Question

1)A student has 10.00mL of a 0.500M solution of vanadium in a 250mL flask. He believe that the vanadium is in the +3 oxidation state. What volume of 0.100M KMnO4 solution should be required to titrate this solution to the endpoint?

2)It turns out the above solution requires 15.0mL of permanganate solution. Calculate the average oxidation state of vanadium in the solution. What does this tell you about the vanadium ions in solution? How can you explain your result?

Answer 1

Reduction reactions :

V³⁺ +e V²⁺ Eo(red)=-0.26V

VO²⁺ +2H+ +e V³⁺ +H2O Eo(red)=+0.36

MnO4^- +8H⁺ +5e Mn²⁺ +4H2O(l) Eo(red)=+1.51V

As Eo(red ) of both the above reaction ,is much less than Eo(red ) of MnO4- So, MnO4- will be reduced and vanadium will be oxidized in the titration.So,vanadium must be oxidized from V3+ to VO²⁺(+4)

redox reaction:

V³⁺ +H2OVO²⁺ +2H+ +e ]*5

MnO4^- +8H⁺ +5e Mn²⁺ +4H2O(l)

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5V³⁺ +MnO4^- +H2O5VO²⁺ +Mn²⁺+2H+

So, vanadium(+3 ) and MnO4- react in 5:1 molar ratio.

mol of vanadium in the flask=0.5 mol/L*0.01 L=0.005 mol

So mol of permanganate required to oxidize it =(1/5)*0.005=0.001 mol

Volume of permanganate required=mol/molarity =0.001 mol/0.1 mol/L=0.01L=0.01L*1000ml/L=10ml