The manganese-containing products of the reduction half reactions for the three

Question

The manganese-containing products of the reduction half reactions for the three pH conditions are: The permanganate ion, MnO_4^-, is violet in color. Under each of the three pH conditions, the product of the reduction of the permanganate ion is a different color than the original permanganate ion (as shown in the table above). Therefore, the manganese-containing species acts as its own indicator during the oxidation-reduction titration. Write balanced half reactions for the reduction of MnO_4^- for each of the three pH conditions. For each of the three pH conditions, write a balanced net ionic equation for the overall redox process by combining each of the oxidation half reactions for the sodium bisulfite (given as equations 1, 2, and 3 above) with the appropriate reduction half reaction for the permanganate ion (question 1). That is, add the acidic oxidation half reaction (equation 1) to the acidic reduction half reaction (from question 1), etc. Recall that, in order to add half reactions, the number of electrons given off in the oxidation process must equal the number of electrons taken up in the reduction process.

Explanation / Answer

Reduction reaction

1. Half reactions,

acidic : MnO4^- + 8H+ + 5e- ---> Mn2+ + 4H2O

neutral : MnO4- + 4H+ + 3e- ---> MnO2 + 2H2O

basic : MnO2 + 2MnO4- + 2H2O --> 3MnO4^2- + 4H+

2. Balanced net ionic equation,

acidic : 2MnO4^- + 16H+ + 10e- ---> 2Mn2+ + 8H2O

            5H2SO3 + 5H2O --> 5HSO3- + 15H+ + 10e-

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2MnO4- + 5H2SO3 + 6H+ --> 2Mn2+ + 5HSO3- + 3H2O

neutral : 2MnO4- + 8H+ + 6e- ---> 2MnO2 + 4H2O

                  3HSO3- + 3H2O --> 3SO4^2- + 6e-

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2MnO4- + 3HSO3- + 8H+ --> 2MnO2 + 3SO4^2- + H2O

basic : MnO2 + 2MnO4- + 2H2O --> 3MnO4^2- + 4H+

               SO3^2- + 2OH- ---> SO4^2- + H2O + 2e-

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MnO2 + 2MnO4- + SO3^2- + H2O + 2OH- ---> 3MnO4^2- + SO4^2- + 4H+

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