A method for determining the quantity of dissolved oxygen in nautral water reequ

Question

A method for determining the quantity of dissolved oxygen in nautral water reequires a seris of redox reactions. Balance the following chemical equations in that series under the conditions indicated: (If you could show a majority of the steps or the "most important" ones it would be appreciated)

A.) Mn+2 (aq) + O2 (g) ---> MnO2 (s) ....Basic solution

B.) MnO2 (s) + I- (aq) ---> Mn+2 (aq) + I2 (s)....Acidic solution

C.)  I2 (s) + S2O2-3 (aq) ----> I- (aq) + S4O2-6 (aq) ...neutral solution

Explanation / Answer

A) 2Mn(2+)(aq) + O2(g) + 4OH(1-)(aq) >> 2MnO2(s) + 2H2O(l)

B)

MnO2 + I- --> Mn 2+ + I2
Mn in MnO2 = 4+ and becomes 2+ as Mn 2+, gains electrons, reduced
I = 1- and becomes 0 as I2, loses 1e-/I- for a total of 2e-, oxidized
MnO2 + 2e- 2+ 4H+--> Mn2+ + 2H2O
2I- --> I2 + 2e-
MnO2 + 4H+ + 2I- --> Mn 2+ + 4H+ + I2 + 2H2O

C)

Write half-reactions and balance them individually:

Oxidation half-reaction:

2 (S2O3)2- ----> (S4O6)2-

All I did above was to balance out the non-hydrogen/non-oxygen atoms. Now, let's figure out the number of electrons being lost. The oxidation state of sulfur in (S2O3)2- is +2 while in (S4O6)2- is mixed, where you have 2 S(V) and 2 S(0). This means that you lose 6 electrons due to formation of 2 S(V) but you gain 4 electrons due to formation of 2 S(0). The net result is that you lose 2 electrons, thus the reaction is:

2 (S2O3)2- ----> (S4O6)2- + 2 e-

The oxygens are balanced on both sides, so this half-reaction is completely balanced.

Now for the reduction half-reaction:

I2 + 2 e- ---> 2 I-

No hydrogen or oxygen atoms to balance here so its done too. Since the number of electrons in both equations is the same, you can add them both together to get the overall balanced equation:

2 (S2O3)2- + I2 ----> (S4O6)2- + 2 I-

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