Questions and Answers for 30

Question:

(Sec. 8.4C BF, Inversion of Configuration) Dr. Hardinger, on the bottom of page 285 it reads: "Substitution with inversion of configuration in one molecule cancels the rotation of one molecule that has not reacted, so that for each molecule undergoing inv ersion, one racemic pair is formed. Inversion of configuration in 50% of the molecules results in 100% racemization." Is this paragraph true for the same reasons that nucleophiles attack from the back (away from the associated leaving group) in SN1 (menti oned at the top of page 285)?

Answer:

Yes.

Question:

(Sec. 8.5, Mustard Gasses) Dr. Hardinger, in the sulfur mustard reaction on page 295, I was wondering how sulfur bonds with the CH2 in the first place. Isn't the ring structure high in energy?

Answer:

This question has already been answered; please scroll down.

Question:

(Ch 8 BF, Skeletal Rearrangement) Dr. Hardinger, I was just wondering if we have to know skeletal rearrangements in substitution reactions. If we do, could you briefly explain the its underlying purpose in SN2 and SN1 reactions?

Answer:

Yes you do. Rearrangement is a fundamental carbocation fate, which occurs to make the carbocation more stable. Rearrangement cannot occur in an SN2 reaction because there are no carbocations.

Question:

On the "substitution and elmination reactions of alkyl halides" page, it says that ideal conditions for SN2 rxns involves a non-polar solvent. (2nd long paragraph). I thought polar solvents with high dielectric constants were favored?

Answer:

On what page of what text?

Question:

(Ch 8 BF, Solvents) Dr. Hardinger, how does polar protic solvents separate charge as opposed to polar aprotic?

Answer:

Protic has nothing to do with separating charges.

Question:

(Ch. 8 BF, Stability) Dr. Hardinger, how is a tertiary carbocation more stable than a primary carbocation (i.e. CH3)? Does this only apply to carbocations?

Answer:

A carbocation is high in energy because it has an open octet and positive charge on carbon. Groups that can stabilize this situation by donating electrons will stabilize the cation. Alkyl groups such as methyl are weak electron donating groups. thu s, a primary carbocation with a single alkyl substituent is less stable than a tertiary carbocation with three alkyl substituents.

Question:

(Ch 8 BF, Stability) Dr. Hardinger, on page it says that iodide is less stable than fluoride. Why is that?

Answer:

What page number?

Question:

Prof, I am really confused as to the trends in basicity. It is my understanding that the best bases are those that do not participate in resonance, are soft, not significantly electronegative and have little inductive effect, all these factors allowing th e base to easily donate electrons. Why then is F- a better base than I-? Thanks a bunch, prof.

Answer:

A good base is hard not soft. The principle reason that fluoride is a better base than iodide is because fluoride is hard and iodide is soft.

Question:

Do we need to know the pKa values from the sheet you gave us in lecture? Also, do we need to know the strong acids, strong bases, good nucleophiles and good leaving groups?

Answer:

Please do not memorize pKa values, although you might find it useful to have a good idea of the pKa values of common acids like H2SO4, hydronium acetic acid, etc. You should be able to look at a structure and figure out if it is a good acid or good b ase, or good leaving group or good nucleophile. We've done numerous examples of this exercise in lecture. There are many more examples in the textbook, concept focus questions, practice problems and old exam questions.

Question:

i have a specific questions: why is HS more nucleophilic than H20, H2S, and HO? why is HOCl more acidic than HOBr? and lastly, why is H20 more acidic than NH3 and CH4? thank you jennifer gardner

Answer:

HS: polarizability and charge. HOCl: inductive effect. H2O: electronegativity.

