Question

1) Using the half-reaction method, balance the following oxidation-reduction reaction which occurs in acidic solution. Be sure to show each step in balancing both half-reactions and how electrons in each half reaction are balanced.

Mn2+ (aq) + IO4– (aq) → MnO4– (aq) + IO3– (aq)

2) Would you expect ions such as Cu2+ (light blue solutions) and Ni2+ (light green solutions) to interfere in the analysis of MnO4–? 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 × 103 L·mol–1·cm–1.

Answer 1

1)

Mn in Mn+2 has oxidation state of +2

Mn in MnO4- has oxidation state of +7

So, Mn in Mn+2 is oxidised to MnO4-

I in IO4- has oxidation state of +7

I in IO3- has oxidation state of +5

So, I in IO4- is reduced to IO3-

Reduction half cell

1 IO4- + 2e- --> 1 IO3-

Oxidation half cell

1 Mn+2 --> 1 MnO4- + 5e-

Balance number of electrons to be same in both half reactions

Reduction half cell

5 IO4- + 10e- --> 5 IO3-

Oxidation half cell

2 Mn+2 --> 2 MnO4- + 10e-

Lets combine both the reactions.

5 IO4- + 2 Mn+2 --> 5 IO3- + 2 MnO4-

Balance Oxygen by adding water

5 IO4- + 2 Mn+2 + 3 H2O --> 5 IO3- + 2 MnO4-

Balance Hydrogen by adding H+

5 IO4- + 2 Mn+2 + 3 H2O --> 5 IO3- + 2 MnO4- + 6 H+

This is balanced chemical equation in acidic medium

I am allowed to answer only 1 question at a time