1) Using the half-reaction method, balance the following oxidation-reduction rea
ID: 872568 • Letter: 1
Question
1) Using the half-reaction method, balance the following oxidation-reduction reaction which occurs in acidic solution. Be sure to show the each step in balancing both half-reactions and how electrons in each half reaction are balanced. Mn^2+ (aq) + IO4 (aq) right arrow MnO4 (aq) + IO3 (aq) 2) Would you expect ions such as Cu2 (light blue solutions) and Ni^2+ (light green solutions) to interfere in the analysis of Mn04? Why or why not? 3) Explain quantitatively how your value for the percent Mn in the unknown steel sample would be affected if your calibration curve produced a molar absorptivity (at 525 nm) of 1.50 X 10^3 L.mol^-1.cm^-1.Explanation / Answer
1) Mn+2 goes to Mn+7 this mean that is oxidating by the action of I-. This means that I goes from +7 to +5 and is reducting. Now Mn gains 5 electrons and I loses 2 electrons. The first half reaction:
H2O + Mn2+ = MnO4- + 5e- + H+ Now we balance the charges: 4H2O + Mn2+ = MnO4- + 5e- + 8H+
H+ + IO4- + 2e- = IO3-+ H2O Balance the charges: 2H+ + IO4- + 2e- = IO3-+ H2O
Balance the electrons: (4H2O + Mn2+ = MnO4- + 5e- + 8H+)x2 and (2H+ + IO4- + 2e- = IO3-+ H2O)x5
8H2O + 2Mn2+ = 2MnO4- + 10e- + 16H+
10H+ + 5IO4- + 10e- = 5IO3-+ 5H2O
Mn2++ 5IO4- + 3H2O = 2MnO4- + 5IO3- + 6H+
2) No, it will not interfere because Cu+2 is a good oxidating agent, and MnO4- cannot be more oxidated, so it will not interfere in the analisys.
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