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Balancing redox equations

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Equation balancing & stoichiometry lectures » balancing redox reactions

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Redox reactions - in which charge is transferred between reagents - can be balanced by inspection, although it can be extremely difficult. But let's start with a very simple example:

Cu + Fe3+ → Cu2+ + Fe2+

At first sight the equation seems to be already balanced, but if we check charges - it is not. There is +3 charge on the left, and +4 charge on the right, so we have to do something about it. Using the inspection method we should select the most complicated molecule first. Let's say it is Fe3+:

Cu + 1Fe3+ → Cu2+ + Fe2+

If so, we can already add 1 in front of Fe2+:

Cu + 1Fe3+ → Cu2+ + 1Fe2+

Trying to balance copper will lead us nowhere (atoms are balanced, but charges will be left unbalanced as they were from the beginning), so let's look at the charge - it can be balanced just like atoms. We have +3 on the left and +2 on the right - so we need additional +1 on the right side. Do you remember that we can use fractions at this stage? Half Cu2+ will do:

Cu + Fe3+1/2Cu2+ + 1Fe2+

What is wrong now is the left side of equation - too much copper at the moment. Once again, half will do:

1/2Cu + Fe3+1/2Cu2+ + 1Fe2+

Final touch - multiply everything by 2 to remove fractions:

Cu + 2Fe3+ → Cu2+ + 2Fe2+

Atoms - balanced (one of copper and two of iron on both sides). Charge - balanced (+6 on both sides). We have just balanced the redox equation by using the inspection method. But in the case of more difficult ones - like this for example:

FeSO4 + KMnO4 +H2SO4 → Fe2(SO4)3 + MnSO4 + H2O

this approach is a waste of time. Much easier and faster ways of balancing redox reactions are oxidation numbers method and half reaction method.


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