Redox Titrations Half & Full Equations

I was presented with the following problem and I am finding it difficult to present the half equations for the answers. 


[![Redox titration](https://i.stack.imgur.com/nmvpl.jpg)](https://i.stack.imgur.com/nmvpl.jpg)


I managed to produce the half-equation for the Iodine redox. 


$\ce{2I- -> I2 + e-}$


I then proceeded to the $\ce{CrO4^{2-}}$ equation.


$\ce{CrO4^{2-} + 3e- -> Cr3+}$


This is because the oxidation number of the Cr in $\ce{CrO4^{2-}}$ is $+6$ and it is reduced to $+3$


However, the answer is given as:


$\ce{CrO4^{2-} + 4H2O + 3e- → Cr3+ + 8OH-}$


I am struggling to get from the equation I made to the one presented in the answer for the $\ce{CrO4^{2-}}$ redox. 


Can someone explain why there are $\ce{4H2O}$ on the left hand side and $\ce{8OH-}$ on the left hand side of the equation?


Please help!
Many thanks. 


