Finding the Oxidation number
About oxidation states, I find it easier to follow the extended definition, as it is explained in this question: Electronegativity Considerations in Assigning Oxidation States

Look at the first compound $\ce{Cr(OH)3- }$. What is the most electronegative element? Assume it gets all the electrons to fill the octet.


   Oxygen, it shall have the oxidation state $-2$  


What about the other atoms?


   Hydrogen can only receive or lose an electron, in this case it is bonded to the much more electronegative element oxygen, it shall have the oxidation state of $+1$
Chromium gets the oxidation state of $+2$, since $\ce{O}(3\cdot -2) + \ce{H}(3\cdot +1) -\ce{e-}(1\cdot-1)=-Ox(\ce{Cr}) = +2$


Look at the hydrogen peroxide ion, apply what you know about Hydrogen to calculate the oxidation state of oxygen.


   Hydrogen gets $+1$ (see above)
 Charge has to remain, i.e. $-1$
 Oxygen gets $-1$, since $\ce{H}(1\cdot +1) -\ce{e-}(1\cdot-1)=-\frac12Ox(\ce{O})=-1$


Now the hydroxyl ion is straightforward.


   Hydrogen gets $+1$ (see above)
 Oxygen gets $-2$


Look at dichromate and apply the above.


   Oxygen gets $-2$, most electronegative
 Chromium gets $+6$, since $\ce{O}(7\cdot -2) -\ce{e-}(2\cdot-1)=-\frac12Ox(\ce{Cr}) = +6$


In total that will give you:


   $\ce{\overset{{+II}}{Cr}~(~\overset{{-II}}{O}~\overset{{+I}}{H}~)3{}^{-} + \overset{{+I}}{H}~\overset{{-I}}{O}_2{}^{-} -&gt; \overset{{+IV}}{Cr}_2~\overset{{-II}}{O}_7{}^{2-} + {}^{-}\overset{{-II}}{O}~\overset{{+I}}{H} }$