hello as im doing the chapter homework, I found that I have a difficulty determining if a certain redox reaction is an acidic or basic solution. specifically looking at number 105.

Macyn_W.....#19.105: Normally the problem will state whether a redox reaction occurs in acidic or basic solution. To be fair, this one should have done so. If not, you can safely assume that the solution is acidic. This analysis would definitely be done in acid. Interestingly, in terns of stoichiometry, which is the major point of this problem, the MnO4(-) / Zn mole ratio will come out the same either way. Try it.

Hi Dr. Bob! I just had a question about the balancing redox equation thing. On the notes, you said if we're balancing in basic solution, then we would add an equal amount of OH- to both sides to "cancel the H+." Wouldn't the OH- and H+ on the same side combine to make water and then we would continue balancing from there?

Davis_L.....You're correct. In basic solution, we add enough OH- ions to both sides of the equation to neutralize the H+ ions that really don't exist in the basic solution. This leaves you with a half-reaction that has some OH- ions on one side and some H2O molecules on the other. Finish balancing it by adding electrons to the appropriate side to balance the total charge. After completing the same process on the other half-reaction, add them together so that the electrons cancel. At that point, you often can simplify things more by subtracting some H2O and/or OH- from both sides.

Hi I am confused on when to write pt(s) for cell notation. HW question 49 b is giving me trouble

Bella....The electrodes must be electronic conductors (i.e., metals). When a half-reaction does not include a metal, Pt is used as the electrode because it is an excellent conductor and it is chemically inert.

How do we know when the electrolysis of water in aqueous solution is significant (such as in problem 95)?

Is anyone getting the book's answers for 115 b & c? I think I have the right process, but I'm not getting the right answers...

Sawyer_D.....When electrolysis is done in aqueous solution, it is possible for H2O to be reduced at the cathode to produce H2 and/or to be oxidized at the anode to produce O2. You should know these possible half reactions. Whether either or both of these reactions will actually occur depends on what other ions (from an ionic solute) are present and whether one or both of them is easier to reduce or oxidize that H2O. To answer this, you need a table of standard reduction potentials. For example, in aqueous solution, H2O will be preferentially reduced instead of Na+ because the reduction potential of H2O (-0.83 v) is higher than that of Na+ (- 2.21 v), i.e., it's easier to reduce H2O than Na+. This general concept is covered in Tro, Section 19.8. Example 19.9 (p 923) is a good illustration of how to predict the products of electrolysis in aqueous solution vs electrolysis of molten salts.

Sawyer_D.....#19.115: The strategy on this type of problem is basically the same as Problem #5 in our extra Chapter 19 review problems. Work through that one again and see if it helps, by analogy, with #115. I think the answers in the Tro appendix are correct on this one.

on problem 95, I see Sawyer already touched on this why do we note write Na+ + e- - Na as the reduction?

not** write^

Bella_S.....In aqueous solution, you always have to consider the possibility that H2O may undergo reduction (at the cathode) and/or oxidation (at the anode). In the case of Na+(aq), H2O is easier to reduce than Na+ because it has a higher reduction potential as shown in a table of standard reduction potentials (Table 19.1 , p 901). A higher reduction potential indicates a greater ease of reduction. The reduction potential for reduction of H2O (-0.83 v) is higher than that for the reduction of Na+ (-2.71 v). Thus, H2O gets preferentially reduced (to form H2 at the cathode) while Na+ does not react.

19.49b

A lot of Dr. P's students are asking about this type of problem, so I've posted the question and the answer in this post.

I am a little confused on #49 in the textbook. When using line notation, are we supposed to keep everything 1:1? Also, I read the section in the book and it says to add Pt (s) when the species are all in aqueous phase, and the example given there makes sense to me, but in part b on #49 the answer says to add Pt when there is a solid and aqueous on the anode side and a gas and aqueous on the cathode side. I am confused about how that part works.

Answer:
Unlike a balanced equation for a reaction, cell notation does not show stoichiometry. It simply lists what is being oxidized and reduced. There is a line between each phase change and a double line between half-cell compartments.

