Ineed ! i will give 25 points
balance the following redox equation, identifying the element oxidized and the element reduced. show all of the work used to solve the problem.
cr2o72- + i-yields cr3+ + i2

If I did this correctly the balanced equation would be:
14H⁺+Cr₂O₇²⁻+6I⁻→3I₂+2Cr³⁺+7H₂O

oxidation half: (iodide was oxidized)
2I⁻→I₂+2e⁻

reduction half: (chromium was reduced)
14H⁺+Cr₂O₇²⁻+6e⁻→2Cr³⁺+7H₂O

H⁺ comes from the solution.  It is in the final reaction since in redox reactions the oxygen is turned into water since it can't just go away.  I multiplied the oxidation half reaction by 3 in order for both half reactions to half the same number of electrons since equal numbers of electrons need to be lost and gained for the reaction to be balanced.

I hope this helps.  Let me know if anything is unclear.

The reaction will shift to the right