Help with a question from 6ch05/01 june 2010 please

Make half eqns for each reaction

Hey! What I've done is written the ionic equation fully and it comes out at a 1:1 ratio? I have written as follows:

Cr2O7 2- + Fe2+ + 14H+ + 5e- GOING TO 2Cr3+ + Fe3+ + 7H2O

The full ionic equation must not contain electrons. One of the half equations must be multiplied by a number so that when they are combined, the electrons cancel. I.e. same number of electrons in the reactants and products

OMGGGG how did I miss that???! THANK YOUUUU so much, it's been a long day for me I'm so nervous for this exam as well

Haha no problem! Good luck with the upcoming exam.

When dichromate ions (VI) ions, Cr2O7 2- , react with iron (II) ions in acidic solution, the products are chromium (III) ions and iron (III) ions. In what ratio do the dichromate (VI) ions and the iron (II) ions react?

A 1:6
B 1:5
C 2:5
D 1:3

I initially thought this would be a simple question, maybe I'm missing something?
Thanks in advance!

These are the half equations:

Cr₂O₇^2- + 14H⁺ + 6e --> 2Cr³⁺ + 7H₂O

Fe²⁺ --> Fe³⁺ + 1e

Now can you see the ratio in which they have to react?