Question:

professor, i am having a hard time following the trends of the periodic table for acidity and nucleophilicity. let me know if i understand them correctly: for electronegativity, the more electronegative, the less acidic, the more nucleophilic? for polarizability, for the rows- as you go from left to right you get more acidic? and for the columns as you go top to bottom, you get harder to softer which means less acidic to more acidic. but what does it do to nucleophilicty? and then, i thought a better base was also a better nucleophile. is my understanding of these trends correct? because i was doing some practice problems and i seemed to contradict myself. please clarify, thank you jennifer

Answer:

The role of a base or nculeophile is to share electrons with a proton or electrophile to make a new covalent bond. Any structural feature that decreases the ability or driving force for that species to donate electrons will decrease it's reactivity. If the atom that is sharing electrons is highly electronegative then it will have less propensity ot share electrons, and the species will be a poorer base or nucleophile. (Please don't just memorize trends, but instead understand the origin of these tr ends. Memorized trends are forgotten, but understood trends are remembered.) It appears that you might be confusing the effects of polarizability and electronegativity.

Question:

For problem 5.16b in BF, isn't aconitic acid Z/trans? Trans indicating that the longest carbon chain is trans and not cis?

Answer:

You are correct.

Question:

I found five mistakes on the study guide and problems book (Brown and Foote). 1. Ch.1 problem 32c, the answer should be C-H

Answer:

6.31c: An iodonium ion is a resonance form of the carbocation shown in the solutions manual. This is not an error. The other four answers you mention all do contain the errors you have noted. Four error bounty points have been recorded for you.

Question:

Hi Dr. Hardinger, Regarding Ch 8.5, internal nucleophiles, can you explain why the highly strained 3-membered ring structure at all? Since that structure is so high energy it seems like an unfavorable configuration, no matter how fleeting it is. Thanks.

Answer:

The ring exists due to a balance of thermodynamic factors. Ring strain is bad, but the new sigma bond that was formed by displacing a leaving group is stronger than the sigma bond that was lost when the leaving group departed.

Question:

Hello Prof. Hardinger, I was wondering on the key for the Stereochemistry CFQ is the definition of a meso compound an achiral molecule with at least two stereocenters or at least one?

Answer:

Either definition works because they both say the same thing. A molecule cannot be achiral if it has a single stereocenter.

Question:

Hi Professor, i just want to clear this up: when considering acidity and nucleophility of a compound (or molecule), would it be correct to say that you would first consider polarizability before electronegativity? thank you.

Answer:

Yes you are correct.

Question:

dear prof H on page 134 of brown and foote it says that the tartaric acid (A) and (b)are enantiomers that are not superposable but on page 138 it claims that they are the same. It says "Enantiomers have identical physical and chemical properties in achira l environments". Am i just understanding it incorrectly or is the book wrong? thanx

Answer:

Enantiomers a & b are not superposable (verify this with models). Mirror images c & d ARE superposable, so they are identical. A molecule that is superposable on it's mirror image is achiral. If a molecule has at least one stereocenter but is achir al it is termed a meso compound.

Question:

i was wonderin for chapter example 4.3 part c. (pg 127 in BF) how do you find the R or S configuration? im not getting it...

Answer:

Make a model exactly as shown, then assign priorities in the usual way. The priorities are OH (highest) > CHO > CH2OH > H (lowest). Now rotate the model so the lowest priority group is in back, and note the direction in which the priorities decrease .

Question:

dr. hardinger-- are fates and carbocations only for SN1 mechanisms?

Answer:

"Everytime you see a carbocation in a reaction mechanism, regardless of where it comes from, we think...." The three carbocation fates apply to any carbocation at anytime, regardless of the origin of that ccarbocation, be it SN1 or some other reactio n we have yet to discuss. There are a number of reactions involving carbocations that you will encounter in your study of organic chemistry.

Question:

Hi Professor, on page 283 of Brown and Foot table 8.5, could you explain why for polar aprotic solvents, F- would be more nucleophilic than I-? thank you

Answer:

This is not a simple explanation, and you are not expected to understand why this is. Here are the key facts to focus on concerning solvent effect and nucleophilicity. A solvent with high dielectric constant does not hiner nucleophilicity because it assists in separating the nucleophile from it's cation. Protic solvents decrease nucleophilicity of negatively charged nucleophiles due to hydrogen bonding, but have little effect on the nucleophilicity of neutral nucleophiles.

