How do you balance $A l + C u C l_{2} \to A l C l_{3} + C u$ $Al + CuCl_2 -> AlCl_3 + Cu$ ?

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Answer 1 · 2018-04-13T06:13:34

Well this is a redox reaction....

Aluminum is oxidized to $A l^{3 +}$ $Al^(3+)$ .

$A l (s) \to A l^{3 +} + 3 e^{-}$ $Al(s) rarr Al^(3+) + 3e^(-)$ $(i)$ $(i)$

And cupric ion is reduced to zerovalent copper....

$C u^{2 +} + 2 e^{-} \to C u (s)$ $Cu^(2+) + 2e^(-) rarr Cu(s)$ $(i i)$ $(ii)$

We cross multiply to remove the electrons.... $2 \times (i) + 3 \times (i i)$ $2xx(i)+3xx(ii)$

$2 A l (s) + 3 C u^{2 +} + 6 e^{-} \to 3 C u (s) + 2 A l^{3 +} + 6 e^{-}$ $2Al(s)+ 3Cu^(2+) + 6e^(-) rarr 3Cu(s)+2Al^(3+) + 6e^(-)$

...to give finally...

$2 A l (s) + 3 C u^{2 +} \to 3 C u (s) + 2 A l^{3 +}$ $2Al(s)+ 3Cu^(2+) rarr 3Cu(s)+2Al^(3+)$

This reaction is spontaneous as written, given that aluminum is higher on the activity table.....

And so see what happens when you add some aluminum foil to a solution of copper nitrate? The foil will disintergrate....

How do you balance Al+CuCl2→AlCl3+CuAl + CuCl_2 -> AlCl_3 + Cu?