CHEMISTRY COMMUNITY

Chem_Mod

The anode side is the product side (reaction starts with lower concentration product), 0.0010 M CrCl 3 (aq). Diluting [P] therefore makes Q << 1, which makes ln Q even more negative, and therefore E more positive. Work through the concentration cell example I did in class. Lower concentration produc...

Chem_Mod

Please post an example by giving full question.

Chem_Mod

As I said in class the more positive potential (cathode) pulls the negatively charged electrons towards it.

The cathode has a compound that 'wants' electrons, meaning more stable gaining electrons.

Chem_Mod

For 6.57 from the hw problems why do we square root the value K? I just want to confirm if my thought is correct that we have to square root K because it is K a *K b and the question is just asking for K a since I got rusty on acids/bases When you balanced the two half reactions, your overall react...

Chem_Mod

In the Sapling week 7/8 #5, part b, this equation is given to be balanced. N2H4(g)+ClO−3(aq)⟶NO(g)+Cl−(aq). It says it takes place in an aqueous solution, but no matter how many times I rebalance it, my OH- keep canceling each other out when I put the half-reactions together. Anyone else have this ...

Chem_Mod

I am not sure how to start this problem as I can't identify which equation I would have to use. Question: What would the potential of a standard hydrogen electrode (S.H.E.) be under the given conditions? [H+]=0.23 M PH2=2.1 atm T=298 K If your cell is at non-standard conditions (ie. concentrations ...

Chem_Mod

I asked Sapling and they responded: Typically this error will come up when a student copies and pastes an answer into the textbox. I would first suggest the student tries to enter the answer manually. If they are having trouble removing the error message, I would suggest, they select the Try Again b...

Chem_Mod

I removed some questions from the syllabus as they were not suitable, and posted an updated file:

https://lavelle.chem.ucla.edu/wp-conten ... 4BSYLL.pdf

Chem_Mod

These should help: Balancing Redox Reactions: Acidic Conditions https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Balancing_Redox_Reactions_Acidic_Conditions.pdf Balancing Redox Reactions: Basic Conditions https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Balancing_Re...

Chem_Mod

I will remove this question as it is too confusing with the negative voltage.

Chem_Mod

Good answer. C 2 H 5 OH Make all the H 1+ and the O 2-. 6 x H = 6+ O = 2- This gives 4+. Since molecule is neutral each C is 2-. C 2 H 4 O Make all the H 1+ and the O 2-. 4 x H = 4+ O = 2- This gives 2+. Since molecule is neutral each C is 1-. I should do an example like this in class. But then some...

Chem_Mod

This question was copied from:
https://www.researchgate.net/post/is_th ... t_reacting

"Is there any electrolyte or ionic liquid that dissolves metal oxides at room temperature without reacting?"

Chem_Mod

Corrosion is the reaction of the surface layer of a metal with oxygen or water in the air to create an oxide or hydroxide layer on top of the metal. Over time, this layer can grow larger until it oxidizes the entire piece of metal. By coating a metal with a different material such as paint, or with ...

Chem_Mod

Good Q and A.

Well done.

Chem_Mod

Since K + (aq) is not part of the redox reaction, don't need to include it. If you do include it shows on the left and right side of the chemical equation and cancels out as it is the same species at the same concentration on both sides of the chemical equation. We saw this in class when calculating...

Chem_Mod

One can measure E o when solutions are 1M, pressure 1atm/1bar, 25 o C. Or One can measure the concentrations of reactant and product at equilibrium and use them to calculate (infer) E o . K and E o are related. The reason a redox reaction has a large K is because the E o is a high positive value. At...

Chem_Mod

This is a harder one to figure out with the information given. If one was given the cell diagram then straight forward. Ag(s) is the same electrode for anode and cathode. What makes the two half-reactions different are the solutions (electrolytes) the electrodes are in. Ag(s) anode is in contact wit...

Chem_Mod

The anode and cathode are both Au(s). What makes the oxidation and reduction half reactions different are the solutions (electrolytes) at each electrode. The anode, Au(s), is in contact with Au 3+ (aq). The cathode, Au(s), is in contact with Au + (aq). Since Au + (aq) going to Au 3+ (aq) is favorabl...

