Chapter 8, Problem 8.98QA

Interpretation Introduction

To find:

The balanced chemical equation for the following reactions:

a) $Mn{O}_4^{-}\left(aq\right)+S^{-2}\left(aq\right)\to MnS\left(s\right)+S_{8\left(s\right)}$

b) $Mn{O}_4^{-}\left(aq\right)+{CN}^{-}\left(aq\right)\to {MnO}_2\left(s\right)+{CNO}^{-}\left(aq\right)$

c) $Mn{O}_4^{-}\left(aq\right)+{SO}_3^{2-}\left(aq\right)\to {MnO}_2\left(s\right)+{SO}_4^{2-}\left(aq\right)$

Answer to Problem 8.98QA

Solution:

a) $16Mn{O}_4^{-}\left(aq\right)+56S^{-2}\left(aq\right)+64H_{2}O\left(l\right)\to 16MnS\left(s\right)+ 5S_8\left(s\right)+ 128O{H}^{-}\left(aq\right)$

b) $2\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{-}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{ }3\mathrm{C}{\mathrm{N}}^{-}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{ }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{M}\mathrm{n}{\mathrm{O}}_2\left(\mathrm{s}\right)+\mathrm{ }3\mathrm{C}\mathrm{N}{\mathrm{O}}^{-}\left(\mathrm{a}\mathrm{q}\right)+2\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{a}\mathrm{q}\right)$

c) $2\mathrm{M}\mathrm{n}{\mathrm{O}}_4^{-}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{ }3\mathrm{S}{\mathrm{O}}_3^{-2}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{ }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{M}\mathrm{n}{\mathrm{O}}_2\left(\mathrm{s}\right)+\mathrm{ }3\mathrm{S}{\mathrm{O}}_4^{-2}\left(\mathrm{a}\mathrm{q}\right)+2\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{a}\mathrm{q}\right)$

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:

$Mn{O}_4^{-}\left(aq\right)+S^{-2}\left(aq\right)\to MnS\left(s\right)+S_{8\left(s\right)}$

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

O.N for $Mn = +7$ in $Mn{O}_4^{-}$ and  $Mn =+2 in MnS$.

So, $∆O.N for Mn = -5$

$O.N of S in S^{-2} =-2 and S = 0 in S_8.$

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 S^2-.

$2Mn{O}_4^{-}\left(aq\right)+{5S}^{-2}\left(aq\right)\to MnS\left(s\right)+S_{8\left(s\right)}$

iv) To balance Mn and S atoms we insert coefficient 2 in front of MnS and $\frac{5}{8}$ in front of S₈.

$2 Mn{O}_4^{-}\left(aq\right)+{5 S}^{-2}\left(aq\right)\to 2 MnS\left(s\right)+\frac{5}{8}{S}_{8\left(s\right)}$

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.

$2 Mn{O}_4^{-}\left(aq\right)+{7 S}^{-2}\left(aq\right)\to 2 MnS\left(s\right)+\frac{5}{8}{S}_{8\left(s\right)}+{8H}_{2}O\left(l\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+{7S}^{-2}\left(aq\right)\to 2MnS\left(s\right)+5/8S_{8\left(s\right)}+{8H}_{2}O\left(l\right)+16{OH}^{-}\left(aq\right)$

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

${16H}_{2}O\left(l\right)+2Mn{O}_4^{-}\left(aq\right)+{7S}^{-2}\left(aq\right)\to 2MnS\left(s\right)+5/8S_{8\left(s\right)}+{8H}_{2}O\left(l\right)+16{OH}^{-}\left(aq\right)$

viii) Therefore, the final balanced chemical equation becomes:

${8H}_{2}O\left(l\right)+2Mn{O}_4^{-}\left(aq\right)+{7S}^{-2}\left(aq\right)\to 2MnS\left(s\right)+5/8S_{8\left(s\right)}+16{OH}^{-}\left(aq\right)$

Finally, to remove fractional coefficient $\frac{5}{8}$ before S₈, we multiply the whole balanced equation by 8.

