17 N 2 H 4 ( g ) ClO 3 - ( aq ) 17 N 2 H 4 ( g ) ClO 3 - ( aq ) 17 18 Solution F

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Find the oxidation numbers for each element in the four compoundsabove [two from each side]. Then compare what happens to thenitrogen atoms and what happens to the chlorine atoms. The N isoxidized, while the Cl is reduced, meaningN2H4 is the reducing agent [that which getsoxidized] and ClO3- is the oxidizing agent[that which gets reduced]. N2H2   + ClO3-   --> NO +Cl- [The N is oxidized from a -2 to a +2. The Cl is reduced from a +5to a -1.] Separate into half reactions Ox: N2H2    --> NO [Not balanced...]      N2H2 +8e- --> 2NO [To balance the nitrogensand the electrons--4e- for each N]       N2H2 + 8e- +2H2O --> 2NO [To balance the oxygens]       N2H2 + 8e- +2H2O --> 2NO +6H+ [To balance the hydrogens]        Red: ClO3-    -->Cl-        ClO3-    -->Cl- + 4e-           ClO3-    -->Cl- + 4e- + 3H2O [To balancethe O]        ClO3- + 6H+ -->Cl- + 4e- + 3H2O [To balancethe H] Now multiply through by two.        2ClO3- + 12H+ --> 2Cl- + 8e- +6H2O [To be able to cancel out thee- when we add the half reactions] Now add the half reactions N2H2 + 8e- +2H2O       --> 2NO +6H+ +                                2ClO3- + 12H+ -->                  2Cl- + 8e- +6H2O _________________________________________________________ N2H2 +2ClO3- + 6H+  --> 2NO + 2Cl- +4H2O     [The electronscanceled out completely. 6H+ canceled out, and2H2O canceled out.] You're not done because this is a basic solution. No H+exists in basic soln's. So you add OH- to eachside.. N2H2 + 2ClO3- +6H+   --> 2NO + 2Cl-+ 4H2O                         +6OH-                          +6OH-                      [H+ + OH- yields H2O] N2H2 + 2ClO3- +6H2O --> 2NO + 2Cl- +4H2O + 6OH- Now 4 of the H2Os will cancel, leaving you with thefinal product-- N2H2 + 2ClO3- +2H2O --> 2NO + 2Cl- +6OH-

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