Chem_Mod

You do not have to memorize all of them, but there are some you should be familiar with. For example, Oxygen is usually -2, Cl is usually -1, H is usually +1. There are exceptions to these rules, which you can find here: http://ocw.umb.edu/chemistry/chem-116-chemical-principles-ii/Handouts%20PDF/oxn...

Chem_Mod

The cell potentials are measured with reference to the standard hydrogen electrode. If electrons flow to the electrode of interest, then it has a positive standard reduction potential. If electrons flow to the hydrogen electrode instead, the electrode of interest has a negative standard reduction po...

Chem_Mod

See my answer at: viewtopic.php?f=142&t=75893&p=325037&sid=c6eb3f6729492305df6a33ea34fb9013#p325037

Chem_Mod

Yes you are correct. I'll update the file on my class website. Equation in question should be: Cl 2 (g) ---> ClOH(aq) + Cl - (aq) [Hypochlorous acid (ClOH) is a weak acid that forms when chlorine dissolves in water, and itself partially dissociates, forming hypochlorite, ClO - . HClO and ClO - are o...

Chem_Mod

Hello everyone,

I attached my Week 8 Worksheet and answer key below. Happy studying :-)

Best,
Kate

Chem_Mod

The metal oxides vary in their chemistry. Depends on what metal oxide. Dissolving a metal oxide like CaO and MgO in regular water is reactive as you are breaking the metal oxygen bonds and making something else with the reactive O. There are some ionic liquid formulations that can dissolve silver ox...

Chem_Mod

Try using the Nernst equation to find E given E naught and the concentrations of the reactants (which you can use to find Q)

Chem_Mod

The lower concentration is on the product because if you are oxidizing a metal into aqueous cations, it would occur on the side with the lower concentration of cations. Which textbook problem are you referencing? For question 6N.3 (b), the equilibrium quotient of the equation has the larger concent...

Chem_Mod

Dylan_K_3B wrote:An electromotive force is any force/work done that is being driven by a difference in potential that results in an electric current.

Good answer.

Chem_Mod

The lower concentration is on the product because if you are oxidizing a metal into aqueous cations, it would occur on the side with the lower concentration of cations. Which textbook problem are you referencing?

Chem_Mod

Do you mean if a diagram were drawn? Or are you referring to the cell diagram notations involving | and , ? If the former, it doesn't matter on which side the cathode is drawn. If the latter, the anode will always be on the left side. Also, note that electrons go from anode to cathode, not ions

Chem_Mod

products/reactants does not mean anode/cathode. Combine the half reactions to get the overall reaction, which will tell you what the products and reactants are

Chem_Mod

See detailed answers here:

viewtopic.php?f=139&p=325283&sid=ea29185e56002e4bd17838c9370b4edb#p325283

Chem_Mod

It wasn't canceled but by my zoom is acting weird. If you still need help send me an email and we can set up another time that works with your schedule :)

-Mel

Chem_Mod

Please see this document on the class website: https://lavelle.chem.ucla.edu/wp-conten ... itions.pdf
You can also balance as in acidic conditions, then neutralize all the H+ by adding the same amount of OH- to each side of the reaction

Chem_Mod

Please include the entire question, especially because the Sapling problems tend to be different for each student. My guess is you had to multiply by 4 to balance the number of electrons in the two half reactions to be the same. This problem also references a basic solution, so you can balance as fo...

Chem_Mod

Find the standard reduction potentials and look for one that is more positive than that of Cr3+ -> Cr but less positive than that of Pb2+ -> Pb. This means it would be reduced, thereby oxidizing Cr, but would not oxidize Pb

Chem_Mod

Your answer is correct.

I updated this file:
https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf

Chem_Mod

See this previous post: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=140&t=5159&p=12748&hilit=use+a+mixed+equilibrium+expression%2C+containing+both+concentrations+pressures&sid=9147d6ac6f16ea9bdc08d91e453db005#p12748 Bars and atmospheres are both OK to use since they are appro...

Chem_Mod

The second method is to balance as if in acidic solution, then at the end, add equal amounts of OH- on each side of the reaction to neutralize all the H+ into H2O. This will leave you with H2O and OH-

Chem_Mod

Your values seem correct but when I plugged them into the equation I got a different cell potential from what you have inputted. Perhaps you made a calculation mistake or are maybe using log instead of ln?

