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To find:
The balanced chemical equation for the following reactions:
a) MnO-4(aq)+S-2(aq)→MnS(s)+S8(s)
b) MnO-4(aq)+CN-(aq)→MnO2(s)+CNO-(aq)
c) MnO-4(aq)+SO2-3(aq)→ MnO2(s)+SO2-4(aq)
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Answer to Problem 8.98QA
Solution:
a) 16MnO-4(aq)+56S-2(aq)+64H2O(l)→16MnS(s)+ 5S8(s)+ 128OH-(aq)
b) 2MnO-4(aq)+ 3CN-(aq)+ H2O(l)→2MnO2(s)+ 3CNO-(aq)+2OH-(aq)
c) 2MnO-4(aq)+ 3SO-23(aq)+ H2O(l)→2MnO2(s)+ 3SO-24(aq)+2OH-(aq)
Explanation of Solution
1) Concept:
The following steps are followed to balance a
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
MnO-4(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 MnO-4 and Mn =+2 in MnS.
So, ∆
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-.
iv) To balance Mn and S atoms we insert coefficient 2 in front of MnS and
v) To balance S and O atoms, we put the coefficient 7 in front of
vi) To balance the charges(-16 on left and zero on right) we add 16OH- on right side.
vii) To balance H atoms we need to put 16 molecules of water on left side.
viii) Therefore, the final balanced chemical equation becomes:
Finally, to remove fractional coefficient
Thus we have:
This is our final balanced chemical equation.
b)
i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:
ii) An analysis of change in oxidation number values(∆O.N) is done for Mn and C.
O.N for Mn = +7 in
Similarly, O.N of C in
∆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
iv) To balance Mn and C atoms, we insert coefficients 2 and 3 in front of MnO2 and CNO- respectively.
v)) To balance O atoms on both sides, we add one water molecule on the right side.
vi) To balance the electrical charges(-5 on left and -3 on right) we add 2OH- on right side
vii) To balance H atoms(zero on left and 2 on right) we need to put 2 molecules of H2O on the left side.
Therefore, the final balanced equation becomes:
c)
i) A preliminary expression of the unbalanced chemical reaction involving reactant and product is:
ii) An analysis of change in oxidation number values(∆O.N) is done for Mn and S.
O.N for Mn = +7 in
∆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-.
iv) To balance Mn and S on both sides, we get,
v) To balance O atoms, we put one water molecule on right side.
vi) To balance the electrical charges(-8 on left and -6 on right) we add 2OH- on right side.
vii) To balance H atoms(zero on left and 4 on right) we add two water molecules on left side of the reaction.
viii) Therefore, the final balanced chemical equation becomes:
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)
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