Question

Complete and balance the following equations for reactions

taking place in acidic solution.

1. \[\mathop {\rm{MnO}}_{\rm{4}}^{\rm{ - }}\left( {{\rm{aq}}} \right){\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right) \to {\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right){\rm{ + SO}}_{\rm{4}}^{{\rm{2 - }}}\left( {{\rm{aq}}} \right)\]

Short answer

1. The balanced equation is as follows:

   \[\mathop {\rm{8MnO}}_{\rm{4}}^{\rm{ - }}\left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right) + 14{{\rm{H}}^ + }\left( {{\rm{aq}}} \right) \to 8{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right){\rm{ + 5SO}}_{\rm{4}}^{{\rm{2 - }}}\left( {{\rm{aq}}} \right) + 12{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)\]

Step-by-step solution

Given information

The reaction is given as follows:

\[\mathop {\rm{MnO}}_{\rm{4}}^{\rm{ - }}\left( {{\rm{aq}}} \right){\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right) \to {\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right){\rm{ + SO}}_{\rm{4}}^{{\rm{2 - }}}\left( {{\rm{aq}}} \right)\]

Step 2: Oxidation number change method

The oxidation number change method is a method by which we can balance the redox reactions. In this method, the oxidation number of each atom is determined first. Then, we multiply the atoms in which the oxidation number is changed with a whole number.

Finding out the oxidation number change

The reaction is given below:

\[\mathop {\rm{MnO}}_{\rm{4}}^{\rm{ - }}\left( {{\rm{aq}}} \right){\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right) \to {\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right){\rm{ + SO}}_{\rm{4}}^{{\rm{2 - }}}\left( {{\rm{aq}}} \right)\]

In this reaction, both oxidation and reduction occurs. The reaction indicated with the oxidation number is given below:

\[\mathop {{\rm{Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + }}\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) \to \mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right){\rm{ + }}\mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right)\]

Manganese reduced its oxidation number from +7 to +2, causing reduction. Sulfur increased its oxidation number from -2 to +6, causing oxidation.

Divide to oxidation half and reduction half

Oxidation-half reaction can be written as:

\[\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) \to \mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 8{{\rm{e}}^ - }\]

The reduction-half reaction can be written as:

\[\mathop {{\rm{Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{e}}^ - } \to \mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right)\]

Balancing the charge

Since the reaction occur in acidic medium, we can add${{\rm{H}}^ + }$ion on the side, where the positive charge is deficient, to balance the charge.

Oxidation-half:\[\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) \to \mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 8{{\rm{e}}^ - } + 10{{\rm{H}}^ + }\]

Reduction-half: \[\mathop {{\rm{Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{e}}^ - } + 8{{\rm{H}}^ + } \to \mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right)\]

Balancing the oxygen atoms

Now, we can add water molecules to balance the oxygen atoms.

Oxidation:\[\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 4{{\rm{H}}_{\rm{2}}}{\rm{O}} \to \mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 8{{\rm{e}}^ - } + 10{{\rm{H}}^ + }\]

Reduction: \[\mathop {{\rm{Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{e}}^ - } + 8{{\rm{H}}^ + } \to \mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right) + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}\]

Equalization of electrons gained and lost

The electrons gained in the oxidation reaction must be same as the electrons lost in the reduction reaction. If they are not same, we must multiply both the reactions with the coefficients.

Oxidation:\[\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 4{{\rm{H}}_{\rm{2}}}{\rm{O}} \to \mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 8{{\rm{e}}^ - } + 10{{\rm{H}}^ + }\;\;\;\; \times 5\]

Reduction:\[\mathop {{\rm{Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{e}}^ - } + 8{{\rm{H}}^ + } \to \mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right) + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}\;\;\;\; \times {\rm{8}}\]

Hence, the reactions become as follows:

Oxidation:\[5\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 20{{\rm{H}}_{\rm{2}}}{\rm{O}} \to 5\mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 40{{\rm{e}}^ - } + 50{{\rm{H}}^ + }\]

Reduction: \[\mathop {{\rm{8Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + 40}}{{\rm{e}}^ - } + 64{{\rm{H}}^ + } \to 8\mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right) + 32{{\rm{H}}_{\rm{2}}}{\rm{O}}\]

Addition of both reactions and its simplification

When we add both reactions together, we get the equation given below:

\[5\mathop {{{\rm{H}}_{\rm{2}}}}\limits^{ + 1} \mathop {\rm{S}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 20{{\rm{H}}_{\rm{2}}}{\rm{O + }}\mathop {{\rm{8Mn}}}\limits^{ + 7} \mathop {{\rm{O}}_{\rm{4}}^{\rm{ - }}}\limits^{ - 2} \left( {{\rm{aq}}} \right){\rm{ + 40}}{{\rm{e}}^ - } + 64{{\rm{H}}^ + } \to 5\mathop {\rm{S}}\limits^{ + 6} \mathop {{\rm{O}}_{\rm{4}}^{{\rm{2 - }}}}\limits^{ - 2} \left( {{\rm{aq}}} \right) + 40{{\rm{e}}^ - } + 50{{\rm{H}}^ + } + 8\mathop {{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}}\limits^{ + 2} \left( {{\rm{aq}}} \right) + 32{{\rm{H}}_{\rm{2}}}{\rm{O}}\]

On simplification, we get

\[\mathop {\rm{8MnO}}_{\rm{4}}^{\rm{ - }}\left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right) + 14{{\rm{H}}^ + }\left( {{\rm{aq}}} \right) \to 8{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right){\rm{ + 5SO}}_{\rm{4}}^{{\rm{2 - }}}\left( {{\rm{aq}}} \right) + 12{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)\]

Therefore, the balanced equation is

\[\mathop {\rm{8MnO}}_{\rm{4}}^{\rm{ - }}\left( {{\rm{aq}}} \right){\rm{ + 5}}{{\rm{H}}_{\rm{2}}}{\rm{S}}\left( {{\rm{aq}}} \right) + 14{{\rm{H}}^ + }\left( {{\rm{aq}}} \right) \to 8{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\left( {{\rm{aq}}} \right){\rm{ + 5SO}}_{\rm{4}}^{{\rm{2 - }}}\left( {{\rm{aq}}} \right) + 12{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{l}} \right)\]