   # Manganese may play an important role in chemical cycles in the oceans. Two reactions involving manganese (in acid solution) are the reduction of nitrate ions (to NO) with Mn 2+ ions and the oxidation of ammonium ions (to N 2 ) with MnO 2 . (a) Write balanced chemical equations for these reactions (in acid solution). (b) Calculate E ° cell for the reactions. (One half-reaction potential you need is for the reduction of N 2 to NH 4 + , E ° = −0.272 V.) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19, Problem 93GQ
Textbook Problem
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## Manganese may play an important role in chemical cycles in the oceans. Two reactions involving manganese (in acid solution) are the reduction of nitrate ions (to NO) with Mn2+ ions and the oxidation of ammonium ions (to N2) with MnO2. (a) Write balanced chemical equations for these reactions (in acid solution). (b) Calculate E°cell for the reactions. (One half-reaction potential you need is for the reduction of N2 to NH4+, E° = −0.272 V.)

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reactions in acidic solutions has to be given.

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding H+ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding H+ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 6. Add the same number of OH- groups as there are H+ present to both sides of the equation.

### Explanation of Solution

Let’s write the reaction between the Mn2+ and NO3-:

Mn2+(aq) + NO3-(aq) MnO(s) + NO(g)

Find the oxidation and reduction reaction half reactions.

Oxidation:  Mn+2(aq)  MnO2(s)Reduction:  NO3-(aq) NO(g)

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.

Oxidation:  Mn+2(aq)  MnO2(s)Reduction:  NO3-(aq) NO(g)

2. 2. Balance O atoms by adding water to the side missing O atoms.

Oxidation: Mn+2(aq) +2H2O(l) MnO2(s)reduction: NO3-(aq) NO(g) + 2H2O(l)

3. 3. Balance the H atoms by adding H+ to the side missing H atoms.

Oxidation: Mn+2(aq) +2H2O(l) MnO2(s)+4H+(aq)Reduction: NO3-(aq) +4H+(aq) NO(g) + 2H2O(l)

4. 4. Balance the charge by adding electrons to side with more total positive charge

Oxidation: Mn+2(aq) +2H2O(l) MnO2(s)+4H+(aq)+2eReduction: NO3-(aq) +4H+(aq) +3e NO(g) + 2H2O(l)

5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Oxidation: 3[Mn+2(aq) +2H2O(l) MnO2(s)+4H+(aq)+2e]Reduction: 2[NO3-(aq) +4H+(aq) +3e NO(g) + 2H2O(l)]___________________________________________________________3Mn2+(aq) + 2NO3- (aq) + 2H2O (l)  3MnO2(s) + 2NO(g) + 4H+(aq)

Therefore, the balanced half cell reaction is as follows.

3Mn2+(aq) + 2NO3- (aq) + 2H2O (l)  3MnO2(s) + 2NO(g) + 4H+(aq)

Let’s write the reaction for oxidation of ammonium ions to N2

(b)

Interpretation Introduction

Interpretation:

The Ecell0 for the reactions has to be determined.

Concept introduction:

Under certain conditions a cell potential is measured it is called as standard potential (Ecello).

Standard potential (Ecello) can be calculated by the following formula.

Ecello=Ecathodeo-Eanodeo

The Ecello value is positive, the reaction is predicted to be product favoured at equilibrium.

The Ecello value is negative, the reaction is predicted to be reactant favoured at equilibrium

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