I think you are right. However, this is a weird question. To construct an electrochemical cell you would normally construct it from two half-cells. Each half-cell would contain a metal electrode in equilibrium with a solution of its own ions. The two half-cells would be connected by a salt-bridge.
This question is describing something very different - a metal electrode in contact with a solution of a different ion. There would be a spontaneous, non-equilibrium reaction and you could not experimentally measure the difference in electrode potential because the reaction would be occurring at thousands of tiny local cells where the oxidation and reduction reactions would be occurring in close proximity.
OK, so lets give the examiner the benefit of the doubt, and assume that they actually mean an electrochemical cell consisting of two metal electrodes.
From examination of column 2 of the table, we conclude that Pb, Y and Z all reduce X2+ ions. Therefore metal X has the highest reduction potential of all the metals under consideration, it is the least active metal (it could be silver, for example Eo=1.76 V). From consideration of column 4 we conclude that X, Pb and Z cannot reduce Y2+ to metallic Y. Therefore Y has the lowest reduction potential in the table (it could be Mg, Eo=-2.36 V), and the best answer is (D).
<- u
Thats like double my trip to burwood
but they all way too far from me i cannot be bothered commuting that far
