# Redox Questions

Discussion in 'MCAT Study Question Q&A' started by Sammy1024, 05.12.14.

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1. ### Sammy1024

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So I was reading the passage below and then doing thing following question:

11. At which electrode is oxygen gas being liberated?
A. 4 only
B. 1 and 2 only
C. 2 and 4 only
D. 2, 3, and 4 only

Can someone explain to me how you know which reaction is occurring at what electrode? I looked at 1 and 2 and since they're in the same solution, how do you know what occurs?

3. ### pilotjoe 2+ Year Member

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Did you use table one in deciding your answer? Also, the second to last sentence in the passage also gives a hint.

4. ### Sammy1024

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I wasn't sure how to use it.

5. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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Is this an AAMC question? Just wondering.

6. ### SSerenity 2+ Year Member

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You don't even need the table to do this problem.

1)How do we form O2 gas from whatever we have in solution? Oxidation of some oxygen containing species. O2 has an oxidation number of zero, so that should be pretty obvious.

2) Where does oxidation occur? At the anodes!

3) Where are our anodes? Its the electrodes where electrons are LEAVING! Thats 2 & 4, as you can see from the arrows. If this was a galvanic cell, you would have to figure out which direction the electrons are moving, but since its electrolytic, the way the battery is connected tells you everything you need.

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7. ### Sammy1024

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So smart!

And no this was a kaplan question.

8. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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I stared and stared at this question and it just bugged me, so I hope you don't mind me reasoning it out.

Aqeous electrolysis is both annoying and tricky, because the salt in either case is in aqeous solution (water), which in itself can be oxidized. So you cannot always assume that O2 will be produced at the anode. You'd need use the standard reduction potentials provided to determine which of the ions present would be oxidized at the anode.

Once you realize that, you then need to compare the oxidation potentials for each. The most positive (or least negative) oxidation potential will be the ionic species that will react at the anode. Because the table provides reduction potentials, you need to reverse those values. Also, because we know anions react at the anode and cations react at the cathode, we can simplify this to make these comparisons easier.

For Cell #1, you have NaOH in H2O.
NaOH dissociates to Na+ and OH-.
H2O dissociates to H+ and OH-.

Here you have 3 ionic species in solution: Na+, H+, OH-.
There is only one anion present, therefore OH- will react at the Anode and produce O2 as illustrated by the half-reaction provided in the table.

For Cell #2, you have CuSO4 in H2O.
CuSO4 dissociates to Cu2+ and SO42-.
H2O dissociates to H+ and OH-.

We have two anions we need to compare: OH- and SO42-. The oxidatoin potential of OH- is simply the reverse of it's reduction potential: -0.40V. For whatever reason, they did not include the reduction or oxidation potential of SO42-. I looked it up and it's value is -2.01V. Because the oxidation potential for OH- (-0.40V) is less negative (or more positive), whichever way you want to look at it, it is the ionic species that will react at the anode and again, it will produce O2 gas.

The third componant of this question requires you to locate both anodes. For electrolysis, we know the anode is positive andd the cathode is negative -- and electrons flow from the anode to cathode (also true for galvanic cells). In this case, electrons will flow from the anode of cell 1 (labeled 4) to the positive potential of the battery, through the battery all the way to the negative potential (cathode) of cell 2 (labeled 1). Therefore, the anode of cell 2 is label 2.

Therefore, choice 4 and 2 is the answer.
What a loaded question for a timed test.

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9. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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Because time is limited, and you just wanted to simplify things down with an educated guess, you can use the reasoning provided above. Because we know we'll go from some negative oxidation state on oxygen to pure elemental O2 (g) where the oxidation state is zero, this represents a loss of electrons (oxidation). Oxidation occurs at anode. Quickly find the anodes in the diagram. You can either choose choice A (#4) or choice C (#2 and #4). It's a 50/50 guess. It just so happened that choice C was right because of the reasoning above.

10. ### Sammy1024

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I didn't understand how you were using the NaOH equation for cell one when it has no Na?

11. ### pilotjoe 2+ Year Member

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I know you don't have to use the table, but I was trying to make a point that with most passages you do not have to know all the information to get the question. Most of what it takes to get an answer correct is in the passage itself. The passage states that the electrons flow as indicated by the arrows. You can then use table 1 to quickly narrow down which electrodes would liberate oxygen by half reactions. Yes it is faster knowing those details, but in a pinch you can use the info given to you.

Reverse the oxygen and water into the hydroxide equation for electrode 2, with sulfate and copper II being reduced at electrode 1. Same logic for cell 2.

12. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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You're right, it doesn't have pure Na to start, but what happens to salts in aqueous solutions? They dissolve into ions: Na+ and Cl-. As electrons pass from the anode to cathode, and the negative charge (buildup of electrons) resides on the cathode, the free Na+ in solution will combine with those electrons to form pure elemental Na. I didn't consider sodium in my explanation because it's the anode (oxidation) we were considering, but if we wanted to analyze things further, at the cathode, it's ionic sodium (Na+) that's being reduced, and so the half-reaction provided above is relevant (it's a reduction potential).

This is the main purpose of electrolysis. They are non-spontaneous and consume electricity (therefore taking up energy), but that energy is used to make pure elements that we cannot otherwise isolate in nature. Most pure sodium is found in NaCl for example. Likewise, this is the main approach to abstracting certain gases like Cl2.

13. ### DrknoSDN

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Well written post, but I'm having a hard time with your cell #2 process. I can't imagine you would need data that is not presented in the problem. The rest of it is solid.

...My Interpretation of why the table helps solve for electrode #2: (electrode 4 is more obvious)

Since you know the 2nd electrode is an anode from the given flow of electrons, then oxidation is occurring.
You also know that the 2nd cell is going to have a negative E* because it was said that both cells are electrolytic.
Additionally the table gives you the reduction potential for the only possible electrode 1 reaction because there is only one cation in solution.
Electrode 1 reduction potential is Cu2+ 2e- > Cu = 0.34
From the previous statement that the cell is electrolytic you know that the oxidation potential for the species at electrode 2 Must Be lower than -0.34, or else the cell would be galvanic.

If you only use the data from the table the possible Oxidation steps are from OH- or H2O+H2SO3 (might be assumed to be present)
To take the oxidation potential for (H2O+H2SO3) you could just take the negative of the reduction potential that is given (0.20). Giving you an oxidation potential of -0.20.
Therefore you can eliminate that Oxidation potential of -0.2 because it is not negative enough.
The only option is the oxidation potential OH- > O2 + H20 + 2e- = -0.4
That reaction would release Oxygen gas as expected by the answer.

Looking up the direct reduction potential for SO42- on Wiki gives the same answer.
SO42- + 4H+ + 2eSO2(aq) + 2H2O = +0.17
Oxidation potential would be -0.17 and not negative enough.

Source: http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

Crazy question, I would have just educated guessed between A and C on a real MCAT.
And looking at the answer choices, three options have electrode 2 and three options have electrode 4, so it's probably both.

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14. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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Excellent approach to analyzing this question. You took what you know about Electrolysis and applied it in different ways given the information provided. You got the right MCAT thinking on lock. I'd give more likes if I could lol.

15. ### DrknoSDN

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Lol, thanks. =)

It does really help when the answer is given, and having excess time to think about it.
This question would be beastly under a time crunch.

16. ### CzarcasmHakuna matata, no worries. 2+ Year Member

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Yeah, but I'm glad you walked me through your thought process because that's the approach I'll take if I see a question similar to this on the MCAT.