Comparing reducing agents

[Glenda Marshall DIS 3M]

In problem 13.25, it asks us to rank the metals given in order of the power of the reducing agent. Part (a) gives us Cu, Zn, Fe, and Cr. I know that the lower the reduction potential the more powerful the reducing agent, but how do we know which reduction potentials to look at to compare their relative strengths. Both have multiple reduction half-reactions listed in the index.
Thanks!

[martha-1I]

You compare the standard reduction potentials for each one. For example, the reaction to form Cu is Cu⁺ + e^- ----> Cu. According to Appendix 2A in the back of the book, the standard potential for the formation of Cu is E_Reduction= +0.34V. You follow the same pattern for the rest of the metals and compare their standard potentials. REMINDER: The more negative the standard potential, the stronger the reducing agent.

[Chem_Mod]

Martha is correct.

The higher the E-red, the more easily the reactant is reduced. The most-easily reduced chemical is equivalent to the best oxidizing agent, or the worst reducing agent. So, turning it around, the best reducing agent corresponds to the lowest E-red.

[Glenda Marshall DIS 3M]

That makes sense, but I am still confused how you know to use

Cu+ + e- ---> Cu Eo =+0.52

as opposed to

Cu2+ + 2e- ---> Cu Eo = +0.34

Because using the lower one would change the answer. Thanks!

[Niharika Reddy 1D]

For this particular question, I don't think the answer is affected by looking at the second Cu half reaction you listed, since both of those reduction potentials are much larger than those of the other metals. When I looked up the half reactions and their standard potentials, I looked at all the half reactions for the M²⁺ cations, where M is the metal, so I used the second one with a standard reduction potential of .34V

[Chem_Mod]

If you have these kind of questions, there must be more information telling you which specific oxidation state you need to compare. Since Cu²⁺ and Cu⁺ are different species as reducing agent.