Mno4- (aq) + Fe+2(aq) -----> Fe+3(aq) + Mn+2(aq) Solution This is a half rxn. So

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Explanation / Answer

This is a half rxn. So you separate the two eqns like so. MnO4- --> Mn 2+ Fe 2+ --> Fe3+ to balance oxygen, you have to add 4 h2o molecules to the product side of the first eqn. MnO4- --> Mn 2+ + 4 H2O now you have 8 H+ on the product side, so you add 8 H+ ions to the reactant side 8H+ + MnO4- --> Mn 2+ + 4 H2O Fe 2+ --> Fe 3+ now you have to make sure that the charges are equal (electrons are never lost or gained.) so you have 8 H+ and one 1- charge, thus 7+ on reactant side. on the product side you have just one 2+. so if you add 6 electrons to the reactant side, your charges will be equal. in the second eqn, you only need to add one to the product side to get them to be equal. 6 electrons + 8 H+ + MnO4- --> Mn 2+ + 4 H2O Fe 2+ --> Fe 3+ + 1 electron now you must multiply the bottom eqn by 6 to get equal number of electrons in each rxn. 6Fe 2+ --> 6Fe 3+ + 6 electrons 6 electrons + 8 H+ + MnO4- --> Mn 2+ + 4 H2O now im assuming this is in an acidic soln. so the next step is to combine the two rxns. 6Fe 2+ + 8H+ + MnO4- --> Mn2+ + 4 H2O

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