Question:

Hi Professor, concerning the previous quesion (as to which compound is more acidic, CF3CO2H (trifluoroacetic acid) or CCl3CO2H (trichloroacetic acid). you said that EN makes the compound with the F a more acidic compound than the one with Cl? I thought that acid streng th with a coloum of the periodic table increases from top to bottom due to decreasing bond strength. isn't this just the opposite of electronegativity? Thank you.

Answer:

Being more electronegative, fluorine reduces the negative charge on CF3CO2- to a greater extend than chlorine reduces the negative charge on Cl3CCO2-. Therefore CF3CO2- is a poorer bases than Cl3CO2-, which means CF3CO2H is a stronger acid than Cl3CO 2H.

Question:

Prof Hardinger, When I submitted a question on Sat Feb 26 I used HTML in my question and I used the

 tag with out including the 

Answer:

The simplest solution would be to avoid using HTML when writing your question. Just type into the VOH box like it was a normal word prcoessor.

Question:

Hi Professor Hardinger, I am not sure but did you cover Fischer Projections? also, on pg 115 of Brown and Foot, question 10d) why is CF3CO2H (trifluoroacetic acid)a more stronger acid than CCl3CO2H (trichloroacetic acid)? Thank you

Answer:

You are not responsible for Fsicher projections. Trifluoroacetic acid is a stronger acid than trichloracetic acid because fluorine is more electronegative and exerts a stronger inductive effect than chlorine.

Question:

Hello Prof. Hardinger, I had a question about a CFQ for Organic Acids And Bases. It states in number 2 for the solutions that "The chloride ion in this case is a Lewis base that is not functioning as a Bronsted base. It is not possible to have a Lewis base that is never a Brons ted base, as any Lewis base can donate electrons to a proton." Is this a contradiction? Thanks

Answer:

This is not a contradiction. A Lewis base can donate electrons to anything, whereas a Bronsted base can only donate electrons to a proton. Thus a Bronsted base is always a Lewis base, but a Lewis base is not always a Bronsted base.

Question:

I would like to know if it's ok when determining the stability of leaving groups after they've left, to use just the resonance stabilizing factor and polarizability only. I know leaving grps are not as important as the nucleophile that's attacking in an Sn2 mech, but will it be ok to just use those two factors in determining the best leaving grps?

Answer:

Yes, use those two factors and some chemical common sense. :)

Question:

This is taken from your Midterm I Solutions Key. 6. (3 points) Write the balanced cathode half-reaction for the CO/PbO2 battery. Answer: Pb4+ =====> Pb2+ + 2e- Since Pb4+ is reduced to Pb2+, shouldn't the half-reaction be: Pb4+ + 2e- =====> Pb2+ Thank you.

Answer:

The correction has been made. Thank you for bringing it to my attention.

Question:

Why is a soft LG good? If it is soft, wouldn't it want to interact with the carbon and therefore be less inclined to leave? Maybe it's because it makes for a more stable transition state?? thanks

Answer:

The leaving group is accepting the pair of electrons that used to be the carbon-leaving group bond. This requires a change in orbitals on the atom that is accepting these electrons. This change in orbitals is partly what we mean when we refer to pol arizability. Thus higher polarizability (i..e, softness) makes for a better leaving group because it assists electron and orbital change.

Question:

Is resonance good for the LG b/c it's more stable after it leaves or b/c it has a weaker partial bond to C and therefore is more inclined to leave?

Answer:

There is no direct connection between resonance and bond strength, so resonance makes a leaving group better by improving it's stability after it leaves.

Question:

Dear Mr. Hardinger,on the electrophiles and nucleophiles tutorial example #2c, i think the solution should read "Each hydrogen atom bears a partial positive charge,so the molecule can behave as an electophile as well"

Answer:

This error has been fixed, and your one point "bounty" recorded. Thanks for bringing it to my attention!

Question:

Dear Mr. Hardinger, in the mechanism fundamentals practice problems, practice problem #5b,I don't understand where one of the methyl groups went in the first drawing (the one with the double bond to oxygen)Could you explain to me where it went and why it isn't shown? Was this just an error?

