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You have to look at how many electrons are transferred in the balanced half-reactions. In other words, find the number of electrons in each balanced half-reaction. If they match, that is n. If they don't match, take the lowest common multiple, and that is n.
You can look at the overall reaction, and split it into 2 half reactions. For example, if one of the half reactions for the total equation is Cr3+ + 1e- -----> Cr2+, then here one electron is being transferred. However, if there are 2Cr3+ and Cr2+, then you know there are 2 electrons transferred because for ever one mol of each of these ions, one electron is transferred.
Yes, you need to work out both half reactions and balance them properly to get the same amount of electrons being gained and lost in both half reactions. The mutual number of electrons in both half reactions is the number for n.
When we get the two half reactions and balance the charges by adding electrons, we have to make the electrons added to each half reaction equal by multiplying each reaction by a number that will do so. Once the number of electrons added to both half reactions is equal, the number of electrons added to the half reactions is the value of n. In other words, the number of electrons that end up cancelling out of both reactions is the value of n.
You proceed through the redox reaction as normal and then count the number of electrons. Remember to multiply the half-reactions beforehand, so the moles of electrons are equal to each other. The number of electrons, then, is "n".
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