Question

Please help me understand?!
Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. Show all of the work used to solve the problem.

CN- + MnO4- --> CNO- + MnO2

Answer 1

@shamil98 do you know this?

Answer 2

@bakonloverk can you help?

Answer 3

i guess not... :(

Answer 4

@thomaster ..... maybe.... ?

Answer 5

@whpalmer4 can you please help?

Answer 6

I once earned the title "Captain Redox" in my high school chem class for getting the highest score ever on the redox test, but now all I remember is that I did so :-)

Let me see if I can figure out how to do it again...

Answer 7

ok :D

Answer 8

Okay, O has an oxidation number of -2 except in a peroxide. MnO4- implies that Mn has an oxidation number of +7, agreed?

Answer 9

Yes, agreed

Answer 10

Also, do you agree that we need 2 CN- and 2 CNO- to make the equation balance?

Answer 11

based on the 4 O's in MnO4-?

Answer 12

This is what I got so far......
CN^- : N usually has an oxidation number of -3, so in order for the overall molecule to be -1 charge, C has to be +2.

MnO4^- : O usually has an oxidation number of -2, since there are 4 O atoms the overall oxidation number of O4 is 4(-2)=-8; in order for the overall molecule to be -1, Mn has to be +7.

CNO-: C+N+O=-1, O is -2 and N is -3, so C has to be +4 for this molecule to have a -1 overall.

MnO2: Mn+2(O)=0, since each O atom has a -2, Mn has to +4

Oxidizing agent is the molecule that gets reduced, so MnO4-, while reducing agent is the molecule that gets oxidized, so CN-

Answer 13

and yeah. I agree

Answer 14

Is there anything left to do?

Answer 15

how can we balance it? I know there's something about adding O and H2O or something like that....

Answer 16

2CN- + MnO4- --> 2CNO- + MnO2

2 C on each side
2 N on each side
1 Mn on each side
4 O on each side

I guess we need to balance charge as well...

Answer 17

yeah....

Answer 18

CN- + MnO4- --> CNO- + MnO2 + H2O (H2O added to use up stray O)
CN- + MnO4- + 2H+ --> CNO- + MnO2 + H2O (2H+ added to provide H2 to use up O)
CN- + MnO4- + 2H+ + e- -> CNO- + MnO2 + H2O (electron added to balance charge)

Answer 19

does that seem plausible to you?

Answer 20

yes! OHHHHHH ok... so you add the H+ and H2O to balance the charges.... right?

Answer 21

Well, the H2O I added to balance the oxygen (there was a stray O leftover), and that required some hydrogen on the left to make the water on the right, and that required an electron on the left to get the charge balanced once the atoms were balanced

Answer 22

Here's the web page I looked at to resurrect some of the dusty old knowledge :-)
http://www.chemteam.info/Redox/Redox.html

Answer 23

sorry about the wrong turn with 2 CN- at first!

Answer 24

OHHHHH.... now I see where I was going wrong.... lol I got my balancing messed up when i did it lol

Answer 25

Thanks! :D