Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 8, Problem 8.98QA
Interpretation Introduction

To find:

The balanced chemical equation for the following reactions:

a) MnO4-(aq)+S-2(aq)MnS(s)+S8(s)

b) MnO4-(aq)+CN-(aq)MnO2(s)+CNO-(aq)

c) MnO4-(aq)+SO32-(aq) MnO2(s)+SO42-(aq)

Expert Solution & Answer
Check Mark

Answer to Problem 8.98QA

Solution:

a) 16MnO4-aq+56S-2aq+64H2Ol16MnSs+ 5S8s+ 128OH-aq

b) 2MnO4-aq+ 3CN-aq+ H2Ol2MnO2s+ 3CNO-aq+2OH-aq

c) 2MnO4-aq+ 3SO3-2aq+ H2Ol2MnO2s+ 3SO4-2aq+2OH-aq

Explanation of Solution

1) Concept:

The following steps are followed to balance a redox reaction in a basic medium:

i) Calculate the change in oxidation number values of elements which are getting reduced and oxidized.

ii) Insert an appropriate coefficient to balance the change in oxidation number values(∆O.N) before the species which are only getting reduced and oxidized. We don’t balance O atoms in this step.

iii) The ionic charges are balanced by adding the appropriate number of OH- ions in the basic solution.

iv) Water molecules have to be added, if necessary, in the last step.

2) Calculation:

a)

i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:

MnO4-(aq)+S-2(aq)MnS(s)+S8(s)

ii) An analysis of change in oxidation number values (∆O.N) is done for Mn and S.

O.N for Mn = +7 in MnO4- and  Mn =+2 in MnS.

So, O.N for Mn = -5

O.N of S in S-2 =-2 and S = 0 in S8.

Therefore, the net change in O.N for each Sulfur atom = +2

iii) To balance these (∆O.N) values we need to put coefficient 2 in front of  MnO4-  and  5 in front of S2-.

 2MnO4-(aq)+5S-2(aq)MnS(s)+S8(s)

iv) To balance Mn and S atoms we insert coefficient 2 in front of MnS and 58 in front of S8.

2 MnO4-aq+5 S-2aq2 MnS(s)+58S8(s)

v) To balance S and O atoms, we put the coefficient 7 in front of S-2 and 8 molecules of water on right side.

2 MnO4-aq+7 S-2aq2 MnSs+58S8(s)+8H2O(l)

vi) To balance the charges(-16 on left and zero on right) we add 16OH- on right side.

2MnO4-(aq)+7S-2(aq)2MnS(s)+5/8S8(s)+8H2O(l)+16OH-(aq)

vii) To balance H atoms we need to put 16 molecules of water on left side.

16H2O(l)+2MnO4-(aq)+7S-2(aq)2MnS(s)+5/8S8(s)+8H2O(l)+16OH-(aq)

viii) Therefore, the final balanced chemical equation becomes:

8H2O(l)+2MnO4-(aq)+7S-2(aq)2MnS(s)+5/8S8(s)+16OH-(aq)

Finally, to remove fractional coefficient 58 before S8, we multiply the whole balanced equation by 8.

Thus we have:

64 H2Ol+16MnO4-aq+56 S-2aq16 MnSs+5 S8s+128 OH-(aq)

This is our final balanced chemical equation.

b)

i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:

MnO4-(aq)+CN-(aq)MnO2(s)+CNO-(aq)

ii) An analysis of change in oxidation number values(∆O.N) is done for Mn and C.

O.N for Mn = +7 in MnO4- and Mn =+4 in MnO2.

Similarly, O.N of C in CN- is +2 and O.N of C in CNO- is +4.

∆O.N for Mn = -3  and ∆O.N for C = +2

iii) To balance these (∆O.N) values we put coefficient 2 in front of  MnO4- and  3 in front of CN-.

2MnO4-(aq)+3CN-(aq)MnO2(s)+CNO-(aq)

iv) To balance Mn and C atoms, we insert coefficients 2 and 3 in front of MnO2 and CNO- respectively.

2MnO4-(aq)+3CN-(aq)2MnO2(s)+3CNO-(aq)

v)) To balance O atoms on both sides, we add one water molecule on the right side.

2MnO4-(aq)+3CN-(aq)2MnO2(s)+3CNO-(aq)+H2O(l)

vi) To balance the electrical charges(-5 on left and -3 on right) we add 2OH- on right side

2MnO4-(aq)+3CN-(aq)2MnO2(s)+3CNO-(aq)+H2O(l))+2OH-(aq)

vii) To balance H atoms(zero on left and 2 on right) we need to put 2 molecules of H2O on the left side.

2H2O(l)+2MnO4-(aq)+3CN-(aq)2MnO2(s)+3CNO-(aq)+H2O(l)+2OH-(aq)

Therefore, the final balanced equation becomes:

H2O(l)+2MnO4-(aq)+3CN-(aq)2MnO2(s)+3CNO-(aq)+2OH-(aq)

c)

i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:

MnO4-(aq)+SO32-(aq) MnO2(s)+SO42-(aq)

ii) An analysis of change in oxidation number values(∆O.N) is done for Mn and S.

O.N for Mn = +7 in MnO4- and Mn =+4 in nO2.

∆O.N for Mn = -3

O.N for S in SO32-= +4 and in SO42- is +6.

∆O.N of S = +2

iii) To balance these (∆O.N) values we put coefficient 2 in front of  MnO4-  and  3 in front of SO32-.

2MnO4-(aq)+3SO32-(aq) MnO2(s)+SO42-(aq)

iv) To balance Mn and S on both sides, we get,

2MnO4-(aq)+3SO32-(aq) 2MnO2(s)+3SO42-(aq)

v) To balance O atoms, we  put one water molecule on right side.