Thus we have:

${64 H}_{2}O\left(l\right)+16Mn{O}_4^{-}\left(aq\right)+{56 S}^{-2}\left(aq\right)\to 16 MnS\left(s\right)+5{ S}_8\left(s\right)+128 {OH}^{-}\left(aq\right)$

This is our final balanced chemical equation.

b)

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

$Mn{O}_4^{-}\left(aq\right)+{CN}^{-}\left(aq\right)\to {MnO}_2\left(s\right)+{CNO}^{-}\left(aq\right)$

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

O.N for Mn = +7 in $Mn{O}_4^{-}$ and Mn =+4 in $Mn{O}_2.$

Similarly, O.N of C in $C{N}^{-}$ is +2 and O.N of C in $CN{O}^{-}$ 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  $Mn{O}_4^{-}$ and  3 in front of CN^-.

$2Mn{O}_4^{-}\left(aq\right)+3{CN}^{-}\left(aq\right)\to {MnO}_2\left(s\right)+{CNO}^{-}\left(aq\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+3{CN}^{-}\left(aq\right)\to 2{MnO}_2\left(s\right)+{3CNO}^{-}\left(aq\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+3{CN}^{-}\left(aq\right)\to 2{MnO}_2\left(s\right)+{3CNO}^{-}\left(aq\right)+H_{2}O\left(l\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+3{CN}^{-}\left(aq\right)\to 2{MnO}_2\left(s\right)+{3CNO}^{-}\left(aq\right)+H_{2}O\left(l\right))+2{OH}^{-}(aq)$

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

${2H}_{2}O\left(l\right)+2Mn{O}_4^{-}\left(aq\right)+3{CN}^{-}\left(aq\right)\to 2{MnO}_2\left(s\right)+{3CNO}^{-}\left(aq\right)+H_{2}O\left(l\right)+2{OH}^{-}\left(aq\right)$

Therefore, the final balanced equation becomes:

$H_{2}O\left(l\right)+2Mn{O}_4^{-}\left(aq\right)+3{CN}^{-}\left(aq\right)\to 2{MnO}_2\left(s\right)+{3CNO}^{-}\left(aq\right)+2{OH}^{-}\left(aq\right)$

c)

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

$Mn{O}_4^{-}\left(aq\right)+{SO}_3^{2-}\left(aq\right)\to {MnO}_2\left(s\right)+{SO}_4^{2-}\left(aq\right)$

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

O.N for Mn = +7 in $Mn{O}_4^{-}$ and Mn =+4 in $n{O}_2$.

∆O.N for Mn = -3

O.N for S in SO₃^2-= +4 and in SO₄^2- 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 SO₃^2-.

$2Mn{O}_4^{-}\left(aq\right)+{3SO}_3^{2-}\left(aq\right)\to {MnO}_2\left(s\right)+{SO}_4^{2-}\left(aq\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+{3SO}_3^{2-}\left(aq\right)\to {2MnO}_2\left(s\right)+{3SO}_4^{2-}\left(aq\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+{3SO}_3^{2-}\left(aq\right)\to {2MnO}_2\left(s\right)+{3SO}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)$

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

$2Mn{O}_4^{-}\left(aq\right)+{3SO}_3^{2-}\left(aq\right)\to {2MnO}_2\left(s\right)+{3SO}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)+2{OH}^{-}\left(aq\right)$

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

${2H}_{2}O\left(l\right)+2Mn{O}_4^{-}\left(aq\right)+{3SO}_3^{2-}\left(aq\right)\to {2MnO}_2\left(s\right)+{3SO}_4^{2-}\left(aq\right)+H_{2}O\left(l\right)+2{OH}^{-}\left(aq\right)$

viii) Therefore, the final balanced chemical equation becomes:

$H_{2}O \left(l\right)+2Mn{O}_4^{-}\left(aq\right)+{3SO}_3^{2-}\left(aq\right)\to {2MnO}_2\left(s\right)+{3SO}_4^{2-}\left(aq\right)+2{OH}^{-}\left(aq\right)$

Conclusion:

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