   Chapter 3, Problem 58PS

Chapter
Section
Textbook Problem

In the following reactions, decide which reactant is oxidized and which is reduced. Designate the oxidizing agent and the reducing agent.(a) Cr2O72– (aq) + 3 Sn2+(aq) + 14 H3O+(aq) → 2 Cr3+aq) + 3 Sn4+(aq) + 21 H2O(ℓ)(b) FeS(s) + 3 NO3–(aq) + 4 H3O+(aq) → 3 NO(g) + SO42–(aq) + Fe3+(aq) + 6 H2O(ℓ)

(a)

Interpretation Introduction

Interpretation:

In the given reaction, the oxidized and reduced reactant has to be decide and oxidizing and reducing agent should be designated.

Concept introduction:

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

Explanation

The given reaction is shown below,

The given compound is not a neutral compound.  Therefore the oxidation state of Cr in Cr2O72- is given below,

Net charge on Cr2O72- is equal to sum of the oxidation number of oxygen atoms. Therefore,

Net charge = Number of oxidation state of Cr atom+Number of oxidation state of oxygen atom -2 = (x) +  7(-2)

(b)

Interpretation Introduction

Interpretation:

In the given reaction, the oxidized and reduced reactant has to be decide and oxidizing and reducing agent should be designated.

Concept introduction:

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

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