2MnO4-(aq)+3SO32-(aq) 2MnO2(s)+3SO42-(aq)+H2O(l)

vi) To balance the electrical charges(-8 on left and -6 on right) we add 2OH- on right side.

2MnO4-(aq)+3SO32-(aq) 2MnO2(s)+3SO42-(aq)+H2O(l)+2OH-(aq)

vii) To balance H atoms(zero on left and 4 on right) we add two water molecules on left side of the reaction.

2H2O(l)+2MnO4-(aq)+3SO32-(aq) 2MnO2(s)+3SO42-(aq)+H2O(l)+2OH-(aq)

viii) Therefore, the final balanced chemical equation becomes:

H2O (l)+2MnO4-(aq)+3SO32-(aq) 2MnO2(s)+3SO42-(aq)+2OH-(aq)

Conclusion:

The given reactions are balanced in the basic medium using the rules for balancing of redox reactions.

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Chapter 8 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 8 - Prob. 8.11QACh. 8 - Prob. 8.12QACh. 8 - Prob. 8.13QACh. 8 - Prob. 8.14QACh. 8 - Prob. 8.15QACh. 8 - Prob. 8.16QACh. 8 - Prob. 8.17QACh. 8 - Prob. 8.18QACh. 8 - Prob. 8.19QACh. 8 - Prob. 8.20QACh. 8 - Prob. 8.21QACh. 8 - Prob. 8.22QACh. 8 - Prob. 8.23QACh. 8 - Prob. 8.24QACh. 8 - Prob. 8.25QACh. 8 - Prob. 8.26QACh. 8 - Prob. 8.27QACh. 8 - Prob. 8.28QACh. 8 - Prob. 8.29QACh. 8 - Prob. 8.30QACh. 8 - Prob. 8.31QACh. 8 - Prob. 8.32QACh. 8 - Prob. 8.33QACh. 8 - Prob. 8.34QACh. 8 - Prob. 8.35QACh. 8 - Prob. 8.36QACh. 8 - Prob. 8.37QACh. 8 - Prob. 8.38QACh. 8 - Prob. 8.39QACh. 8 - Prob. 8.40QACh. 8 - Prob. 8.41QACh. 8 - Prob. 8.42QACh. 8 - Prob. 8.43QACh. 8 - Prob. 8.44QACh. 8 - Prob. 8.45QACh. 8 - Prob. 8.46QACh. 8 - Prob. 8.47QACh. 8 - Prob. 8.48QACh. 8 - Prob. 8.49QACh. 8 - Prob. 8.50QACh. 8 - Prob. 8.51QACh. 8 - Prob. 8.52QACh. 8 - Prob. 8.53QACh. 8 - Prob. 8.54QACh. 8 - Prob. 8.55QACh. 8 - Prob. 8.56QACh. 8 - Prob. 8.57QACh. 8 - Prob. 8.58QACh. 8 - Prob. 8.59QACh. 8 - Prob. 8.60QACh. 8 - Prob. 8.61QACh. 8 - Prob. 8.62QACh. 8 - Prob. 8.63QACh. 8 - Prob. 8.64QACh. 8 - Prob. 8.65QACh. 8 - Prob. 8.66QACh. 8 - Prob. 8.67QACh. 8 - Prob. 8.68QACh. 8 - Prob. 8.69QACh. 8 - Prob. 8.70QACh. 8 - Prob. 8.71QACh. 8 - Prob. 8.72QACh. 8 - Prob. 8.73QACh. 8 - Prob. 8.74QACh. 8 - Prob. 8.75QACh. 8 - Prob. 8.76QACh. 8 - Prob. 8.77QACh. 8 - Prob. 8.78QACh. 8 - Prob. 8.79QACh. 8 - Prob. 8.80QACh. 8 - Prob. 8.81QACh. 8 - Prob. 8.82QACh. 8 - Prob. 8.83QACh. 8 - Prob. 8.84QACh. 8 - Prob. 8.85QACh. 8 - Prob. 8.86QACh. 8 - Prob. 8.87QACh. 8 - Prob. 8.88QACh. 8 - Prob. 8.89QACh. 8 - Prob. 8.90QACh. 8 - Prob. 8.91QACh. 8 - Prob. 8.92QACh. 8 - Prob. 8.93QACh. 8 - Prob. 8.94QACh. 8 - Prob. 8.95QACh. 8 - Prob. 8.96QACh. 8 - Prob. 8.97QACh. 8 - Prob. 8.98QACh. 8 - Prob. 8.99QACh. 8 - Prob. 8.100QACh. 8 - Prob. 8.101QACh. 8 - Prob. 8.102QACh. 8 - Prob. 8.103QACh. 8 - Prob. 8.104QACh. 8 - Prob. 8.105QACh. 8 - Prob. 8.106QACh. 8 - Prob. 8.107QACh. 8 - Prob. 8.108QACh. 8 - Prob. 8.109QACh. 8 - Prob. 8.110QACh. 8 - Prob. 8.111QACh. 8 - Prob. 8.112QACh. 8 - Prob. 8.113QACh. 8 - Prob. 8.114QACh. 8 - Prob. 8.115QACh. 8 - Prob. 8.116QACh. 8 - Prob. 8.117QACh. 8 - Prob. 8.118QACh. 8 - Prob. 8.119QACh. 8 - Prob. 8.120QACh. 8 - Prob. 8.121QACh. 8 - Prob. 8.122QACh. 8 - Prob. 8.123QACh. 8 - Prob. 8.124QACh. 8 - Prob. 8.125QACh. 8 - Prob. 8.126QACh. 8 - Prob. 8.127QACh. 8 - Prob. 8.128QACh. 8 - Prob. 8.129QACh. 8 - Prob. 8.130QA
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