CHEMISTRY COMMUNITY

ValerieChavarin 4F

If there is no conductive solid already participating in the redox reaction, then you would need to [censored] an inert metal such as Pt(s)

ValerieChavarin 4F

Yes, they are the same. The other kind of cell that was mentioned but not studied in class was the concentration cell.

ValerieChavarin 4F

State functions, such as enthalpy, entropy, and Gibb's free energy, are dependent on the initial and final values. However, in the case of heat (q) and work which are pathway functions, the 'path' taken or how much is done is important.

ValerieChavarin 4F

Start off by balancing the oxygen atoms by using H20. From there depending on the pH, you use H+ for acidic solutions and OH^- for basic solutions to balance out the Hydrogens.

ValerieChavarin 4F

No, H+ is only used to balance a redox rxn in acidic conditions. In the case of a basic solution, you MUST use OH^-

ValerieChavarin 4F

Pt(s) is used when there is no conductive solid participating in the reaction. It is an inert metal, therefore will not affect the redox rxn.

ValerieChavarin 4F

E° refers to standard conditions, such as 1M, 1atm, 25°C (This is found using the cathode-anode), while E is not under these same conditions and can be found using the Nernst equation. As long as you have one of them known, the Nernst equation can be used to find the other.

ValerieChavarin 4F

It's on the outline, so it can appear on the final. However, I think as long as you understand the basics, you should be fine.

ValerieChavarin 4F

enthalpy is the change in heat at constant pressure (at which is can be equal to ), while q is measure of heat under varying conditions.

ValerieChavarin 4F

You'll need to use the standard enthalpy of formation and the ∆Hr° = ∆H° $\sum$ (products) - ∆H° $\sum$ (reactants)

ValerieChavarin 4F

If the values are not given, you might have to use the ∆G° = ∆H° - T∆S°equation to calculate enthalpy. If not, then the value might be zero. I don't believe there should be much confusion regarding this on the final.

ValerieChavarin 4F

changing the size does not affect the cell potential

ValerieChavarin 4F

First, make sure you balance the charges of each half reaction (for the most part you should need electrons to balance the half rxns). If then, you notice that both half rxns have the same charges then you should be fine as the electrons eventually cancel out.

ValerieChavarin 4F

Since all the standard cell equations are written as reduction (which coincide with the cathodes), then the anode must be flipped to get the oxidation half reaction.

ValerieChavarin 4F

Yes, H2O is not included in the cell diagram

ValerieChavarin 4F

The effect of the dilution depends on where it is occurring. If the cathode is diluted then the cell potential decreased. However, if the anode is dilution there is an increase in cell potential.

ValerieChavarin 4F

Changing the concentration of the electrode might result in an increase, however for the mass it should not affect cell potential.

ValerieChavarin 4F

Kaylee Clarke 1G wrote:what of the elements such as platinum that seem to be attached to a cell diagram?

Pt(s) is used as a conductor when there is no conductive solid participating in the redox reaction. It is added to the cell diagram but does not participate.

ValerieChavarin 4F

Yes. Since Pt(s) is an inert metal, it is commonly used in the absence of a conductive solid.

ValerieChavarin 4F

for basic solutions, you balance out the oxygens with H2O and then use OH^- to balance the hydrogens

ValerieChavarin 4F

the oxidation number helps identify whether a species is oxidized or reduced. Using the oxidation number, you can determine how many electrons are necessary to balance the charges of a half reaction.

ValerieChavarin 4F

the anode is the side with the lower concentration as it is the side where oxidation occurs.

ValerieChavarin 4F

Eº refers to standard conditions while E is not and therefore is affected by changes such as temperature, concentration etc.

ValerieChavarin 4F

Yes, we will be given that information as it is necessary to determine how to balance the redox rxns. In some cases, we might be given the pH or even pka or pkb to determine pH.

ValerieChavarin 4F

You get the half reactions by looking at the species used in the galvanic cell. It helps to think about AnOX RedCat. The left side or anode is where the oxidation reaction occurs (a loss of electrons) and the right is where the reduction occurs (gaining of electrons).

ValerieChavarin 4F

Pt(s) is an inert metal, so it won't affect the reaction but is necessary to serve as conductor.

