When balancing a Redox reaction, Im confused on balancing the charges , See Replies for equation :)

Question

anonymous · Answer

$$CrO _{7}^{2-} + 14H ^{+} + 2I ^{-} \rightarrow Cr + H _{2} O + I _{2}$$
Which then gets balanced to $$CrO _{7} ^{2-} + 14H ^{+} + 12I ^{-} \rightarrow Cr + 7H _{2}O + 6I _{2} $$   But as for balancing the charges, Has it balanced itself already? 14H+ is +14, take away 2 from the Chromium, and 12 form the Iodine, or am I wrong in assuming that the Oxidation number is not dependant on the Amount of the substance ?

anonymous · Answer

if you need electrons included its +6 for Cr and +2 for Iodine, to which that changes it ?

anonymous · Answer

and there be 12 electrons as it is the Lowest common multiple ?

aaronq · Answer

do you mean $$Cr _{2}O _{7}^{2-}$$

the oxidation on the Cr is +6