-2
$\begingroup$

I have been trying to balance this chemical reaction in an "acidic" medium without using oxidation number or ion electron method: [$\ce{Fe^2+ + Cr2O7^2- -> Cr^3+ + Fe^3+}$].

Though I can balance number of atoms on both the sides $[\ce{Fe^2+ + Cr2O7^2- + 14H+ -> 2 Cr^3+ + Fe^3+ + 7 H2O}]$, but charge is still unbalanced. I am not sure why? I don't have any intuition about what's going on here.

$\endgroup$
  • 2
    $\begingroup$ I think you mean $\ce{Cr2O7^2-}$. $\endgroup$ – Mathew Mahindaratne Aug 4 '19 at 19:12
  • 1
    $\begingroup$ This gives a pretty good step-by-step guide on how to go about the task: ck12.org/book/CK-12-Chemistry-Concepts-Intermediate/section/… $\endgroup$ – Buck Thorn Aug 4 '19 at 21:29
  • $\begingroup$ $\ce{6Fe^2+ + Cr2O7^{2-} + 14H+ -> 2 Cr^3+ + 6Fe^3+ + 7 H2O}$ $\endgroup$ – Adnan AL-Amleh Aug 5 '19 at 4:38
  • $\begingroup$ You need not to use oxidation number, they are just convenient guidance. Good chemists can balance equation by their intuition, as they can easily see the proper way to do it. Other need to follow 2 fundamental rules of nature: 1/ Law of the total charge conservation 2/ Law of the element mass conservation. $\endgroup$ – Poutnik Aug 14 '19 at 16:16
2
$\begingroup$

A balanced equation must have a mass balance i.e., the masses should be equal on both sides and charges must be balanced as well. Oxidation number is just a way of book-keeping. Nothing fundamental there.

$$\ce{Fe^2+ + Cr2O7 + 14H+ -> 2 Cr^3+ + Fe^3+ 7 H2O}$$

What are you forgetting? It is mass balanced. Does $\ce{Cr2O7}$ exist? Hint: It is potassium dichromate.

$\endgroup$

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