How to tell if its being reduced or oxidized

[aphung1E]

i get confused on how to figure out which one is being reduced and oxidized in redox rxns. Does anyone know how I can tell the difference for example MnO4^- to Mn^2+. Since it is going from a negative to positive charge in the end wouldn't it mean that it is being oxidized?

[Selena Yu 1H]

No because the charge of Mn in MnO4- is +7 and so Mn is going from a +7 charge to a +2 charge which means that it is gaining electrons for the charge to be less positive. So Mn is being reduced. You have to look at the charges individually if it is a compound and not the overall charge of the compound.

[Katie Kyan 2K]

Take a look at the charge of the Mn in each. For MnO4- we know that the charge of Mn is 7+ whereas the charge of Mn2+ is simply 2+. If Mn is going from 7+ to 2+ where there is a decrease in its oxidation number, we know that electrons must be gained for this decrease to occur. Thus, a gain in electrons tells us that Mn is being reduced.

[Juana Abana 1G]

i get confused on how to figure out which one is being reduced and oxidized in redox rxns. Does anyone know how I can tell the difference for example MnO4^- to Mn^2+. Since it is going from a negative to positive charge in the end wouldn't it mean that it is being oxidized?

Yes, I also get confused on how to identify which one is being reduced and oxidized in redox reactions.

[Anthony Hatashita 4H]

You have to look at the charge of Mn, not the entire compound. In this case, Mn goes from +7 to +2 and is being reduced.

[andrewcj 2C]

When looking at redox reactions or electrochemical cells, I find it helpful to label each element with its own charge, as it clears up any confusion that may arise, as you stated. If you label Mn with +7 in MnO₄^-, it becomes clear that it is being reduced, as it goes from a +7 charge to a +2 charge.

[Jacob Motawakel]

To find the charge of Mn, you would first find the charge of oxygen in that compound, which is -2, and then multiply it by the number of oxygen atoms. So for MnO4- the oxygen part of the compound is -2*4, which is -8. Finally, to find the charge of Mn, you set the charge of Mn plus the charge of O4 equal to the charge of the compound, which is -1. So for MnO4-, Mn + -8 = -1, which gives us the charge for Mn as +7. And in Mn2+, the charge of Mn is just +2. So based on these findings, going from +7 to +2 is reduced.

[Angela Prince 1J]

That makes sense but when I tried it for 6K.3 on the hw, I got confused again. How would you write the half reactions for part a?

Cl2 + S2O3^2- goes to Cl^- + SO4^2-

[Ying Yan 1F]

Another way you can find if the reaction is undergoing oxidation or reduction is through just balancing the half reactions. If you see that electrons needs to be added to the reactant side, then that half reaction is being reduced (Gain electron reduction); if you see that electrons needs to be added to the product side of the other half reaction, then that half reaction is being oxidized (Lose electron oxidation). I hope this helped!

[Madeline Phan 1E]

Mn goes from +7 to +2 which means it gained electrons and diminishes its positive charge, indicating that it is reduced.

[Abby Soriano 1J]

It's important to look at the charge of the element you're working with rather than that of the entire compound when deciding if it is being oxidized or reduced. So although MnO4- has a 1- charge on the outside, that refers to the charge of the entire compound. Since oxygens always have a 2- charge (unless they are in the form O2 (g) in which their charge is 0), the oxygens have an overall 8- charge. To make it so that the entire compound does have a 1- charge, the Mn must have a 7+ charge. And thus, Mn is actually being reduced (not oxidized) from 7+ to 2+.

[Diana Andrade_4F]

You have to look at Mn only, and we see that it goes from 7+ to 2+, thus being reduced

[Shutong Hou_1F]

When the oxidation number of an element in a reactant increases when it becomes a product, it is oxidized; when the oxidation number of an element in a reactant decreases when it becomes a product, it is reduced. The previous sentence holds because being reduced means gaining electrons while being oxidized means losing electrons.

[Emily Chirila 2E]

I find it easier to remember "OIL RIG": oxidation is losing electrons, reduction is gaining electrons
Oxidation half rxns are losing electrons so the e- are a product, reduction half rxns are gaining electrons so e- are a reactant

[Omar Selim 1D]

You have to differentiate between the ions you are looking at. One of the elements gains electrons while the other loses. the mnemonic to help you remember the definitions of oxidation and reduction if that was your confusion is OIL RIG: Oxidation Is Losing; Reduction Is Gaining.

[Radha Patel 4I]

I think it is best to look at the oxidation numbers of the atoms that are being reduced or oxidized so you can see what is happening to the reaction. In this case the Mn in MnO4^- is +7 because the x+4(-2)=-1 where -1 is the overall charge of the molecule and 4 times -2 is the oxidation number of oxygen. We find that x= +7 which is the oxidation number of Mn. If we do this to see the oxidation number in Mn^2+ , we see that Mn's oxidation number is +2. Hence, it is being reduced as the oxidation numbers go from +7 to +2, showing that the Mn is gaining electrons.