[XxslbabesxX]

I am having a hard time balancing this redox problem:

CH₃OH + Cr₂O₇^-2 ---> CH₂O + Cr⁺³

First I found the oxidation #s then I separated them, balanced them, and added the necessary  H₂O and H⁺ so that it looked like this.

CH₃OH ---> CH₂O  + 2H⁺

Cr₂O₇^-2 + 14H⁺ ---> 2Cr⁺³ + 7H₂O

When I went to find the change of electrons for each I found that Carbon had a change of 2, but Cr did not change at all because of the coefficient  2 before Cr⁺³.  2Cr⁺³ and Cr₂O₇ ended up equaling to six. I have no clue what to do from here. I would really appreciate a steer in the right direction. Thanks!

[Borek]

Add electrons to both half reactions to balance charge. Rest should be easy.

[mike]

Cr₂O₇^2- (Cr is +6)

Cr³⁺ (Cr is +3)

[XxslbabesxX]

Cr₂O₇^2- (Cr is +6)

Cr³⁺ (Cr is +3)

But the coeffiecient before Cr³⁺ is 2 so that means the number of electrons is 6 which is the same as Cr₂O₇^2-....or those that not matter?

[Borek]

But the coeffiecient before Cr³⁺ is 2 so that means the number of electrons is 6 which is the same as Cr₂O₇^2-....or those that not matter?

There are 2*3+ and 2*6+, not 2*3+ and 6+ - you have two Cr on both sides of equation.

[mike]

one Cr⁶⁺ gains 3 electrons to form one Cr³⁺

two Cr⁶⁺ gains 6 electrons to form two Cr³⁺

etc...