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Peelee
2014-04-02, 12:19 AM
I'm going back to school at 29. I'm married and have a full-time job, so university-funded tutors are inaccessible to me, and I can't afford one on my own. I really like science and math, and am pretty good at it once I understand it. Problem is, I can't understand it; my current Gen Chem II professor, despite being obviously very good at the math of it, is terrible at teaching conceptually. I'm barely passing, and I dread Organic Chem next semester if I don't grasp what's going on right now. So I turn to you, kind sciencey-kings of The Playground, to help me understand this better so I can get a better life for my future family!

Most pressing issue at the moment:

Electrochemistry

*I'm going to start bringing up examples from the book, likely from some homework. I don't care about the answer, other than to possibly check my own answer. I care about how to get the answer. Not trying to cheat, just trying to understand*
Example given:
Write the shorthand notation for the following equation.

Pb(s) + Br2(l) --> Pb2+(aq) + 2Br-(aq)
Pb2+(aq)+2e- -->Pb(s) E*= -0.13 V
Br2(l)+2e- --> 2Br-(aq) E*= +1.07 V

Solution:

Pb(s) | Pb2+(aq) || 2Br-(aq) | Br2(l) | Ti(s)

Where did the Titanium come from? If this question was asked on a test, would the fact that Titanium is an electrode be given, or would I have to figure that out, and if so, how?

Why does the titanium not go into the standard chemical reaction notation? Is it excluded due to being a spectator?

Can anyone explain how buffers work in acid/base equilibrium reactions? Why don't they affect the pH before actually buffering against acids? Why do buffers work like they do?

Anyone who wants to help, you have my dearest thanks.

Don't be afraid to treat me as if I'm not a smart man.

Livius
2014-04-02, 04:12 AM
Here is a short overview of acid-base buffers.

Assumptions:
The reactions are done in aqueous solution (H2O concentration is always large).
Mixing is instantaneous (the solution is always completely homogenous).
Reactions happen instantly (the solution is always at equilibrium).

For the reaction {A+B -> C+D} there is a k which is a constant (at constant temperature) based on the properties of A, B, C, and D.
At equilibrium, k = [C][D]/[A] (k is larger if the reaction happens more). This changes slightly in form if heat is any of A,B,C,D but not in overall effect.
[B]LeChatelier's Principle is based on the fact that k is constant. It says that if any component of the reaction is changed externally, the equilibrium concentrations change to oppose the new change. This means that if you add A to a mix of A, B, C, and D without changing anything else, some of the extra A will react with B to make more C and D.

Buffers are generally a mix of a weak acid/base and one of its salts (examples: NH3 + NH4NO3, H2CO3 + NaHCO3, HCH3CHO2 + NaCH3CHO2)

Relevant Equilibria:
1) H2CO3 + H2O <-> HCO3- + H3O+
2) HCO3- + H2O <-> H2CO3 + OH-

Relative Concentrations of reagents:
H3O+ ~= OH- << H2CO3 ~= HCO3- ~= Na+ << H2O

Initial effect: Increase [H3O+].
Reactions: Equilibrium 1 shifts to the left to oppose (but not entirely negate) the change in [H3O+]. [HCO3-] decreases slightly. Equilibrium 2 shifts slightly to the left to oppose the change to [HCO3-]. [OH-] decreases slightly.
Net result: Solution becomes slightly more acidic (but much less than if there were no buffering reactions).

Initial effect: Increase [OH-].
Reactions: Equilibrium 2 shifts to the left to oppose (but not entirely negate) the change in [OH-]. HCO3- increases slightly. Equilibrium 1 shifts slightly to the left to oppose the change to HCO3-. [H3O+] decreases slightly.
Net result: Solution becomes slightly more basic (but much less than if there were no buffering reactions).

Brother Oni
2014-04-02, 07:26 AM
GENCHEM II

Can anyone explain how buffers work in acid/base equilibrium reactions? Why don't they affect the pH before actually buffering against acids? Why do buffers work like they do?

pH is the concentration of H+ ions in solution for acids (OH- for bases after a conversion step). Buffers work by mopping up small amounts of these ions so the pH isn't overly changed by the addition of H+/OH-.

Livius covered the mechanics of how and why they do this, although I would have mentioned pKa as well.

