Is this redox reaction balanced?

How to determine the stoichiometry for the redox reaction between iodate ions and iodide ions in acid solution?

• I can't figure out how to do this question, I have been trying it for hours. I have already completed doing the lab, I just can not figure this part out determine the stoichiometry (ie the balanced equation) for the redox reaction between iodate ions and iodide ions in acid solution in other words determine the values of a,b,c,d and e in the equation below. *a*IO3-(aq) + *b* H+ + *c*I-(aq) ---> *d* I2 (aq) + *e* H20 (l) This is done by reacting a known amount of potassium iodate with excess acidic potassium iodide and determining the amount of molecular iodine produced (thereby first determining the ration of d:a) The latter will be found by titrating the iodine formed against a given solution containing 0.1000N sodium thiosulfate. The balanced equation for the reaction of sodium thiosulfate with molecular iodine is: 2S2O3-(aq) + I2(aq) --> 2I-(aq) + S4O6 (2-) (aq) note: molecular iodine I2(s) will react with iodide I-(aq) to form stable triiodide I3- (aq) in aqueous solution For the first part of the question... I used 6ml of potassium iodate(0.015M) and 3ml of potassium iodide(1.0M) mixed with 3ml of HCl(1.0M) For the second part of the question... I used on average 4.63 mL of sodium thiosulfate (0.1000N) for the titration of the above solution (I used starch indicator) ANY help at all would greatly be appreciated!!!!!

• Answer:

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