Chem_Mod

Using the standard reduction potentials is a good idea :)

Chem_Mod

It is because for these ion selective electrodes, you would use an electrode containing the same ion as whatever solution whose ion concentration you are trying to measure. For instance, in the pH electrode, this electrode contains H+ ions and would be used to measure [H+] of different solutions

Chem_Mod

Yes, you are correct. Include Pt(s) when it is needed as an inert electrode

Chem_Mod

Balance the Cl by adding stoichiometric coefficeint of 2. Then add H2O to balance oxygen. Finally add H+ then electrons.
For the other half-reaction, balance H by adding H+, then balance electrons

Chem_Mod

n is the number of moles of electrons that are transferred per reaction. Ag --> Ag+ involves the transfer of 1 mole of electron. Zn--> Zn2+ involves the transfer of 2 moles of electrons. You can see by the change in charge of the substances that are involved. E naught is 0V because it is the cell po...

Chem_Mod

Is there a specific homework or textbook problem that corresponds to this question? If so, please indicate which one it is

Chem_Mod

First, add up the total charges on each side of the equation. Without the e-, you have -2+14=+12 on the left and 2(+3)=+6 on the right. You will then add 6e- to create a charge of +6 on the left so that the charges are the same on both sides.

Chem_Mod

Great answer above. As there are many questions on 6L.3 I'll add additional comments to assist everyone with similar and related questions. In cell diagrams the left side represents the anode (oxidation half-reaction) and the right side the cathode (reduction half-reaction). The conducting electrode...

Chem_Mod

Changing the concentration of the electrolyte alters the resistance. Lower concentration correlates with higher resistance. Higher resistance means a lower current flow, for a given applied voltage. So lowering the concentration will lower the current flow.

Chem_Mod

2H2O + 2e- = > H2 + 2OH- cathode
H2O => 1/2 O2 + 2H+ + 2e- anode

Chem_Mod

E = Eo- 0.05916 V/n(log Q) or E = Eo- RT/nF(ln Q ) are the two versions that can be used interchangeably when calculating for a reaction not at equilibrium based on the Q value.

Chem_Mod

Hi everyone!

Happy Week 8 :)

Here's the worksheet for tonight's session. See ya'll at 6!

*TYPO: The first reaction given in 3b should have 2H2O.

*Answers are posted!

-Matt

Chem_Mod

Another way of thinking about it is looking at the oxidation number: if it goes down, it is reduction. For example, from 6+ -> 3+, the number is going down, so the number is reducing. Similarly, if you are going form 1- -> 3-, the number is reducing.

Chem_Mod

The redox reactions will be done in basic or acidic conditions, and the question should specify this in some way. Sometimes it will tell you directly that it is in acidic conditions, or sometimes it will have an acid in the equation. In each case, you can use the H + ions to balance. Similarly for b...

Chem_Mod

The general idea of electroplating is just that you are using the base metal (in the example, Fe) as a passive electrode to donate electrons to reduce the plating metal cations (in the example, Cr) to metal. For example, if using trivalent chromium, Cr^{3+}(aq) +3e^{-}\rightarrow Cr(s...

Chem_Mod

Since both pressure (1 atm approximately equal to 1 bar) and concentration are referenced (standard reduction potentials) with respect to standard conditions (gases 1 bar, solutions 1M), one can use (bar or atm) and molarity in the Nernst equation.

Chem_Mod

The [H 3 O] + concentration difference in the electrode and in the solution generate a voltage difference which is measured, and using the Nernst equation the concentration of [H 3 O] + is calculated. More specifically the pH (log [H 3 O + ]) is calculated. Re-watch my lecture discussing the applica...

Chem_Mod

These ion-selective electrodes measure the concentration of the product or the reactant. For example, the pH meter (glass electrode) measures the H+ concentration, where all variables of the Nernst equation are known except for the concentration of the product or reactant.

Chem_Mod

Bethany Yang 2E wrote:Oh man, sorry to hear about the internet. If you are in the chem groupme, there is a drive with the UA worksheets that people post in! I find it helpful when I can't make it to a session to do it after on my own. People often post the answers too so it helps!

Great. Thank you. This also helps.

Chem_Mod

Most likely tomorrow, Thursday. CCLE does not allow bonus points. It forces all calculations out of total points (84). For example CCLE calculates 80/84 as 95%. I am implementing a work around so that everything is calculated out of 80 points. I will explain in class email Thursday. CCLE will show t...

