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=sthese are the correct answers.

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.

Ionic compounds dissociate into component ions dissolved into a solution creating substances called "electrolytes".hope my answer has you : )

The answer is ax3. hope this