class example

Moderators: Chem_Mod, Chem_Admin

805329408
Posts: 49
Joined: Wed Nov 20, 2019 12:21 am

class example

Postby 805329408 » Sun Feb 16, 2020 12:56 pm

Just to clarify, in class Lavelle used the example:
8H+ + MnO4- + 5e- = Mn+2 + 4H2O where he said Mn is reduced from +7 to +2

Mn is initially +7 because O4 is is -8 and in order for the overall molecule to have a negative charge, Mn has to be +7 right? Thanks in advance!

505316964
Posts: 95
Joined: Thu Jul 11, 2019 12:17 am

Re: class example

Postby 505316964 » Sun Feb 16, 2020 1:00 pm

That's correct. You know Mn has to be +7 in MnO4- because O4 has an overall -8 charge and the molecule has a -1 charge.

It is then being reduced to Mn 2+, as it's oxidation number decreases.

Sidharth D 1E
Posts: 98
Joined: Sat Aug 24, 2019 12:17 am

Re: class example

Postby Sidharth D 1E » Sun Feb 16, 2020 7:23 pm

As a good rule of thumb, remember that oxygen in a compound other than O2 will have an oxidation number of -2 and hydrogen in a compound other than H2 will have an oxidation number of +1.


Return to “Balancing Redox Reactions”

Who is online

Users browsing this forum: No registered users and 4 guests