Answer:

The problem asks you to write a version of the equation that is correct. Removing this erroneous methyl group is one such solution.

Question:

Mr. Hardinger,could you explain an electron donating induction effect with an example? in CF3CH2OH, the fluorines have a withdrawing induction effect, so does the carbon have an electron donating effect because of its lower electronegativity?

Answer:

Fluorines are highly electronegative and thus electron withdrawing. Atoms that are of low electronegativity can be electron donating.

Question:

I had a quick question about one of the midterm solutions. Should the numerator and denominator be switched in number 10 for Q?

Answer:

The answer key appears to be accurate. Q is a quotient of the product of all the product concentrations (stuff on the right side of the equilibrium) over the product of all the reactant concentrations (stuff on the left of the equilibrium).

Question:

Hello, I am having a hard time finding the link to the Winter Midterm I key. Could you please e-mail it to me? Thanks, Michelle

Answer:

The link is on my Chem 30 page, near the top. http://web.chem.ucla.edu/~harding/30_w00_e1.html

Question:

In the specific rotation tutorial, part c, why is the concentration of 1.24 g/mL used instead of the diluted concentration of 1.00g/mL?

Answer:

Thank you for pointing out this error. It has been fixed.

Question:

prof, in working through stereochemistry, I have found that the book has discrepencies with the way you teach rotation. Using the book's way (using atomic numbers) I get opposite rotations than when I use your way, atomic weights. Is this my problem, or s hould I just ignore the book's answers and do it your way? Thanks

Answer:

The only time that amotic weight vs atomic number can lead to discrepencies is in the cases of certain isotopes. If you are getting wrong configurations, then there is some more serious error. Please come and visit me and we'll work it out.

Question:

are the solutions for the midterm going to be posted on the web? if it already is, where is it? thanks

Answer:

They are posted. Look for the midterm link near the top of my Chem 30 web page. --Dr H

Question:

Mr. Hardinger on the specific rotation tutorial, could you please explain part c; Focusing on the concentration portion of it? Thank you

Answer:

Could you make this question more specific? I don't know what you need for me to explain.

Question:

HI Dr Hardinger, Something is kind of confusing. When choosing cathode and anode to calculate standard cell potential we are supposed to choose the algebraicall bigger E0 as cathode right? If we have 2 -ve E0 say: -1.2 and -0.5/0.7, -0.5/0.7 is a larger # than -1.2 and according to OGN should be chosen as cathode and -1.2 as anode, but in your examples, you choose -1.2 as cathode, as if you are taking the absolute value that is bigger. This is contrary to OGN's method, please tell me which is the right way? Thanks

Answer:

The half reaction with the largest absolute value gets to do what it wants,.i.e., proceeds in a direction that gives that half-cell a positive voltage. Postive voltage corresposnds to negative deltaG (thermodynamically spontaneous).

Question:

Do we name compounds? For anode/cathode problems, which one is written first in the parentheses, (Cl l Cl) ll (Ag2+ l Ag+)? Which comes 1st in each parentheses? What goes through salt bridge? What is it's purpose?

Answer:

Naming: You are responsible for basic nomenclature of common molecules, but will not be asked to name complex things. About anode/cathode: i cannot tell what your cell is, so I cannot answer your question directly. We always put the anode on the lef t and write it as an oxidation. The cathode always goes on the right and it is always written as a reduction. About salt bridge: Ions migrate throuhg the salt bridge to balance the charge that builds up in each half-cell during an electrochemical proces s.

Question:

Prof. Hardinger, Do we need to know the short hand for Benzene (ie Ph)? Also which causes more Van Der Walls Repulsion (due to electron repulsion) a benzene ring or a -CH3, and also benzene vs -CH2CH3.

Answer:

About Ph: No, but feel free to use it. About van der Waals repulsion: Generally speaking, larger groups have greater repulsion. Since methyl is smaller than ethyl and both are smaller than phenyl, methyl would have the least steric demands (repulsio n), and phenyl would have the greatest steric demands.

