What is the smallest whole number coefficent of $\ce {Cr^{2+}}$ when the equation is balanced?
$$\ce {Cr^{2+}_{(aq)} + Al_{(s)} -> Cr_{(s)} + Al^{3+}_{(aq)}}$$

I don't understand, how can this be balanced? And why would 3 be the smallest coefficient?

  • 2
    $\begingroup$ Currently, the charges are not balanced. You need to balance the charges to have a balanced equation. $\endgroup$ – LDC3 Oct 30 '14 at 0:34
  • $\begingroup$ You need to balance the reaction, that's it. But you should know step wise method, because if you use try and error method you might waste your lot of time! $\endgroup$ – Freddy Oct 30 '14 at 6:56

You need to factor in the electrons that are being moved around

Chromium is doing this:
$\ce {Cr^{2+}_{(aq)} + 2 e- -> Cr_{(s)}}$

Aluminum is doing the reverse of this:

$\ce {Al^{3+}_{(aq)} + 3 e- -> Al_{(s)}}$

Now, like a math equation flip around the aluminum reaction.
And balance out the electrons

$\ce {[Cr^{2+} + 2 e- -> Cr_{(s)}] \times 3}$

$\ce {[Al_{(s)} -> Al^{3+} + 3 e- ] \times 2}$

$\ce {3Cr^{2+} + 6 e- + 2 Al_{(s)} -> 3Cr_{(s)} + 2 Al^{3+} + 6 e-}$

Hope this helps


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.