I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O +10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ----> 2HIO3+10NO2 +4H2O Hope this helps, Steve Solution I2 + 6H2O ---> 2HIO3 +10H+ +10e- HNO3 + 1H+ +1e- --->NO2 + H2O In order to balance electrons, we have to multiply the reductionhalf-reaction by 10. This gives us: I2 +6H2O +10HNO3 +10H++ 10e- ----> 2HIO3 + 10H+10NO2 +10H2O + 10e- We can cancel things that are on the left and right side, likewater and protons. This gives us: I2 +10HNO3 ----> 2HIO3+10NO2 +4H2O Hope this helps, Steve.