Chem_Mod

Fortunately I organized over 60 hours a week of zoom sessions. Attend UA and TA hours that are available when your internet is on. Also many post their worksheets, etc., on Chemistry Community which is available 24/7. My lectures and Sapling Learning are also available 24/7. Hope this helps and thin...

Chem_Mod

Charlotte Adams 1A wrote:
Arya Adibi 1K wrote:As stated by Chem_Mod, Whenever a conducting electrode is needed, use Pt.

You would need a conducting electrode whenever either the reactants or products are aqueous and don't have a solid conductor.

Just to clarify, can a conducting electrode be an anode or a cathode?

Yes

Chem_Mod

Anyone working on this. :-)

Chem_Mod

Correct.

To determine a favorable process need:

Total entropy > 0
or
Gibbs free energy system < 0

Chem_Mod

Thank you so much for the part 2! Reading this is really inspiring and motivating, especially right now during week 8. Wishing you the best of luck!! Thank you so much! Once I graduate in April, I'm hoping to have a bit of time where I can do a part 3 and detail the process of residency application...

Chem_Mod

I was wondering what particular classes made you interested in medicine/ particular clubs? As an MCDB major, I really enjoyed many of the upper division courses. My favorite was a stem cell bio course. To me it's just fascinating to study the pluripotency of various cells. I think going into colleg...

Chem_Mod

I'll use the example I gave in class.

You have two beakers. One with 1L water. The other empty.
When does the maximum difference in water volume occur?

Ans. When no water is flowing.

Chem_Mod

The Appendix in the textbook list the standard reduction values and half-reactions.

Chem_Mod

For a particular redox reaction, BrO− is oxidized to BrO 3 - and Ag+ is reduced to Ag . Complete and balance the equation for this reaction in basic solution. Can someone explain how to balance the half reaction BrO- --> BrO 3 - for this problem? Balancing the half reaction with Ag was similar to t...

Chem_Mod

We are covering electrochemistry this week.

Next week I will discuss 1st order, 2nd order, and zero order reactions.
When I cover reaction mechanisms it will be clear why we focus on these three in general chemistry.
Hint, they are the most common.

Chem_Mod

For a second-order reaction, the half-life is inversely related to the initial concentration of the reactant (A). t 1/2 = 1/k[A] o As the reactant concentration decreases the half-life is longer (slower reaction as there are fewer bimolecular collisions due to lower reactant concentration). Converse...

Chem_Mod

Hi Marlene, Not a resource, but I can try to explain the thought process I use. I'll go through one of the book self-tests to help illustrate this. "Copper reacts with dilute nitric acid to form copper(II) nitrate and the gas nitric oxide, NO. Write the balanced net ionic equation for the react...

Chem_Mod

Most of these replies are correct. The salt bridge shuffles ions between the two sides to balance out the charge of the ions being generated/reacted away on either side of the cell. For example, at the anode during the discharge of a battery, positive ions are being created through oxidation of the ...

Chem_Mod

Correct, the moles used in the Nernst equation are the moles of electrons being transferred from the oxidized species to the reduced species. When forming a balanced redox equation, you are looking to balance the number of electrons so an equal number are donated by the species being oxidized and ac...

Chem_Mod

Yes, whenever Q<K the reaction will be spontaneous. We can show this by relating standard Gibbs free energy to the equilibrium constant K, \Delta G ^\circ = -RTlnK and the equation for the nonstandard Gibbs free energy \Delta G = \Delta G^\circ +RTlnQ you get the equation \Delta G = RTln(\frac{Q...

Chem_Mod

It is experimental observation. A battery can do work (chemical energy in a redox reaction converted to electrical energy): delta G is negative for the redox reaction. When a battery is recharged the reverse redox reaction occurs: delta G is positive. Here electrical energy is converted to chemical ...

Chem_Mod

To measure a voltage difference one needs two electrodes. Cannot measure a voltage difference with one electrode. Correct, I give all the standard reduction half reactions and their E o (standard voltage) values that were measured under standard conditions while connected to the standard hydrogen el...

Chem_Mod

The most common reference electrode is the standard hydrogen electrode (SHE). This is the reference electrode I discuss in class. There are other reference electrodes: https://en.wikipedia.org/wiki/Reference_electrode The calomel electrode uses mercury, but given its toxicity and environmental impac...