As for why buffers don't affect the pH, the assumption is that the reaction is performed in the buffer solution. If you had a solution and dumped a load of buffering reagent in, then the pH would change. It's also possible to add so much acid/base that the buffering agent is overwhelmed.

Buffer solutions (http://en.wikipedia.org/wiki/Buffer_solution)
pKa (http://en.wikipedia.org/wiki/Acid_dissociation_constant)

Knaight
2014-04-02, 07:20 PM
I'm going to break this down to a pretty simple, conceptual level. Strong and weak acids can be treated as if they completely disassociate (this simplification stops working after Gen Chem II, but whatever). So, you have a system.

BH<--->B- + H+
Where B is a base, and BH is the conjugate acid.

Alternately, you have
AOH<--->A+ + OH-
Where A is an acid and AOH is the conjugate base.

By LeChatellier's principle, if you add something on one side of the reaction, the reaction shifts to the other side. So, lets look at the scenarios here. Note that I am using Bronsted Lowry bases and acids here, as there are some Lewis Bases that cause problems. Fortunately, said Lewis Basis tend not to show up in Gen Chem. Arhennius acids and bases are needed for the AOH system.

Acid to BH system: You add an acid, in H+. This causes the system to shift towards BH, using up some acid. Thus, it mitigates the effect.
Base to BH system: You add a base. It reacts with H+, dropping the amount on that side, so some more BH breaks up in reaponse.
Acid to AOH system: You add an acid, and it reacts with the hydroxide. With that down, AOH disassociates some more, partially restoring the original PH
Base to AOH system. You add a base, which inserts some hydroxide, so it reacts with the loose A to form more AOH, again moving back towards the original PH.

Now, lets break those buffers. First, we have the BH system. Lets add a strong base, in force - enough to use up all of the H+ ions that were attached to BH. Suddenly, the buffer isn't doing so much. Then there's the AOH system, where you can do a similar thing with a strong acid.

Peelee
2014-04-03, 03:22 PM
Hooray! I really needed that simplification. For the moment, I don't care about non-Gen Chem II or non-Bronstead-Lowry acid/bases, so that was perfect. Livius and Brother Oni, yall were also hugely helpful; I'm gonna need those "proper" bits for study. And Livius, I really liked the list of assumptions.

Can you go into a little more detail about strong vs weak acids/bases when reacting with buffers?

Also, new focus: starting on more thermodynamics starting tomorrow. What the crap is Gibbs Free Energy? How does it relate to entropy and enthalpy?

Knaight
2014-04-03, 03:43 PM
Hooray! I really needed that simplification. For the moment, I don't care about non-Gen Chem II or non-Bronstead-Lowry acid/bases, so that was perfect. Livius and Brother Oni, yall were also hugely helpful; I'm gonna need those "proper" bits for study. And Livius, I really liked the list of assumptions.
Honestly, buffer equations are typically not what Lewis Acids and Lewis Bases are really used for (except for the Bronstead Lowry subset). If you're not tracking electron movements, you probably don't need them, and that doesn't pick up majorly until O-Chem.

Can you go into a little more detail about strong vs weak acids/bases when reacting with buffers?
The short version is that with weak acids and bases you're balancing two equilibrium equations with a shared variable, which makes things more difficult. With strong acids and bases, you just flood the entirety of the OH- or H+ in, ignore the rest of the molecule completely, and go from there. Basically, take your buffer. You can treat it as k[A][B]=[C][D] (again, there are simplifications here). With a strong acid or base, [D] changes, and you then change the rest of them to accommodate the equation, getting k[A-x][B-x]=[C+x][Dnew+x]. It's a pretty simple equation. With weak ones, you get a whole lot of fun instead.

Also, new focus: starting on more thermodynamics starting tomorrow. What the crap is Gibbs Free Energy? How does it relate to entropy and enthalpy?
Gibbs Free Energy is basically a measure of thermodynamic potential. As for how it relates to enthalpy and entropy, G=H-TS, where T is temperature, and everything else should have a delta in front of it. The big things affected are spontaneity, as well as coming up in equilibrium constants. You'll also find it coming up again in electrochem.

Livius
2014-04-03, 05:36 PM
Also, new focus: starting on more thermodynamics starting tomorrow. What the crap is Gibbs Free Energy? How does it relate to entropy and enthalpy?