ValerieChavarin 4F

The salt bridge is essentially meant to neutralize solutions and balance the charges which occur due to the transfer of electrons in a redox reaction.

ValerieChavarin 4F

Familiarize yourself with the LEO (loss of electron, oxidization) and GER (gains electrons, reduction) acronyms, they help identify what is being reduced and oxidized. From there you also have to know that the species that is being reduced in the 'oxidizing agent' and the one being oxidized is the '...

ValerieChavarin 4F

You take a look at the change in oxidation number. The molecule that is being reduce deceased in the oxidation number such as Cu2^+(aq)→ Cu(s), while the species being oxidized increases in the oxidation number Zn(s) → Zn2+(aq) *These examples can be found in RedOx reactions Part 1 sheet with furthe...

ValerieChavarin 4F

for half reactions you have to play attention on whether the solution is acidic or basic, this will tell you if you should use H+ and H2O (for acidic solutions) or OH- and H2O (for basic solutions). Start off with adding H2O to the side that needs it to balance out the O atoms. From there, depending...

ValerieChavarin 4F

You don't really KNOW which is T1 and K1, you more assign given values. Despite this, whichever you choose to be T1 and K1 must correlate with one another. I usually label the initial temp and k as T1 and K1, then the second temp as T2 to find K2 (the unknown value). As long as you keep the related ...

ValerieChavarin 4F

In the case of ∆G° = ∆H° - T∆S°, which indicates standard condition yes. The combustion of gas will be spontaneous due to the negative enthalpy value, the release of heat through an exothermic reaction, and a positive (increasing) entropy.

ValerieChavarin 4F

You have to look at the number of electrons transferred in the redox reaction. (Essentially, the number of electrons that balances the combined half-reactions.)

ValerieChavarin 4F

It is important to note where the dilution is occurring as it will affect cell potential differently if it is in the cathode or anode. You can use E˚(cell) = E˚(cathode) - E˚(anode) to get a better understanding. When the cathode is diluted, cell potential decreases due to the lower cathode. Therefo...

ValerieChavarin 4F

The "o" indicates standard conditions, (in temperature, pressure and concentration. ∆G° can be used to find ∆G and vice versa in the equation ∆G = ∆G° + RT ln Q

ValerieChavarin 4F

Enthalpy is considered a state function as it is dependent on initial and final states.

ValerieChavarin 4F

w=-P $\Delta$ V
The equation for work has a negative. If the change in volume is also negative, then work will be positive. If not, then it will be negative.

ValerieChavarin 4F

qv is at constant volume, with changing pressure, while delta H is heat transfer at constant pressure.

ValerieChavarin 4F

q refers to the energy, as heat, transferred due to a change in temperature. $\Delta H$ is the heat transferred at constant pressure

ValerieChavarin 4F

They are equal in the case of solids and liquids but change for gases due to compression and expansion.

ValerieChavarin 4F

Both can be used for a gas, however they do not yield the same answer. The p subscript indicates constant pressure while the v is for volume.

ValerieChavarin 4F

0°C= 273.15 K. When calculating change in temp, if you have 283.15K-276.15K or 10°C-3°C, the change in temperature is essentially the same.

ValerieChavarin 4F

C is a closed system as the clariometer does not allow the transfer of energy to the suroundings.

ValerieChavarin 4F

A lower pKa indicates a stronger acid and a lower pKb is a stronger base, as pKa + pKb = 14

ValerieChavarin 4F

They aren't included, since you wouldn't include them in the K expression.

ValerieChavarin 4F

You take a look at the K value. If it is less than 10^-3, then you can approximate. If the K value is 10^-3, it can get a bit tricky, so I would still do the full calculation.

ValerieChavarin 4F

Since K=products/reactants. A small K value indicated that the numerator is being divided by a large denominator. So there are more reactants.

ValerieChavarin 4F

This in relation to concentration (n/V). It's used mainly to calculate for the unknown variable.

ValerieChavarin 4F

No, test one is only for chemical equilibrium and acids and bases

ValerieChavarin 4F

It depends on what asked and given. If you are calculating molar concentration then you would use Kc.

