This table is of reduction potentials. I chose answer C thinking if you flipped the Pd^2+ half reaction, the reduction potential for Pd would -0.99, which would be the on that is least reduced of the options and most oxidized and thus the best reducing agents. Apparently that's wrong and all the answer key says in regards to C is that Pd is the product of a reduction half rxn with a voltage of +0.99. Why can't we flip that half rxn to get the voltage for Pd? I thought that +0.99 is the voltage to reduce Pd^2+ not Pd as well. The correct answer is A.
Reduction potentials
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This table is of reduction potentials. I chose answer C thinking if you flipped the Pd^2+ half reaction, the reduction potential for Pd would -0.99, which would be the on that is least reduced of the options and most oxidized and thus the best reducing agents. Apparently that's wrong and all the answer key says in regards to C is that Pd is the product of a reduction half rxn with a voltage of +0.99. Why can't we flip that half rxn to get the voltage for Pd? I thought that +0.99 is the voltage to reduce Pd^2+ not Pd as well. The correct answer is A.
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I see how you're tripped up, it took me a while to think of a good reason why C is right! But I think I have one.
When you flip the reaction, you're right that your E sign will get flipped, but I don't know if the lower E after flipping means that it's a better reducing agent.
Think about lithium at the top there. We KNOW that's a great reducing agent. Its got a lone electron in its valence, so that's going to easily come off. So if we flip that reaction to be an oxidation of Li, it's going to have an E of +3.05V. Flip the Zn reaction and you get +0.76V. Flip the Pd reaction and you get -0.99V.
It's kind of confusing because the reactions are written as reductions, but we're interested in the oxidation reactions actually.
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I think it should be C as well, so I disagree with the answer key. Think about it: what does a good reducing agent do well? Lose electrons. That's the same as saying a low ionization energy. Pd definitely has a lower ionization energy than Fe and Zn, and because of the increased effective charge (Zeff) that results in ionization to Pd2+ (and thus increased electrostatic force), it would be much easier to remove an electron from Pd than Pd2+. I vote that the answer key is wrong but I'm hoping someone can tell me I'm wrong 
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The correct answer is A. Reducing agents get oxidized, so you need to consider the reverse reactions of what are listed in the Table (which means you need to reverse the sign on the emf).
Zn going to Zn2+ will have Eo = +0.76 V (most favorable oxidation half reaction listed)
Fe going to Fe2+ will have Eo = +0.44 V
Pd going to Pd2+ will have Eo = -0.99 V
Choice A is the most favorable oxidation half reaction of the choices.
Choice D is not possible, because Pd has already lost two electrons and will not lose any additional electrons without the input of a significant amount of energy, making it high unfavorable.
If you happen to have the latest BR general chemistry book, look at examples 10.7 and 10.8, as they explain the concept really well.
@David513, you should consider that the periodic trends you are describing apply to the main group elements, not the transition metals. Zn upon losing two electrons still has a filled d-shell, which accounts for its relatively low first and second ionization energies.
Zn going to Zn2+ will have Eo = +0.76 V (most favorable oxidation half reaction listed)
Fe going to Fe2+ will have Eo = +0.44 V
Pd going to Pd2+ will have Eo = -0.99 V
Choice A is the most favorable oxidation half reaction of the choices.
Choice D is not possible, because Pd has already lost two electrons and will not lose any additional electrons without the input of a significant amount of energy, making it high unfavorable.
If you happen to have the latest BR general chemistry book, look at examples 10.7 and 10.8, as they explain the concept really well.
@David513, you should consider that the periodic trends you are describing apply to the main group elements, not the transition metals. Zn upon losing two electrons still has a filled d-shell, which accounts for its relatively low first and second ionization energies.
Yeah, I'm thinking the answer key might be wrong, as well.I think it should be C as well, so I disagree with the answer key. Think about it: what does a good reducing agent do well? Lose electrons. That's the same as saying a low ionization energy. Pd definitely has a lower ionization energy than Fe and Zn, and because of the increased effective charge (Zeff) that results in ionization to Pd2+ (and thus increased electrostatic force), it would be much easier to remove an electron from Pd than Pd2+. I vote that the answer key is wrong but I'm hoping someone can tell me I'm wrong
But if you flip the reactions don't you get the reduction potentials for the reverse rxn? For example, you flipped the Zn2+ rxn and got Zn going to Zn2+ and that +o.76 V would be the reduction potential, meaning it would be more likely to be reduced. As a parallel, flipping the Pd2+ rxn will give you the reduction potential for Pd to Pd2+ (-0.99) and a negative reduction potential means it gets oxidized easier which means it is a better reducing agent.
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What you're saying here isn't true. A is right. The answer key is not wrong.
Remember that a (-) E value means the half reaction will not occur as written. (+) means it will occur as written. All of the reactions are written as reductions, we care about the oxidations, so we have to flip the sign of all the choices. After doing that, you have to pick the option that gives the most positive E. Why? We just made all of the reactions oxidations. In order for the half reaction to occur as written, the E must be (+).
Pd is not right because it gives a (-) E value when written as an oxidation. That means it's NOT going to give up electrons spontaneously for oxidation, and therefore a poor reducing agent.
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This makes sense, and I now agree with A as the answer in light of what Doctor Dream said about needing a positive Eº for the oxidation reactions, but still... Pd has a lower ionization energy than Zn (about 800 vs. 900). I thought ionization energy was shorthand for "how easily will an element give up electrons" with higher ionization energies meaning elements will give up electrons less easily and lower ionization energies meaning elements will give up electrons more easily. Therefore I'm confused since, based on that, it seems like Pd should give up electrons more easily than Zn, i.e. it would be a better reducing agent. But since we're given the Eº in this problem I guess I should just focus on that. CURSE YOU PHYSICAL SCIENCES ! !
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Correct answer is A. A strong reducing agent get oxidized very easily. Because given is the reduction potential, the opposite sign number will be the oxidation potential. The larger the number, the better it get oxidized. In this case, Zn got 0.76 ( the highest oxidation potential). D is not on the table but can be eliminate because it's an cation, which is very tough to remove any e-.