Chem_Mod

The Na⁺ and Cl^- diffuse through the salt bridge.
The salt bridge is a high water content gel-like material that enables ion diffusions.

Good questions. In class I focus on the main concepts. I think I mentioned the salt bridge is gel-like.

Chem_Mod

In a redox reaction, does the electrolyte used make a difference in the reaction? Or do electrolytes serve the same purpose and thus it is not important to know which electrolyte is used? Since electrolytes are not directly involved in the redox reaction (not oxidized or reduced) they are less impo...

Chem_Mod

Think of "n" as the number of moles of electrons that would be transferred during this reaction. To do this, you can break down the reaction into two simple half reactions. In doing so, you will see that two moles of electrons are being used to reduce 2 Ce 4+ to 2 Ce 3 +. In general, it is...

Chem_Mod

Chemical reactions produce electrical energy via reactions such as redox (reduction and oxidation) reactions. In these reactions, the species that is oxidized gives up electrons at the anode, where it is then transferred to the cathode where species gain electrons to be reduced. It is through the tr...

Chem_Mod

Audra Mcleod 3G wrote:I unfortunately still have a fair amount of work to do over the weekend for Chem 14BL, but right after the midterm I went out to eat with my family which really helped unwind and forget about chemistry!

But good food is good chemistry.
:-)

Chem_Mod

May you give a little more information about the problem, so that the text does not need to be referenced please. Thanks :)

Chem_Mod

Yes they are.

The ignition is to overcome the energy barrier (which is kinetics), it does not change the thermodynamics (favorable reaction).

Chem_Mod

Please post all jokes here: viewtopic.php?f=156&t=4291

Thank you

Chem_Mod

Exothermic reactions release heat. If the reaction is done in a calorimeter, the heat will be absorbed by the calorimeter, which will then increase in temperature. So if given the change in temperature of the calorimeter and its heat capacity, you should be able to find how much heat was released by...

Chem_Mod

If the substance is changing phase, it is not at equilibrium. It will be at equilibrium once the process is complete

Chem_Mod

Don’t get hung up on the negative sign; it’s the same as giving a positive answer when one is asked how much energy is released

Chem_Mod

At constant volume, dU=q and q=nCvdT

Chem_Mod

Check out the free energy of formation for each molecule. If formation is spontaneous than the reverse reaction of decomposition is not spontaneous, meaning the molecule is stable relative to its constitutive elements, which is what it would decompose into

Chem_Mod

(a) are all aligned in the same direction There is one unique structure: all 64 molecules are aligned in the same direction degeneracy, W = possible positions N = 1 64 (b) lie in any one of four orientations with the same energy each molecule can have one of four possible orientations degeneracy, W ...

Chem_Mod

Enthalpy is q at constant pressure, so in b you can think of finding enthalpy as finding how much heat you would have had to add at constant pressure instead of constant volume to reach 373K. dU= q+w = dH-PdV at constant pressure dH= dU + PdV then using PV=nRT, dH= dU + nRdT At constant volume w=0, ...

Chem_Mod

Adiabatic means heat does not leave the system. If your reaction is occurring in an adiabatic calorimeter, the temperature of the calorimeter’s contents can still change, the heat will just not be lost to the surroundings

Chem_Mod

Please include the whole problem in your post

Chem_Mod

Just consider the moles of gas because you are using the ideal gas law, which is only for gases.

Chem_Mod

Yeah you're on the right track. For this question you want to assume that these events occur independently and not at the same time. So you can use Cv when calculating the change in entropy for a temperature change because now you assume that the volume is constant.

Chem_Mod

There is a temperature and a volume change, but for this question you want to assume that these events occur independently and not at the same time. So you can use Cv when calculating the change in entropy for a temperature change because now you assume that the volume is constant.

Chem_Mod

Hello everyone!

I attached my midterm review worksheet below and the answer key as well. I hope it is helpful and best of luck on your midterm tomorrow! You guys got this :)

Kate

Chem_Mod

TA office hours are as scheduled (unless there is overlap/delay with proctoring for that TA).

UA sessions are as scheduled and not impacted by proctoring.

Chem_Mod

These topics were covered in MT1 (and are not covered in MT2):

• Calculate an overall reaction enthalpy from the enthalpies of the reactions in a reaction sequence (Hess's law/approach).

• Use bond enthalpies to calculate an unknown reaction enthalpy.

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