ValerieChavarin 4F

When it comes to alphabetizing, you look at the beginning of the compound itself, not the prefixes used.

ValerieChavarin 4F

I believe that the ligands on the table are the principle ones that may come up during the final.

ValerieChavarin 4F

The subscript of two shows that there is double of what is inside the parentheses, therefore 2 N. The third one is the N that is on the outside in the HN.

ValerieChavarin 4F

It also includes bidentate complexes, as chelating ligands refers to those that can make multiple (2 or more) bonds to the central metal atom.

ValerieChavarin 4F

iron is a naming exception when it comes to adding the -ate suffix, instead of ironate. We use the roman name of ferrum, then add the suffix.

ValerieChavarin 4F

I'm confused on these two terms. The textbook uses the form cyanido, but Lavelle wrote cyano for a naming example in lecture. Can they be used interchangeably, or are they different?

ValerieChavarin 4F

-ate is used with there is a negative charge on the complex and is added to the end of the TM

ValerieChavarin 4F

3. I don't think there is a preference, but the textbook uses OH2 a lot, so i would stick with that format.

ValerieChavarin 4F

Yes since the regions of electron density correlates with the hybridization. 2 regions=sp and so on like without lone pairs.

ValerieChavarin 4F

You begin with the Lewis structure that will provide you with the regions of electron density of the center atom. Form their you can determine the hybridization. 2 regions corresponds with $sp$ ; 3= $sp^{2}$ and so on.

ValerieChavarin 4F

For the right carbon you use $sp$ hybridized orbitals while the left carbon uses $sp^{3}$

ValerieChavarin 4F

I was a little confused about that too.

ValerieChavarin 4F

Sydney Myers 4H wrote:
Daniel Martinez 1k wrote:Si2F2

Is this due to the shape? because Si2F2 would have a more linear structure, compared with a round structure of SiF4, and higher surface area makes for higher intermolecular forces, specifically London Dispersion Forces.

I believe this question is dependent on polarity, no shape for size.

ValerieChavarin 4F

This question has to do with polarity, not size. Since AsF3 is polar, it has stronger dipole-dipole interaction than AsF5, a nonpolar molecule.

ValerieChavarin 4F

A hydrogen bond can occur for one lone pair of N,O, and F. Therefore, a nitrogen atom with 2 lone pairs can form 2 hydrogen bonds.

ValerieChavarin 4F

The question is asking what are the IMF of each molecule. For H2SeO4, due to its polarity, it has dipole-dipole and induced-induced. On top of that, it can also form hydrogen bonds.

ValerieChavarin 4F

London forces depends on the size of a molecule. Since larger molecules are more polarizable they form stronger London forces.

ValerieChavarin 4F

Butanol has a higher melting point, due to its ability to form hydrogen bonds.

ValerieChavarin 4F

Dipole-dipole interactions occur in polar molecules, while an induced dipole has to do with the uneven distribution of charges, resulting form a shift in electrons.

ValerieChavarin 4F

I'm confused between the melting point and the boiling point of a compound? Is this referring to the same thing?

ValerieChavarin 4F

Hydrogen bonds are formed between H and N,O,F atoms that have lone pairs.

ValerieChavarin 4F

It's good to start off drawing the Lewis Structure. From there you can determine whether the molecule is polar or nonpolar and capable of forming a hydrogen. If a molecule is polar then there is a dipole-dipole interaction. All molecule tend to have a induced-induced interaction.

ValerieChavarin 4F

The dipole-dipole interaction in the solid phase is stronger

ValerieChavarin 4F

In my discussion we did a problem that was essentially the following:
Calculate the ionization energy of RB given that a radiation with the wavelength of 58.4nm produces electrons with a velocity of 240km/s, when it hits RB.

Why do we solve for the work function?

ValerieChavarin 4F

The noble gasses do not follow the trends of IE and EA due to their full valence shell which makes them pretty much unreactive.

ValerieChavarin 4F

$m_{l}=-l,...0,...l$ , so if the l= 1 then the magnetic numbers are -1,0,1

ValerieChavarin 4F

If you take a look at the magnetic quantum numbers $m_{l}$ for the d, they are -2, -1, 0, +1, +2, which are 5 in total. Each orbital is able to hold 2 electrons, therefore the d-orbital can hold a total of 10 electrons.

