So Zn +2HCl-> ZnCl2 + H2 + Cl2
Zn is (0 to 2+)->>>oxidized
H is (1+ to 0)->>>reduced
2Al + 6HCl-> 2AlCl3 + 3H2
Al is (0 to 3+)->>> oxidized
H is (1+ to 0)->>> reduced
I don't know what I was thinking when I chose C because Cl doesn't change it's oxidation state. (I think I some how formed Cl2)<<--dumb
I think I would have gotten it correct through POE, but I don't understand why the key says a new solid is formed when Al is mixed with Zn2+ (It did not say Zn and Al was in the same beaker).
Why is Al more susceptible to oxidation than Zn? Because of the 3+ charge rather than 2+? If I was given Ecell values, I think I would have concluded this.
Did I need to use this table at all?