14.3
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Re: 14.3
You add 5e- because you need to balance out the charge. Since the left side is Mn7+ and the right side is Mn2+, you need 5 electrons to make the charges on both sides equal. You know it is Mn7+ because oxygen has a 2- charge, so O4 should have an 8- charge. However, MnO4 only has a 1- charge overall, so you need Mn7+.
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Re: 14.3
To balance the half reaction MnO4- ---> Mn2+ we must first balance all atoms (except oxygen and hydrogen) which is already done since there is one Mn on the reactants side and one Mn on the products side. Next, we have to balance the oxygens by adding H2O. Since there are 4 oxygens on the left side and none on the right side, we add 4 H2O to the right (MnO4- ---> Mn2+ + 4H2O). Next, we have to balance the hydrogens by adding H+. There are 8 hydrogens on the right side and none on the left side, so we have to add 8H+ to the right (MnO4- + 8H+ ---> Mn2+ + 4H2O). Next, we have to balance the charge. Currently, on the left side we have a +7 charge and on the left we have a +2 charge. In order to balance the charge, we must add 5 electrons to the left side, in order to make both sides have a charge of +2 (MnO4- + 8H+ +5e- ---> Mn2+ + 4H2O).
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