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

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