The answer should be around 2+, but I can\'t figure out how to get the moles of

Question

The answer should be around 2+, but I can't figure out how to get the moles of electrons. and I can't figure out how to put the hints together to find the oxidation of Vanadium once it's been reduced by the KMnO4.

The problem says perform the calculations necessary to determine the average oxidation state of vanadium for the reduction step (We know that the initial oxidation state is 5+ and the end oxidation state is 5+. But what oxidation state did we reach from the reduction/start of the titration?)

Hints: Use the concentration of V2O5 (0.0500M) and volume (0.025 L) to calculate the number of moles of V5+ present in the initial.

Use the concentration (0.03948 M) and volume (0.03486 L) of KMnO4 to calculate the moles of KMnO4 consumed in the ititration.

Find the moles of electrons gained by manganese and the moles of electrons lost by vanadium.

Explanation / Answer

Number of moles of KMnO4 = 0.03486 L x 0.03948 mol/L = 0.00137 moles

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O (5 electrons involved)

0.00137 moles x 5 = 0.00685

No. moles of V2O5 = 0.0500 mol/L x 0.025 L = 0.00125 moles

2e- + V2O5 ---------> 2VO2 + H2O

0.00125 moles x 2 = 0.0025

so, 0.0025 x ? = 0.00685

moles of electrons gained by manganese and the moles of electrons lost by vanadium = 0.00685/0.0025 = 2.74

approximately +3.

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