Please Help me answer this question i really need to understand it for a fast approaching exam.
50 mL of a 0.0435 mol dm-3 solution of V2O5 was pipetted into a flask and then reduced with a zinc-mercury amalgam. The solution was later titrated by oxidation with standardized KMnO4 solution (0.0500 mol dm-3). The average titre was 34.97 mL of KMnO4.
Calculate the oxidation state of vanadium after this reduction. Show all steps in your calculation and write a balanced equation for the reaction with permanganate.
Here is my attempt:
meq KnO4- = 0.0500 mmol/mL x 5 meq/,,ol x 34.58 mL = 8.74 meq MnO4-
mmol V = 50.0 mL x 0.0435 x 2 V/V2O5 = 4.35 mmol V
8.74 meq / 4.35 mmol = 2 meq/mmol V thus V was reduced from +5 to +3
4 MnO4- + 10 V+3 + 9 H2O -----> 4 Mn++ + 5 V2O5 + 18 H.
thank you very much for your help.