Question:

Dr. H, For two molecules to react they must collide with the proper orbital orientation and with sufficient kinetic energy to overcome...? Is it electorn density repulsion or does it have something to do with bond energy changes? I'm confused. Peter

Answer:

To overcome the energy barrier in the transition state. We will address this issue when we talk about the SN2 reaction in perhaps a week.

Question:

professor, I was wondering, did you want us to memorize all the functional groups, or just specfic ones. Thanks for your time, dave

Answer:

You are responsible for all the functional groups on the handout I gave in class. You should be able to draw them, pick them out of a structure, and understand why we study functional groups. Please do not waste your time by attempting to memorize s pecific examples.

Question:

professor, just to make sure I'm not missing anything. the miterm will cover redox, electrochemistry, molecular dynamics, and ??acid base stuff?? thanks for your time, dave

Answer:

The midterm covers everything up to and including molecular dynamics but not stereochemistry.

Question:

when counting valence electrons for lewis structures, we use a periodic table. should we bring one to the midterm? also, will we be able to have a cheat sheet?

Answer:

You will be supplied with a periodic table. You MAY NOT have a cheat sheet. (You don't want to see a Hardinger exam that lets you use one!)

Question:

Professor Hardinger, I was curious as to whether or not we would be able to bring our model kits into the midterm exam and if so, how necessary they would be to carry out the exam. Thank you for your time.

Answer:

You may bring any or all of the following to the exam: pencil, pen, eraser, model kit. A model kit might be useful for this exam. Stereochemistry has been moved to midterm II, so your model kit will definately be useful there.

Question:

professor, i understood S = counterclockwise and R = clockwise, but i'm having trouble to figure them out in exercises. could you explain easier way to define S&R configuration? thank you.

Answer:

My experience with this kind of question suggests that the problem is one of visualization of the three-dimensional molecular structure. Are you using models? If not, try using them. If you are, then bring this question to my office so we can figur e out the exact problem.

Question:

Will electrons of highest energy always occupy the valence shell of a particular atom? Is it correct to define valence shell as the space that contains electrons of highest energy for a particular atom instead of saying the "outtermost" electron shell?

Answer:

The valence shell is DEFINED as the highest energy shell, so the answer to your question is yes.

Question:

Professor, I'm still a little confused about the redox fom the beginning of the course. Like in the example given on the first redox handout, how can you tell what is and isn't a spectator ion? thanks for your time. Dave

Answer:

Spectator ions are ions in which there are no oxidation state changes.

Question:

Hi Prof. Hardinger, I have a question regarding resonance and molecular orbitals. Taking the example of the CO3 carbonate ion from the 1/24 lecture, I understand that the contributing structures don't really exist but that in reality all three C-O bonds are equivalent. If th is is the case then what do the hybridized molecular orbitals look like? Is it just that the overlap of the pi-bonding orbitals is not as great as it would be in a "true" pi-bond? Or would it be better to think in terms of probability densities? Thanks, Bernie.

Answer:

Carbon of carbonate has three attachments and is therefore sp2 hybridized. The C=O pi bonds in carbonate are weaker than normal C=O bonds because they don't have a full compliment of two pi electrons.

Question:

Dear Professor Hardinger: When drawing resonance structures, should we consider structures which have higher formal charges on some atoms(ie -2 as opposed to -1) as viable contributing structures? Thanks.

Answer:

Any resonance contributor is possible as long as it does not violate the normal rules from drawing Lewis structures. If everything else is equal, then the resonance contributor that has the least number and/or smallest formal charges is more importan t. For more details on resonance structure, work through Chapter 3 of the Weeks paperback, and explore the resonance tutorial on my home page.

Question:

Dear Professor Hardinger: How familiar should we be with the topic of molecular orbitals (bonding and anti-bonding orbitals,etc.) discussed in Chapter 16? Thanks.

Answer:

I assigned Chapter 16 mostly as review. This is material that you should have had in Chme 20A. For an exam, you'll be expected to know and fully understand the level of detail discussed in lecture.

