Figuring out which one is oxidation and reductionand 5e-

[Cristian Cortes 1L]

Lecture #17 shows an equation and break down of which one is oxidation and reduction. From first glance and even time trying to see how or why, I am still having trouble, can anyone help? Also, where does the 5e- come from and move around.

[Ryan Hoang 1D]

Yeah, so think of OIL RIG when you first do oxidation and reduction. OIL RIG stands for: oxygen is loss, and reduction is gain (in electrons). So just try seeing how the electrons interact within the context of the equation to produce this oxidation/reduction.

[Adrienne Chan 1G]

Hi Cristian, are you talking about the permanganate and Fe 2+ in water equation? Lavelle breaks down the redox equation into its two components: the oxidation and reduction components. In oxidation: 5 Fe²⁺ -> 5 Fe³⁺ + 5e- we see that each Fe atom is long an electron, for a total loss of five electrons total. Then, we proceed to look at the reduction half-reaction. Due to the different charges present, in order to get from our reactants to our products, we need some more electrons. In MnO4^-, each O has a -2 charge, so that's a -8 charge total from oxygen. To achieve the -1 charge listed, Mn must be +7. Then, looking at the products side, our Mn now has a charge of +2. Mn has been reduced from +7 to +2 charge, meaning that 5e- have been added. Hope this helps, let me know if you need any further clarification!

[Lorena_Morales_1K]

I believe the 5e- is for trying to cancel out both sides of the reaction.