What is the coefficient...in the redox reaction..?HELP

Given the following unbalanced redox reaction:

Fe^2+ + Cr2O7^2- --> Fe^3+ + Cr^3+

What is the coefficient of Fe^3+ in the balanced equation in acidic solution?

A. 1
B. 2
C. 3
D. 6

So the answer is D....but I have no idea how the answer is D... I got C : /
Explain please!

Fe^2+ is being oxidised to Fe^3+

oxidation half equation
Fe^2+ ------------> Fe^3+ + e

Now for the tricky part

Cr2O7^2- is being reduced to Cr^3+
Cr2O7^2- ----> Cr^3+

first balance the Cr atoms
Cr2O7^2- ----> 2Cr^3+

balance the O atoms by adding H2O to the side with no O
Cr2O7^2- ----> 2Cr^3+ + 7H2O

balance the H you added into the equation in H2O by adding H+ (acidic solution remember) to the other side
Cr2O7^2- + 14H+ ----> 2Cr^3+ + 7H2O

now, balance the charge to fiigure out the number of electrons involved
total charge on Right hand side = 2 x Cr^3+ = 6+
LHS = 1 x CrO7^2- + 14xH+ = 12+
Now add enough electrons to the most positive side to balance the charge... ie add 6 e to the LHS
LHS = (1 x CrO7^2-) + (14 x H+) + 6e = 6+ = charge on RHS

balanced reduction half equation
Cr2O7^2- + 14H+ +6e ----> 2Cr^3+ + 7H2O

Now, the electrons need to balance as well, but the reduction half equation transferrs 6 electrons, and the oxidation half equation only transferrs 1.. So you need 6 x oxidation half equations to match the number of electrons transferred in the reduction half equation.

6 x (Fe^2+ ----> Fe^3+ + e)
gives
6Fe^2+ ------- 6Fe^3+ + 6e
Cr2O7^2- + 14H+ +6e ----> 2Cr^3+ + 7H2O
---------------------------------------… add them together

6Fe^2+ + Cr2O7^2- + 14H+ ----> 6Fe^3+ + 2Cr^3+ + 7H2O
This is the balanced equation

you can see the coefficient in front of the Fe^3+ is 6
= D

Separate into two half-reactions:

Oxidation

Fe(2+) ----> Fe(3+) + e-

Reduction

2Cr(+6) + 6e- ---> 2Cr(3+)

Multiply the first by 6 to balance charge

6Fe(2+) ----> 6Fe(3+) + 6e-

add the half reactions

6Fe(2+) + 2Cr(+6) + 6e ----> 6Fe(3+) + 6e- + 2Cr(3+)

cancel the e-

6Fe(2+) + 2Cr(+6) ----> 6Fe(3+)- + 2Cr(3+)