }^{ - }\quad \rightarrow \quad 3{ S }^{ 0 } (oxidation)$

$2{ N }^{ V }\quad +\quad 6{ e }^{ - }\quad \rightarrow \quad 2{ N }^{ II } (reduction)$

In this reaction the oxidant is${ HNO }_{ 3 }$ and the reductant is ${ H }_{ 2 }S$"

Answer 2

Explanation:

A reducing agent is an element that loses electrons.

The reducing agent means to lose electrons; it is said to have been oxidized.

The element which undergoes oxidation (gets oxidized) is called reducing agent.

H

2

is reducing agent.

2H

2

+O

2

→2H

2

O

In the given reaction , 2H

2

is oxidized to 2H

2

O by gaining oxygen atom. Thus , 2H

2

is reducing agent as it undergoes oxidation.