TPR SW general chem passage 83 #3 (Electrochemistry)

[faith hopelove]

I get the math but I fail to understand how they came to the answer.

If 6 moles of permanganate anion are required to completely oxidize a sample of vanadium (II), how many moles of permanganate ion would be required to completely oxidize an equivalent molar sample of vanadium (IV)?
A. 2
B. 3
C. 12
D. 18

The answer is A. I chose B because I was thinking going from vanadium (II) to vanadium (IV) is 2 electron difference so 6/2=3. For those who have the book can you help me? How did you understand this problem. I feel that explanation in the back is not too clear.

---

Members don't see this ad.

 

[Cawolf]

I don't have the book, but I feel like some passage information must be missing in the question.

Did the passage say how far you are oxidizing it? I don't know to what state the vanadium ion is going? I am assuming if it oxidizes V (II) and V (IV) then it goes to V (V) or V (VI)?

If it goes to V (V) then the answer of 2 makes sense.

V (II) to V (V) requires a 3 mol e-/mol loss, and with 6 mol MnO4-; that means each mol of MnO4- is accepting 1 mol e- and our sample is 2 mol. (you could find that math to any factor of the same numbers)

V (IV) to V (V) requires a loss of 1 mol e-/mol - and with a sample of 2 mol, that is 2 mol e-. Since we determined each mol of MnO4- can accept 1 mol e-, we would need 2 moles - A.

This is just a hypothetical way to make the problem fit choice A.

TLDR; Must be missing info.