balancing redox equations

Balance the half-equation for the reduction of dichromate(VI) ions.
Cr2O72– + ......H+ + ..... e− → ..... Cr3+ + ..... H2O

I got:
Cr2O72– + 7H+ + 4e− → 2Cr3+ + 3.5H2O

which is completely wrong...
the correct answer is:
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O

how would you approach this question, and how can i get that answer?

hopefully this should be my last chem question today..
many thanks

Cr2O7 2- ==> Cr3+ balancing the atoms gives

Cr2O7 2- ==> 2Cr3+ now add waters to the RHS to balance oxygens

Cr2O7 2- ==> 7H2O + 2Cr3+ and add hydrogens to LHS to balance 7H2O

14H+ + Cr2O7 2- ==> 7H2O + 2Cr3+ and then add the electrons, we have a 6+ charge on the RHS and a 12+ charge on the LHS so we need to take six off the LHS so add 6 electrons

6e- + 14H+ + Cr2O7 2- ==> 7H2O + 2Cr3+

If reduction has happened why did this equation start with cr3+?
Isn't that only if it's lost electrons? ?

I know this is 8 years late, but it will help me anyways

Its all about balancing.

First you have to balance the Cr
Because it is Cr2 on the LHS it will be 2Cr on the RHS
this means the charge will be 2- to 6+
As there are 7 oxygen's, the other side must have 7H2O in order to also have 7 oxygens with a neutral charge as 2 H+ balances 1 O-2.
Then that means we have 14 H+ which needs to be added onto the other side.
Finally balance the electrons.
6+ charge on the RHS
12+ charge on the LHS
that means the LHS needs to add 6 electrons which have a negative charge. +6e- to balance the everything out.

I am doing my a-levels now, so plz dont hate that this is a late response.