Charge balancing with half-reactions

Step 2 Balance each half-reaction with respect to atoms and then charges, ox Fe2+ - Fe3+ + e ... 

Step 2 Balance each half-reaction with respect to atoms and then charges. 

Balance the charge of the half-reactions with respect to each other by multiplying the reactions so that the total number of electrons is the same in each half-reaction. 

Balance the charge of the half-reactions with respect to each other. 

To balance a reaction involving oxidation and reduction, we must first identify which element is oxidized and which is reduced. We then break the net reaction into two imaginary half-reactions, one of which involves only oxidation and the other only reduction. Although free electrons never appear in a balanced net reaction, they do appear in balanced half-reactions. If we are dealing with aqueous solutions, we proceed to balance each half-reaction, using H20 and either H+ or OH-, as necessary. A reaction is balanced when the number of atoms of each element is the same on both sides and the net charge is the same on both sides. ... 

Now, balance both half-reactions for charge by adding electrons (e-) to the side with the greater positive charge. The oxidation half-reaction must have 2e added to the product side, and the reduction half-reaction must have 6 e- added to... 

Step 5. Balance each half-reaction for charge by adding electrons to the side with greater positive charge, and then multiply by suitable factors to make the electron count the same in both half-reactions. 

Step 5. Now balance each half-reaction with respect to charge by adding electrons to the side with excess positive charge. The iron half-reaction has a charge of -1-2 on the left, and +3 on the right, so we add one electron to the right-hand side ... 

Balance each half-reaction for charge by adding electrons. For the alnminnm half-reaclion, we need to add three electrons on the product side to balance the charge. With a loss of electrons, this is an oxidation. 

Both half-reactions are balanced with respect to charge by adding electrons as needed. 

Step 7 In the final step, you re going to combine the two half-reactions. After you have done this, you want to double check to make sure that all charges are balanced and that the number of atoms is balanced. Once you have done that, remove the electrons, and you re left with the balanced net ionic equation ... 

In the right compartment the active components in their standard states are 1.0 M Mn04, 1.0 M H+, and 1.0 M Mn2+, with appropriate unreacting ions (often called counter ions) to balance the charge. The half-reaction in this compartment is... 

The nest step is to balance each of the half-reactions in order to make sure that the charges fit. The balancing is once again done with electrons. For the oxidation reaction we have ... 

The two half-reactions are now balanced with respect to atoms, but not with respect to charge. 

The situation at this stage is that both equations are completely balanced with respect to atoms and charge. It is now necessary to recognize that there must be equal exchanges of electrons in a redox reaction numbers of electrons consumed must be equal to the numbers of electrons produced. The oxidation of 1 mol of Fe " produces 1 mol of electrons. The reduction of 1 mol of Mn04 requires 5 mol of electrons. This is accounted for by multiplying the iron half-reaction by 5 ... 

In order for the electron exchange to be equal, we must multiply the oxidation half-reaction by 2. The equations are now balanced with respect to atoms and charge, and may now be added together, canceling all terms which appear on both sides ... 

Balance the half-equations with respect to charge. In most biogeochemical reactions, ionic balance can be achieved by adding either H+ or 0H to the left or right side of equations. Mostly, is added to balance hydrogens in the equation. 

Both half-reactions are already balanced for O, so we balance charge with electrons... 

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