When the reactants or products are gases or solids, rather than species in solution, there must be a way for electrons to get into and out of the gas or solid. In those cases there needs to be some solid metal to act as the conduit for the electrons. Often a Pt electrode is used because it will serve as a way to allow electrons to move in or out of the solid or gas, and it is inert and will not react itself.

In 49b, I2 is a solid so an electrode must be pushed into the “pile” of solid sitting in the bottom of the half-cell. On the other side ClO2, is a gas, and it also needs some way for electrons to move out of the gas, so again an inert platinum electrode is used.

Cell notation is done as follows, though unfortunately the book is far from consistent or correct. We'll try to update the Corrections to Homework on both D2L and OneDrive soon.

elelctrode (one line) component oxidized, oxidation product (two lines) component reduced, reduction product, (one line) electrode. Different species in the same solution are separated by commas.

Be sure to look at the corrections to Tro homework page on D2L or in the One Drive for corrected answers.

Hello! On #67 in the book problems, is it best advised to use the (Delta G)=-nFEcell equation to solve for the equilibrium constant, or can I also use Ecell=(0.0592/n)logKc ? I did both ways and for some reason only the (Delta G) works. Maybe it is a calculator error, but I just wanted to make sure.

19.67

It's better to skip the delta G step! Your answer is closer to correct than first calculating delta G, rounding, and then calculating K.

By the way the answers both ways are still the same thing. The difference is to tiny to matter especially when the power of 10 is 75.

Thanks to a Dr. P. student for using the Discussion Board. Time for others in my class to step up!

Thank you!

Additional HW Problems #6

Hey! For standard cell potential I get 0.23v and then use this as part of the Ecell=(0.0592/n)logKc to calculate the Ksp and do not get the right answer. I use n=2. May be calculator error or I'm just not doing this right?

#6 Additional problems.


Be careful that you write out the equation for the reaction for Ksp. It is not simply adding those two equations. Since Ksp is not a favorable reaction, K must be small. If it's not a favorable reaction, then E zero is going to be negative. Again, the key step is writing the right Ksp reaction.

Additional Problem #3:

I am having trouble showing work for this. I showed the half oxidation reaction for Pb and the half reduction reaction for Au3+. What calculations do I need to show, and how does SO42- factor in?

Thank you.

Is it correct to do the inverse of log by 10^(-#)?

Dr. Patty's Additional Problem #3

What would the balanced chemical reaction be? I know that Au 3+ goes to Au (s), but what do you do with the Pb (s), PbSO4, and SO4 2-? Is Pb (s) the only reactant?

Thanks!

Additional Problem #3:

Look at the fourth half-reaciton in the right column on Appendix page14. This may help with the half reaaction needed for PbSO4.

Jadan --

Mathematically, that's what the inverse log is as we showed in earlier chapters. How you accomplish getting an inverse log depends on which type of calculator you are using. Try an internet search for directions for your calculator. Also try taking the log of a number and then figure out how to do the inverse log of that.

Dr. Patty's class.

Read above! #67 has been answered. Throw your answer books in the garbage. I'd say burn them, but we should never burn books!

Can you explain cell notation more please? I know how to read it when it's a simple two reactants and two product equations, but I get confused when there is multiple of each.

Is #119 referring to the Nernst equation? I am solving for E0cell, but I got -0.71V (instead of the correct answer, 0.71V).

Alexis,

Look at the examples in the homework, and in the corrected homework. As stated above, if there is more than one thing in solution in a half cell, e.g., H+ and H2O, then only a comma goes between these two things. If there is something in a difference phase, gas in a liquid or solid in a liquid, one line goes between the two phases.

Bottom line is read the simple ones, study the more complex ones and then go on to something more important.

Everyone is missing the forest by only looking at one tree.

Additional #3:

I have two reactions now, but both Pb and Au seem to be reduced. Is there an intermediate equation with Pb, or is the PbSO4 equation switched around?

Thank you.

Emme,

Everything is 1.0 M, so there is no Nernst equation. Glad to see you've looked ahead to see what that is though.

Here is what I should have posted from another student, answered by email so it was difficult to relocate.

