How do you balance #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 #Al^(3+)#.

#Al(s) rarr Al^(3+) + 3e^(-)# #(i)#

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

#Cu^(2+) + 2e^(-) rarr Cu(s)# #(ii)#

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

#2Al(s)+ 3Cu^(2+) + 6e^(-) rarr 3Cu(s)+2Al^(3+) + 6e^(-)#

...to give finally...

#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....