Question

Determine the oxidizing and reducing agents after answering the following questions. First, determine the oxidation number of every atom on both sides of the arrow.

KMnO4 + H2S + HCl <--> KCl + MnCl2 + H2O + S8
Did Cl change oxidation number? No
Did Oxygen change oxidation number? No
Did H change oxidation number? No
Did Mn change oxidation number? Yes
Did S change oxidation number? Yes
oxidizing agent=Mn
reducing agent=S
THESE ARE THE CORRECT ANSWERS.

Answer 1

Answers:

Did Cl change oxidation number?

No

Did Oxygen change oxidation number?

No

Did H change oxidation number?

No

Did Mn change oxidation number?

yes

Did S change oxidation number?

yes

oxidizing agent =

Mn

reducing agent =

s

Answer 2

Oxidation states of each element on left side of Arrow,

In KMnO₄;

O.N of K = +1
O.N of Mn = +7
O.N of O = -2

In H₂S;

O.N of H = +1
O.N of S = -2

In HCl;

O.N of H = +1
O.N of Cl = -1

Oxidation states of each element on right side of Arrow,

In KCl;

O.N of K = +1
O.N of Cl = -1

In MnCl₂;

O.N of Mn = +4
O.N of O = -2

In H₂O,

O.N of H = +1
O.N of O = -2

In S₈;

O.N of S = 0

Result:

As the Oxidation Number of Mn (in KMnO₄) has changed from +7 to +4 due to gain of 3 electrons, so it is reduced and has act as Oxidizing agent.

And the Oxidation Number of Sulfur (in H₂S) has changed from -2 to 0, hence it has lost electrons and is oxidized. Therefore it is a Reducing agent.