Yes, agreed with spectre. Looking at the table on the left, the metal ions in solution are being reduced. So you see that when the aluminum electrode is in a solution of zinc ions, they get reduced to a new solid (solid zinc). The reaction there is Zn2+ + 2e- -> Zn (S). At the same time, the solid aluminum is getting reduced, and putting aluminum ion into the solution (Al --> Al3+ + 3e-). In order for this to happen, this entire reaction must be spontaneous.
If we look at what happens under opposite conditions, when a solid zinc electrode is placed in a solution of aluminum ions, we see no change. If no reaction occurs, this reaction is not spontaneous. This means that between the two metals, solid aluminum has a higher oxidizing potential than solid zinc, and Zn2+ has a higher reducing potential than Al3+. It will also help to know that Zn2+ generally has one of the higher oxidizing potentials of the common ions you see. This means that Zn2+ does NOT like getting reduced (a poor reducing potential). In order for that first reaction to be spontaneous, solid Al must have a really really favorable oxidizing potential in order to compensate for the unfavorable reducing potential of Zn2+. Knowing this means you don't even have to look at the other experiment to know that Al has a higher oxidation potential than Zn.
Most of the time, the thing being oxidized will be a solid, like the thing in the passage. So the thing to realize is that the question in is telling you that the HCl (or some part of it) is the thing being oxidized. You should automatically know that HCl in water will completely dissociate to H+ and Cl-. The question also tells you that in the presence of aluminum ion, H2 is formed. Answering this question requires you to think intuitively about this.
The H+ is being reduced to H2. The half cell reaction for this is: 2H+ + 2e- --> H2 . This is what's called the reference potential, for which the standard cell potential is 0. The question tells you that H2 IS formed, which means that the OVERALL reaction (reduction + oxidation) does go forward, which means that the overall standard cell potential must be greater than 0. Knowing that the reduction half potential is 0, all we need for the overall potential to be >0 is that the oxidation half potential is >0, or favorable.
Because H2 is formed, we know that Zn has an oxidizing potential greater than 0 (and Zn2+ is also being formed). We ALSO know that aluminum has a higher oxidizing potential (more favorable) than Zn, so the reaction must go forward, and we will form H2 & Al3+.
