Balance these reduction potential half equations first:
Ox. Cr+3(aq) + 3e => Cr(s) -0.74v
Red. Cr(OH)3(s) + 3e => Cr(s) + 3OH-1(aq) -1.30v
Ok first of all the reaction you write with Ox next to it is actually a reduction as written but that is fixed easily as :
Ox. Cr+3(aq) + 3e < = Cr(s) ...
This solution includes formula, calculations and explanation which the student can use to easily find K.