How could you balance this redox equation involving Cr2O7 2- and H2O

Hi I know that balancing redox equations you first work out the separate half equations first and then combine them however I'm not sure how to balance this one? 


$$\ce{Cr\_2O\_7^{2-} + H^+(aq) + e^- \rightarrow Cr^{3+}(aq) + H\_2O(l)}$$


I've arried at $\ce{Cr\_2O\_7^{2-} + 3e^- \rightarrow Cr^{3+}}$ however for $\ce{H^+ \rightarrow H\_2O}$ I'm not sure what to add because the oxidation for state of hydrogen is the same?




To balance a redox reaction (in acid), 


1. First balance the atom being reduced or oxidized:


$$\ce{Cr\_2O\_7^{2-} \rightarrow 2Cr^{3+}}$$


2. Then balance the oxygens by adding $\ce{H\_2O}$:


$$\ce{Cr\_2O\_7^{2-} \rightarrow 2Cr^{3+} + 7 H\_2O}$$


3. Then balance the hydrogen by adding $\ce{H^+}$:


$$\ce{Cr\_2O\_7^{2-} + 14 H^+ \rightarrow 2Cr^{3+} + 7 H\_2O}$$


4. Finally, balance the total charge by adding $\ce{e^-}$:


$$\ce{Cr\_2O\_7^{2-} + 14 H^+ + 6e^- \rightarrow 2Cr^{3+} + 7 H\_2O}$$


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