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.

CH3OH + Cr2O72- yields CH2O + Cr 3+

Question

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.

CH3OH + Cr2O72- yields CH2O + Cr 3+

anonymous · Answer

I really need help with this. It's my last question and I have an exam on it tomorrow

anonymous · Answer

@tester97

tester97 · Answer

im still looking sorry for the wait

anonymous · Answer

Oh okay. No problem

tester97 · Answer

Sorry i cant find it :\

tester97 · Answer

it must have gotten deleted in my notes or something

anonymous · Answer

Ahh. It's okay

anonymous · Answer

CH3OH ---.>CH2O. 
First balance this reaction. In order to balance the H on either sides,   put in an H+ on the right hand side . 
CH3OH ---.>CH2O +H+ 
But now the right hand side has an excess charge of +1 , so put an e-(electron) on rhs to balance that.
CH3OH ---.>CH2O +H+ e-

anonymous · Answer

Is that it?

anonymous · Answer

noo ... wait

anonymous · Answer

Now do so for (Cr2O7)2- ion
(Cr2O7)2- --->2Cr3+ 
(Cr2O7)2- --->2Cr3+ +7H2O ....to balance oxygen atom on either side
Cr2O7)2- + 14H+ --->2Cr3+ +7H2O ... to balance H+
Cr2O72- + 14H+ + 6e- ---> 2Cr3+ 7H20

anonymous · Answer

Two half reactions are CH3OH ---.>CH2O +H+ e-  and Cr2O72- + 14H+ + 6e- ---> 2Cr3+ 7H20 ............ multiply the first equation by 6 (so that the electrons equal in both reaction) and add

anonymous · Answer

So multiply 6 by CH3OH ----> CH2O +H + e-

anonymous · Answer

yes

anonymous · Answer

So it would be 6CH18OH ---> 6CH12O +6H + 6e-

anonymous · Answer

yaa ... now add those 2 equation ,and that the electrons on either side  will cancel each other

anonymous · Answer

6CH18OH ---> 6CH12O +6H + 6e- + Cr2O72- + 14H+ + 6e- ---> 2Cr3+ 7H20 

6CH18O72-H +

that's as far as I got. I'm a little confused on how to write it

anonymous · Answer

I'm not exactly sure what I did either

anonymous · Answer

just write the 2 equations one below the other ...
6CH3OH                      --->  6CH2O + 6H+ + 6e-
(Cr2O7)2- + 14H+ 6e----> 2Cr3+ + 7H2O
----------------------------------------(+)
Cr2O72- + 6CH3OH + 8H+ --- > 6CH2O + 2Cr3+ + 7H2O 

got it ?

anonymous · Answer

I think so. So Cr2O72- + 6CH3OH + 8H+ --- > 6CH2O + 2Cr3+ + 7H2O would be the answer?

anonymous · Answer

yaa

anonymous · Answer

Omg thanks so much! You are a life saver!!

anonymous · Answer

thanks ... and Cr2O7 2- wil be oxidising agent cos its oxidation state changed from +6 to +3 .... and so CH3OH will be reducing agent