-1
$\begingroup$

How do we calculate the equivalent weight of a compound, in a case where certain fraction of an element of the compound is getting reduced, while the other fraction is unaffected (no change in oxidation state)? For example, consider the following reaction:

$$\ce{Zn + K4[Fe(CN)6] -> K2Zn3[Fe(CN)6]2 + K}$$

I would like to calculate the equivalent weight of potassium ferrocyanide, given its molecular weight $M$. Here some potassium gets reduced to oxidation state (0) from (I).

$\endgroup$
  • 3
    $\begingroup$ "Here, some potassium gets reduced to oxidation state (0) from (I)" That reaction equation makes no sense whatsoever. $\ce{K+}$ does not get reduced that way. $\endgroup$ – Gert Dec 4 '17 at 14:55
  • 2
    $\begingroup$ Your reaction isn't even balanced. Please provide a reference for it. There no reducer strong enough in there to reduce $\ce{K+}$ to elemental K. The only reducing agent present is $\ce{Zn}$. $\endgroup$ – Gert Dec 4 '17 at 15:08
  • 2
    $\begingroup$ @MaxW: to be fair, he didn't specify aqueous medium. $\endgroup$ – Gert Dec 4 '17 at 15:18
  • 2
    $\begingroup$ I found numerous sources which suggest the following reaction: $$\ce{3Zn^2+ + 2[K4Fe(CN)6] -> K2Zn3[Fe(CN)6]2 + 6K+}$$ which is also performed in aqueous media only. Sorry, no potassium metal among products, not by a chance. $\endgroup$ – andselisk Dec 4 '17 at 16:23
  • 3
    $\begingroup$ @andselisk: yes, I saw these too. OP's refusal to provide a reference speaks volumes. $\endgroup$ – Gert Dec 4 '17 at 16:26
-2
$\begingroup$

This should do your work:

3Zn + 2K4[Fe(CN)6] = K2Zn3(Fe(CN)6)2 + 6 K

Now you can calculate the n factor.

On a second thought Zn may be ionic with +2 Oxidation state releasing K+.

$\endgroup$
  • 2
    $\begingroup$ I don't understand why this got upvoted. All you've done is balance a non-existing reaction. We can all do that! $\endgroup$ – Gert Dec 4 '17 at 18:09
  • $\begingroup$ As I see, he wanted to solve some question and the exams many a times commit this mistake. So, it's better to ignore it. $\endgroup$ – CodeBlooded Dec 6 '17 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.