This is not the kind of question that is immediately obvious. It is essential the reverse of problem #6 on the additional problems.
I started with writing out the reaction for Ksp of CuI --> Cu+ + I-. From Ksp, I could get the E zero for that reaction. Since Cu+ and Cu are also involved in the cell, I looked up that E zero. That was one half of the reaction that needed to be added to another half reaction to get to the Ksp reaction. Once you determine what that other reaction is (look at #6 additional problems), you can determine that its E zero is 0.188. It turns out that the last value is not even needed.
The next thing I did was write the cell half reactions from the way the cell notation is written. (They look very familiar to those above). Add these together and you see that you have the reverse of the Ksp reaction, so the answer is essentially the reverse E zero.

How do you determine Q for book #75b? I thought that it was 0.100 M/0.500 M, but I got the wrong answer

Hi, for finding the K constant, I have been using G=-RTlnK but for question 69, I used the same equation but have not been getting the right answer. I looked at the solution manual and saw that they didn't use the negative and wanted to know why?

75
First write and balance the overall reaction:
3 Pb + 2 MnO4^- + 8 H+ --> 3 Pb^2+ + 2 MnO2 + 4 H2O

This shows that yo ualso have ot considre the H+ concentration, as well as the exponents for Q, i.e. [Pb^2+] is cubed, etc.

69

Because electrolysis has not yet been covered, the problems in the book are assuming every cell is galvanic, so E zero will be positive.

Skip the entire delta G step and use the E zero = (0.0592/n)log K equation. It's less math, less errors and a MUCH MUCH MUCH better answer that the crazy TRO book! See some of the posts above on 65 and 67. The answer book needs to be buried in your garbage pail. It is terribly misleading and often wrong. We never recommend that students buy it! Indeed, I would prevent anyone from purchasing it if I could.

Hi Dr.Patty, I need help on question 115 b. I understand having to get the new molarity of Cu & Mg and then plugging it into the Nernst equation but I dont think im doing it correctly

115b

Have you watched the video showing a very similar problem at the end of Chapter 19? That may help.

Just to make sure you have the correct final concentrations they are [Cu2+] = 0.754 M and [Mg2+] = 0.746 M.

Let me know if this doesn't help, or perhaps email me a photo of your work.

Hi, on #85 I am having some trouble writing the half reactions and solving for n

19.85

You don't actually need the half reactions here. Just write the balanced reaction for combustion of oxygen, CH4 + 2 O2 --> CO2 + 2 H2O. Then find delta G from the thermodynamic data. Finally, the relationship between delta G zero = -nFE zero. Look at the oxidation numbers of C in both CH4 and CO2 or the sum of the number of electrons gained by four oxygen atoms to get n, the number of electrons transferred.

Hello,

I am having trouble writing the reactions for additional problem #1 and electrolysis problems in general. I just know the final reaction must be nonspontaneous. Looking at the reactions on page 20 of Dr. Patty's notes, I see that Al3+ cannot be used in solution because water is more easily reduced, but Al3+ is in the reaction.

Thank you.

Emme,

Does the problem say the electrolyis is done in water? Nope. Lots of things, including molten (melted) sodium chloride are electrolyzed.

I saw the earlier discussion on 19.95. For 95c, I understand why the reduction with H2O occurs, but for the oxidation side of things... isn't the SO4 also a reduction reaction as given by the equation in the potential table? Since SO4 is gaining electrons?

If this is so.... is this why we choose to use the oxidation of water? Because it is the only oxidation occurring? Originally I would have thought to use SO4 as the cathode because it has the more negative potential, but I feel as though it isn't oxidation at all...

Hope this makes sense!

Caitlyn,

I'm giving up on this problem. People are spending too much time on it. It's no longer on the list of problems.

Ha! Thank you, sounds good!

Hi! I have a question on quiz 8 #2

I am not understanding how the Anode equation is formed and how to calculate the E cell for it.
could you help me?

DThomas.....The cell notation shows a Pb anode in contact with solid PbO and a basic solution. The unbalanced anode (oxidation) reaction must therefore be: Pb ==> PbO. When balanced in basic solution, you have the oxidation half-reaction in the answer key. To work the numerical part of the problem, you'll need to look up the reverse reaction (i.e., a reduction) in a table of standard reduction potentials. Of course, finding the reaction in the table can also help to figure out the correct half-reaction.