ValerieChavarin 4F

I got that far, as to find the Energy of the ejected electron, but how do I cancel the mol^-1 of the work function to be able to add it to the $E_{k}$ ?

ValerieChavarin 4F

I was confused about this too, since he went over it really quick. The textbook provides a bit more information on it.

ValerieChavarin 4F

The sampling isn't mandatory but it is highly recommended as it provides you with extra practice and questions that are useful to study for the exams.

ValerieChavarin 4F

I was confused about this too, but its the first equation with the 4pi.

ValerieChavarin 4F

I don't believe we have to know what frequencies correspond to what, but i do think it's good to be familiar with the visible light spectrum. This questions was more based on vaguely knowing the Balmer and Lymen series.

ValerieChavarin 4F

No this will not be tested, since it was just briefly mentioned.

ValerieChavarin 4F

8. B) A newly designed laser pointer with a certain frequency is pointed at a sodium at a sodium metal surface. An electron is ejected from the metal surface with wavelength 1.10nm. What is the frequency of the light from the laser pointer? The work function of sodium is 150.6 kJmol^1. I understand ...

ValerieChavarin 4F

I think it's something that's meant to be more conceptual and help us understand what we are learning, not really something we have to focus a lot on since we didn't have questions on it.

ValerieChavarin 4F

It could be due because your velocity is not squared.

ValerieChavarin 4F

A photon does not have mass which is required for the de Broglie's equation where $\lambda =\frac{h}{mv}$

ValerieChavarin 4F

The new formula is derived from solving for the frequency in $c=\lambda \nu$ which gives you $\nu = c/\lambda$ . You can then plug in this new value of $\nu$ into E=h $\nu$ to get $E=\frac{hc}{\lambda }$

ValerieChavarin 4F

Wavelength is measured in meters or other prefixes of meters such as nano and pico. It's easy to remember that units should cancel, c=λν which essentially is (m/s)=m x s^(-1)

ValerieChavarin 4F

Increasing the intensity of light will not change the frequency of a wave, it will only change the amplitude of the wave.

ValerieChavarin 4F

An equation sheet will be provided during every test, so i don't think you have to worry much about memorizing the equations but more about how to apply them.

ValerieChavarin 4F

I don't believe we have to memorize the exact ranges, but possibly their placement based on wavelength. I do think you should know the range of visible light because it is common.

ValerieChavarin 4F

The actual yield is less than the theoretical yield due to impurities and side reactants. I don't believe it's a concept you'll have to justify in the lab. You just need to familiarize yourself with it.

ValerieChavarin 4F

I'm still a little confused as to when I should use scientific notation or not. For example, the textbook solution for Fundamentals E.23 part a is 0.0134 mol Cu^2+, while the solution for E.23 part b is 8.74x10^-3. Why did they decide not to use scientific notation for part a? I've noticed that sci...

ValerieChavarin 4F

When there is only one reactant in an equation, there is no limiting reactant. When calculating the limiting reactant of a reaction with multiple reactants, you need to use mass and molar mass. Then you find the theoretical/maximum yield.

ValerieChavarin 4F

The question usually specifies which unit is preferred in the answer. If there is a certain question you are referencing, try posting it verbatim, in order to clear up the confusion.

ValerieChavarin 4F

This question requires the M1V1=M2V2 as it is a dilution problem. The 5.00g is necessary to find the moles of KMnO4. From there, you can calculate the M1, divide your answer by .15L (after unit conversion of mL). The 20.00mL or .02000L is your V1, with the new volume 250mL or .250L as your V2

ValerieChavarin 4F

I was confused on that question too, but you have to make sure to do a unit conversion of mmol to mol and to use the M=n/v. This question also required the use of the ratio of Na:Na2CO3, which is 2:1 and CO3:Na2CO3, 1:1.

ValerieChavarin 4F

The unit conversion is necessary for dimensional analysis. In a calculation that requires the use of a molecule's molecular weight, we cannot simply go from mg to moles. The mg nor the g would cancel out.

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