   Chapter 3, Problem 89IL

Chapter
Section
Textbook Problem

The following reaction can be used to prepare iodine in the laboratory.2 Nal(s) + 2 H2SO4(aq) + MnO2(s) → Na2SO4(aq) + MnSO4(aq) + I2(g) + 2 H2O(ℓ)(a) Determine the oxidation number of each atom in the equation.(b) What is the oxidizing agent, and what has been oxidized? What is the reducing agent, and what has been reduced?(c) Is the reaction product-favored or reactant-favored?(d) Name the reactants and products.

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each atom in the given reaction should be identified.

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,

2NaI(s)+2H2SO4(aq)+MnO2(s)Na2SO4(aq)+MnSO4(aq)+I2(g)+2H2O(l)

The oxidation state of the reactant is given below,

Oxidation state of Sodium is +1, iodine -1, oxygen is -2

Oxidation state of sulfur in sulfuric acid is given below

The given compound is neutral compound. Therefore the oxidation state of S in H2SO4 is given below,

Net charge on H2SO4 is equal to sum of the oxidation number of oxygen atoms, hydrogen atom and sulfur atom. Therefore,

Net charge = Number of oxidation state of sulfur atom+Number of oxidation state of oxygen atomNumber of oxidation state of hydrogen atom 0 = 2 +(x) +  4(-2)0 = 2 +(x) +  4(-2)0 = x- 66 = xX = 6

Net charge on MnO2 is equal to sum of the oxidation number of oxygen atoms, manganese atom. Therefore,

Net charge = Number of oxidation state of manganese atom+Number of oxidation state of oxygen atom 0 = (x) +  2(-2)0 = (x) +  2(-2)0 = x- 44 = xX = 4

The oxidation state of the product is given below,

Oxidation state of Sodium is +1, iodine is 0, oxygen is -2

The given compound is neutral compound

(b)

Interpretation Introduction

Interpretation:

Oxidizing agent and reducing agent has to be identified and oxidized and reduced agent should be identified in the given reaction.

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.

(c)

Interpretation Introduction

Interpretation:

The reaction favored whether reactant or reactant should be identified.

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.

(d)

Interpretation Introduction

Interpretation:

Name of the reactant and product in the given reaction should be identified.

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