total number of electrons transfered in following oxidation reduction reaction?

Question

2Al (s) + 3Cu2 + (aq) +6Cl- (aq) ----------- 2Al3 + (aq) + 6Cl- (aq) + 3Cu (s)

Answer 1

The 2 x Al atoms lose 3 each, and the 3 x Cu2+ ions gain 2 each. So the total is 6.

Answer 2

Start by identifying the oxidation states

oxidation states:
Al (s) = 0
Cu^(+2) = +2
Cl^(-1) = -1
Al^(+3) = +3
Cl^(-1) = -1
Cu(s) = 0

so Al loses electrons to become oxidized
Cu gains electrons to become reduced

remember "OIL RIG"
"oxidation is loss reduction is gain"
Al(+0) ----> 2Al^(+3) + 2(3e-)
3Cu^(+2) ----> 3Cu(+0) + 3(2e-)


so 2 Al atoms give up a total of 6e- which are transferred to 3 Cu^(+2)ions to form Cu solid.