Posted by Jen on Friday, August 8, 2008 at 6:58pm.
I looked up two half reactions form a "Standard Reduction Potentials" table. The 2nd half reaction occurred in reverse as an oxidation. I doubled this reaction so that the number of electrons gained and lost are equal. Then I combined the two half reactions into an overall reaction. This is needed to see how many electrons are transferred from H2O2 to Fe^+2. [Unless you have studied this material in your chemistry class it may be confusing.]
The two half reactions are:
H2O2 + 2H+(aq) + 2e- --> 2H2O (reduction)
2Fe^+2(aq) --> 2Fe^+3(aq) + 2e- (oxidation)
The overall reaction is:
H2O2 + 2H+(aq) + 2e- + 2Fe^+2(aq) --> 2H2O + 2Fe^+3(aq) + 2e-
or
H2O2 + 2H+(aq) + 2Fe^+2(aq) --> 2H2O + 2Fe^+3(aq)
The above show 2 electrons going to a molecule of H2O2 from two Fe^+2 ions which oxidize to form two Fe^+3 ions.
Balance the equation: MnO4¡V + H+ + Fe2+ Mn2+ + H2O + Fe3
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