# Chromate Confusion **needs help**

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(I am a 9th grade student. ) Acidified potassium dichromate (VI) is used to test for the presence of reducing agents. How do I derive the 6 electrons in the following ionic equation during which the test of a reducing agent like sulphur dioxide is carried out?

Cr2 O7 2- (aq) + 14H+ (aq) +6e- ==> 2Cr3+ (aq) + 7H2O (l)

How is dichromate (VI) ion reduced to chromium (III) ion?

Cr2 O7 2- ---> 2Cr 3+

How can reduction result in a negative charge changed to a positive charge?

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Cr2O7 is a complex which is negativly charged but the chrome ion itself is positivley charged . If you count the oxidsining value you will know why the complex is negativley charged.

Cr(+6) *2 = +12

O(-2) *7 = -14 => if you count the charges you will get the -2 , although chrome is still positivley charged.

(hope i got your question right)

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You are talking about acidified solutions of dichromate. Then you have the following unbalanced equation:

The dichromate requires acid to be present in order to oxidize, so you'll have H(+) at the left. The dichromate is an oxidizer, so the half reaction has electrons at the left.

The reaction products are water and Cr(3+) ions.

Cr2O7(2-) + H(+) + e --> Cr(3+) + H2O

Now, how do we come to the precise reaction equation? It is just a matter of good bookkeeping.

At the right, we have Cr(3+) ions. At the left we have two Cr-atoms in the dichromate ion, at the right we need two Cr(3+) ions.

Now, we are left with the oxygens from the Cr2O7(2-). There are seven of them, so we need 14 H(+)'s in order to compensate for all of them. The result is 7H2O at the right.

Now, the equation is balanced for Cr, H and O. We still have to balance for charge.

At the left, we have a charge, equal to +12 for the Cr2O7(2-) ion and the 14 H(+) ions. At the right we have a charge, equal to +6, due to the two Cr(3+) ions. At both sides we need the same charge, so we need to add 6 electrons at the left (charge of -6) in order to make the equation correct for charge as well.

Try to do this as an exercise for the reaction of pervanadyl ion to vanadium (III) ion. If you can derive the correct half-equation for this, then you understand the concept.

VO2(+) + H(+) + e --> V(3+) + H2O

Now, the question, why a negatively charged ion can change to positively charged ions due to reduction.

This answer is not that difficult. The charge of the ions involved does not really say anything about the possibility of reaction. Look at the left-hand side of the equation. A lot of H(+) ions go in as well. The electrons are not really present as free entities, they are just introduced for bookkeeping purposes. In reality, they are passed immediately from reductor to oxidizer and there are multiple mechanisms. Google for "outer-sphere" and "inner-sphere" redox reaction mechanism. With these concepts, a lot more will be explained on the mechanisms and I think things will become more clear to you.

Here follows a final example, which may be interesting to try. It also goes from the yellow positive VO2(+) ion to the bright blue positive VO(2+) ion:

VO2(+) + H(+) + e --> VO(2+) + H2O

Try to balance this one. If you can balance this correctly, you really understand this . I've done this reaction many times and it is very neat to watch it, going from deep yellow through all kinds of green until a final bright blue color is obtained.

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