[pistaciolow]

question: Use the oxidation number method to balance the following. (Use the lowest possible coefficients.)
K2Cr2O7(aq) +HI(aq) -> KI(aq)+ CrI3(aq) +I2(aq) +H2O(aq)

I was using this to try to figure it out. http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

So far, I  have come up with Cr being reduced from +6 to +3 and I being oxidized from -1 to 0.

K2Cr2O7(aq) +HI(aq) -> KI(aq)+ CrI3(aq) +I2(aq) +H2O(aq)
+1 +6 -2       +1 -1       +1 -1    +3 -1       0         +1 -2

According to the webpage, I put 3 in front of K2Cr2O7 and 1 in front of HI to balance the electrons.

3K2Cr2O7(aq) + 1HI(aq) -> KI(aq)+ CrI3(aq) + I2(aq) + H2O(aq)

Am I in the right direction? I'm finding it difficult to balance the oxygens and hydrogens.

[orgoclear]

So far, I  have come up with Cr being reduced from +6 to +3 and I being oxidized from -1 to 0.

K2Cr2O7(aq) +HI(aq) -> KI(aq)+ CrI3(aq) +I2(aq) +H2O(aq)
+1 +6 -2       +1 -1       +1 -1    +3 -1       0         +1 -2

Till this point you were on the right track.

After that you have gone wrong.
Think like this.
How many Cr atoms are reduced in K2Cr2O7? What is the total no. of electrons involved in this?
So how many electrons should be involved in oxidation of I- ? How many electrons are involved in oxidation of 1 mole of I- to I?

Then the method should automatically strike you

[Borek]

Go net ionic - some of the iodine gets oxidized, some does not. You can't change coefficient before HI to balance electrons, as you will have not enough iodides on the right.