Chemistry redox question!.

How many moles of electrons are required to reduce one mole of Cr2O7^-2 to two moles of Cr^3+?

2 e-

3 e-

6 e-

8 e-

I have figured out the individual charges of the Cr2O7^-2 molecule. Cr has a charge of -6 and O has the charge of -2 and when you add them all up you get -2. But I don't know how to reduce it, can someone help? Thanks!

Write and balance the half-reaction.

Cr2O7^2- ==> 2Cr^3+ . . .Add 7H2O to the right side to balance O.
Cr2O7^2- ==> 2Cr^3+ + 7H2O . . .Add 14H+ to the left side to balance H.
Cr2O7^2- + 14H+ ==> 2Cr^3+ + 7H2O . . .Add 6e- to the left side to balance the charge.
Cr2O7^2- + 14H+ + 6e- ==> 2Cr^3+ + 7H2O

You can see that to go from Cr2O7^2- to 2Cr^3+ is a 6-electron change.

Cr₂O₇²⁻
==> let charge on Cr in Cr₂O₇²⁻ be = x
2x + 7(-2) = -2
2x - 14 = -2
2x = 12
x = 6

1 mole Cr₂O₇²⁻ has 2 moles of Cr⁶⁺ = 12+
2 moles Cr³⁺ = 6+

2Cr⁶⁺ + 6e⁻ = 2Cr³⁺
```````````````````````
6 moles of electrons are required