Question:

Professor, is it ok to say that resonance is to maximize amount of bonding between atoms, or would i have to say that resonance is one or more lewis structures?

Answer:

A molecule is DEFINED to have resonance if it can have more than one acceptable Lewis structure. Resonance doesn't really "occur" because individual resonance contributors do not exist. Neighboring atoms will share electrons if possible to increase bonding, and this is consist with the resonance model.

Question:

Professor: regarding the electrochemistry practice problems, is question 2C a -3 charge ion as is on the question, or O3 subscript as in the answer key? also on question 5 of the same electrochemistry problems, on the question it was written as Ce as th e anode and aluminum as the cathode, but on the answer key it was written the other way around which is correct? because I got the answer of -3.316V as the standard potential but the answer key read +3.316V which would change everything....

Answer:

The errors in the electrochemistry practice problems have been fixed. Visit the page again, and use your "refresh" or "reload" button to see the newest version of the page.

Question:

I have a few questions that are loosely related with VSEPR which states that electrons within a shell will arrange in the least energetic arrangeent, i.e. far away from each other as possible. My question is this: why do atoms bond at all? I mean, it seems that if 2 atoms present their negatively charged electrons to one another then as they approach one another they should be repelled. My guess is that the desire for electrons and the desire t o fulfill the octet (electronegativity) overcomes this repulsion? A related question is how come there are electron orbitals at all? Why don't the electrons collapse inwards towards the positively charged nucleus? Finally, why is there a nucleus at all? Shouldn't the positively charge protons separate violently? I have heard of the strong and weak nuclear forces but still haven't been able to reconcile this yet in my mind. Thanks. Bernie.

Answer:

Atoms bonds because the resulting electronic and orbital arrangement has lower energy than if the atoms did not bond. The remainder of your questions are best answered by someone who knows more quantum physics that I do!

Question:

hi, i'm just wondering how much of ch.12 in oxtoby we should read?

Answer:

According to the reading schedule on the web page, all of Chapter 12 was assigned, so I expect you to read all of it. However, you should focus on the material covered in lecture.

Question:

For practice problem 1a on the oxidation/reduction set, how do you know it's a basic solution (using step 5--to add OHH and H20 instead of H30 and H20)? Also for 2C on the same set--I got the half reaction of Cr6+ --> Cr3+, but I don't know how to get the other half reaction of CH3CHO--> CH3CO2H? Does the CH3CO2H have a negative charge? What is being oxidized?? And--is the Faraday's Laws handout going to be posted on the web? Thanks

Answer:

About 1a: Because the problem says it's in basic solution. About 2c: The oxidation states for carbon were discussed in the first day of lecture. You can use the rules in Oxtoby to figure it out. Alternately, start by noting that all atoms in their elemental state (C for carbon) have an oxidation state of zero. Then adjust the oxidation state depending upon the bonds. Each bond between carbon and an element more electronegative than carbon counts as an oxidation, so the oxidation state is made one unit more positive. Each bond between the carbon and another carbon or any atom less electronegative than carbon counts as a reduction, and so we make the oxidation state one unit more negative. Thus for the carbonyl carbon of acetic acid (CH3CO2H), th e oxidation state is +2. About the Faraday's Laws handout: I have no plans to post this on the web. I do this to encourage students to attend lecture. A copy of the handout can be obtained from me in my office. PS: In the future, please only ask one question per submission. It makes it easier to see the answers. Thanks!

Question:

professor, I was wondering about the nerst equation. Where does the n come from? Thanks.

Answer:

In the Nerts equation, n is the number of moles of electrons passed between the two half-cells in the balanced redox equation. Example: Cu + 2Ag+ ---> Cu2+ + 2Ag The Cu ---> Cu2+ is a two electron process. Ag+ ---> Ag is a one electron process, but the Ag bears a coefficient of 2 in the balanced equation. Thus, the silver transfers two moles of electrons in the balanced equation. For this reaction, n = 2.

Question:

Why do some rechargeable batteries have memory problems and others do not?

