How do you determine the reducing/oxidizing agent from standard potentials?

[Jay2910]

Hi everyone,

Here is the question:


What I am confused about, on #14, is that, how can we compare I(-) with the rest?
Similarly, how can we compare Pd on #15?

I've been stuck on this for couple of days, so any help would be great!

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[Dr. Doctor]

I believe what you are asking is why they mention I-(aq) and Pb(s) as answer choices when they appear in the products side of the reaction. You just need to apply the idea that using these molecules would simply be the opposite of the E given for each reaction(because they are on the products side) so for I-(aq) it would be E= -.535. I hope that answered your question.

 

[Labrat07]

I'm not sure what your question is but this is how I answer the 2 questions. Please let me know if I got wrong.

All of them are reduction rxn, so the more positive = the more likely to be reduced = would be the best oxidizing agent. So
14/ C
15/ C

please let me know the answer . i want to make sure. Thank you.

 

[Altius Premier Tutor]

Reducing agents reduce the OTHER species with which they interact, so they must GIVE electrons to the other species (losing them themselves, which = oxidation). That will happen if they "want" electrons less than the other species.

Oxidizing agents oxidize OTHER species (gaining electrons themselves, which = reduction), and will only do this if they "want" electrons more than the other species.

To decide the "agent" you will ALWAYS have to be comparing species. You will compare the reduction potentials (E values) off of the table, recognizing that a large positive E means something likes to be reduces WAY more than hydrogen (the reference compound) and a negative E means it is harder to get that specie to accept electrons than it is for hydrogen.

So...the one with the largest (most positive) E is going to be reduced (oxidizing agent) because it wants those electrons MOST. This could be the difference between -0.24 and -0.04. In other words, neither may be positive at all, but, in this hypothetical example, -0.04 is "more positive" than -0.24, so when these guys exchange electrons the one with the -0.04 value is always going to get them (oxidizing agent) and the other one is always going to give them up, or be oxidized (reducing agent).

Don't memorize this, but it helps to make mental note that they are opposites...oxidizing AGENTS actually get REDUCED and reducing AGENTS actually get OXIDIZED. You have to keep careful mental track of "Me vs. The Other Guy." =)

 

[Labrat07]

The concept sound really complex for such a simple idea. Jesus why do all the scientist create such words play??

 

[RH8448]

14. C
15. B This is what I found. A higher Ecell is more favorable reaction regardless of the sign. They are judged in reference to each other only.


Please let me know if I did this correct.

 

[Altius Premier Tutor]

14. C
15. B This is what I found. A higher Ecell is more favorable reaction regardless of the sign. They are judged in reference to each other only.


Please let me know if I did this correct.

You're right, C/B; Labrat07 was wrong. One way to look at #15 is to say that as a reducing agent the specie is actually being oxidized itself, so you flip the signs, to represent oxidation, or the reverse of the equations shown, and then the one with the largest positive value is Mg.

 

[amy_k]

To summarize some of the points made here, the strongest oxidizing agent has the largest positive value for electrode potential in a reduction half-reaction, all of the reactions shown are reduction half reactions already (reactants are gaining electrons) so we are looking for the REACTANT species in the reduction reaction with most positive electrode potential are for #14:

A) -0.535V for I-(aq) it cannot be the oxidizing agent because it is on the product side
b) -0.789V for Hg(l) it cannot be the oxidizing agent because it is on the product side
C) +0.789V for Hg2 2+(aq)
D) +0.535V for I2(s)

Answer C +0.789V is the largest positive value so Hg2 2+ is the strongest oxidizing agent. this makes sense because Hg2 2+ gains electrons from and oxidizes another species.

For #15, we are looking for the largest electrode potential for an oxidation half-reaction to give the species which is the best reducing agent. All of the reactions shown are reduction reactions so we need to reverse all the reactions and the E potential signs to find the oxidation potentials. We are then looking for the REACTANTspecies which has the most positive oxidation electrode potential to be the strongest reducing agent:

A) - 0.9V for Pd(s)
b) +1.18V for Mn(s)
C) 0.00V for H+
D) -1.18V for Mn2+(aq) (cannot be the reducing reagent because it is on the product side)

So the answer is B Mn(s), this makes sense because Mn(s) is oxidized to form Mn2+, which releases electrons to reduce another species.

 

[Labrat07]

Thank you.