Answer:

Batteries have different designs, and that can lead to differces in the details of how they work. Other than that, I do not know. I would suggest a web search. If you find anything, please let me know!

Question:

Hello Prof. Hardinger. I have a quick question about one of the practice problems for the Electrochemistry section. For number 4 part a I was wondering when you solve for n the number of moles, why is the answer 8 moles of electrons and not 16 moles? Thanks.

Answer:

Ooops! That typo has been fixed. Thanks for bringing it to my attention.

Question:

for the midterms and final are we allowed to have a 3x5 index card for a cheat sheet?

Answer:

At this time you may not have a "cheat sheet" for exams.

Question:

mr hardinger,in #4 of the electrochem web practice problems, the work you have in the key would give the answer of 16.0 moles of e-, and not 8.00 moles of e- (as you have written). Could you please tell me why it is 8.00 and not 16.0?

Answer:

I do not understand what you are asking here. Could you clarify the question, or come to my office and ask it again? Thanks.

Question:

mr hardinger, in the electrochemistry web practice problems, problem #2b is written with Pb2+ then Pb as the anode. if this is being oxidized, then shouldn't it be written as Pb then Pb2+?

Answer:

You are correct. Thanks for pointing out this typo, which has been fixed.

Question:

Professor, in an abbreviated battery, anode goes on the left, and cathode on the right, correct? In #2 of the electrochemistry practice problems on the web, you have in the answer the Fe3+ and Fe2+, which were on the left in the abreviation, taking the ca thode spot in the equation to find the reaction potential. Is this correct or a mistake?

Answer:

By convention, the anode is written on the left and the cathode on the right (Oxtoby page 407). The problem contains an error which has been fixed. Thanks for bringing this to my attention.

Question:

Dr. Hardinger, Can you please show me an example how to balance a reduction or oxidation reaction in an "organic" system? Do we balance the electron base on the difference between the oxidation states of carbon? How do we put the H20/H30+ or 0H-/H20 pairs into the half- reaction? Thank you.

Answer:

The procedure for balancing redox reactions is the same in inorganic as well as organic systems. Use the oxidation states on carbon to figure out the electron change. Use H3O+ or HO- in the usual way.

Question:

(Example 12.6, Electrical Work) Dr Hardinger: I'm familiar with the concept of work, so I'm assuming that I'm just not reading the question correctly. Anyway, in this example, the question states that the battery is delivering the steady current, so I ass ume that the battery is performing the electrical work; the work should then be negative. But the last sentence of the example states, "This [-40.5 kJ] is the work done on the battery, and so the work done by the battery is the negative of this, or +40.5k J." Can you please clarify this for me? Thanks.

Answer:

You are correct. The battery is doing the work on the external circuit. Sometimes another current source is doing work on the battery, for example, when a battery is being charged.

Question:

(Sec. 12.2, Galvanic Cell) Dr Hardinger, looking at the set up of the galvanic cell, I was wondering if we didn't need silver nitrate to form copper(II) ions because I don't see how the silver nitrate in one beaker can interact with the copper in the othe r beaker. Thanks again.

Answer:

The silver and copper solutions do not interact directly. The electrons produced or consumed flow through the external circuit (or are carried by ions through the salt bridge).

Question:

(Sec. 12.2, Salt Bridge) Dr. Hardinger, I was curious as to the purpose of preserving charge neutrality within each beaker. Wouldn't the electrons be more likely to travel from the positively charged solution to the the external circuit? Thanks for your t ime.

Answer:

A system is most stable when it has the least amount of charge. If charge neutrality was not maintained, the energy of the system would rise. It would quickly rise so high that charge production would cease, and the current flow would stop.

Question:

Dr. Hardinger: Sec. 12.2: Galvanic Cell I'm a little unclear on exactly what happens at the metal-solution interface when electrons reach the silver cathode. Can you help me, please?

Answer:

When electrons leave Ag to form Ag+ and an electrons, the Ag+ ion goes into solution as Ag+(aq), and the electron travels through the silver electrode and from there through the rest of the circuit, and eventually over to the other half cell.