Gibbs Free Energy (G) determines the spontaneity of the reaction. A reaction is "spontaneous" when (delta) G < 0, which means it will happen automatically if the reactants are present.
As Knaight said, the equation for Gibbs Free Energy is (delta) G = (delta) H - T* (delta) S, where H is the enthalpy, T is the temperature, S is the entropy. (The deltas are there since it applies to the changes from the reaction itself; I'm going to drop them for the rest of this explanation for simplicity).

Enthalpy is the "heat" of the reaction; if H < 0, the reaction releases energy to the surroundings; if H > 0, the reaction absorbs energy from the surroundings. Example: C2H4 + 3O2 -> 2CO2 + 2H2O has H = -1411 kJ/mol.

Entropy is the "disorder" of the reaction; generally, more molecules and more complicated molecules have higher entropy. 2NI3 -> N2 + 3I2 has S > 0.

If H < 0 and S > 0, the reaction is always spontaneous.
If H > 0 and S < 0, the reaction is never spontaneous.
Otherwise, the spontaneity of the reaction depends on the temperature of the reactants.

Knaight
2014-04-03, 06:17 PM
Gibbs Free Energy (G) determines the spontaneity of the reaction. A reaction is "spontaneous" when (delta) G < 0, which means it will happen automatically if the reactants are present.
To expand on this - it means it will happen automatically, eventually, provided the conditions hold. For instance, (delta) G < 0 for change in carbon bonds for elemental carbon to change from diamond to graphite. This doesn't mean that anyone has to worry about their diamonds forming graphite. Also, I'm about 90% sure that this example is stolen from Chemistry: The Central Science, so I'm throwing that in as credit.

Entropy is the "disorder" of the reaction; generally, more molecules and more complicated molecules have higher entropy. 2NI3 -> N2 + 3I2 has S > 0.
More precisely, it's a measure of the number of distinct micro-states. Entropy increases with phase change, because the way molecules can be arranged increases with each change in phase. Breaking a molecule apart increases entropy, because the number of arrangements increases with the number of particles.

Peelee
2014-04-05, 12:41 AM
Quick moratorium on the ΔG, ΔS, and ΔH (though all that is becoming a lot clearer to me, I'm still pretty sure I'm gonna find a question or two when I start in on the homework)....

Head of the science department is teaching Analytical Chem next semester. From what I hear, it's only done every few years, the requirements are same as Organic Chem (>C in Gen Chem II), and the professor is really good at explaining things in a way I easily understand. Also, it sounds kinda fun. Assume there are no issues with scheduling. What advice do you guys have on taking
a.) both Organic and Analytical,
b.) Organic only, or
c.) Analytical only?

Brother Oni
2014-04-05, 06:36 AM
Quick moratorium on the ΔG, ΔS, and ΔH (though all that is becoming a lot clearer to me, I'm still pretty sure I'm gonna find a question or two when I start in on the homework)....

Head of the science department is teaching Analytical Chem next semester. From what I hear, it's only done every few years, the requirements are same as Organic Chem (>C in Gen Chem II), and the professor is really good at explaining things in a way I easily understand. Also, it sounds kinda fun. Assume there are no issues with scheduling. What advice do you guys have on taking
a.) both Organic and Analytical,
b.) Organic only, or
c.) Analytical only?

In my experience, most chemists loathe organic chemistry with a passion reserved for child molesters and people who talk in theatres.

Organic chemistry makes several assumptions about your basic chemistry knowledge - here's a good rundown of what's required (http://www.masterorganicchemistry.com/2010/06/28/gen-chem-and-organic-chem-how-are-they-different/). If you're a bit fuzzy on the fundamentals required, then you're going to struggle.

Analytical chemistry from the sounds of it, is mostly going to be focused on the techniques and methodology but some knowledge of the compound's chemistry is required to understand how the analysis work.

Knaight and Livius have more knowledge than me of the actual classes (British degree courses are significantly different to US ones), so I'd take whatever they say over me.

Knaight
2014-04-05, 05:48 PM
In my experience, most chemists loathe organic chemistry with a passion reserved for child molesters and people who talk in theatres.

O-Chem is divisive. There's the chemists that decide to specialize in organic chemistry and adore it, and then there's the ones that absolutely detest the subject. I can't say I've seen all that much in between the two.

I'm also on the side of adoring it - Organic Chemistry I and II are, hands down, by far, the favorite classes I have ever taken in any subject. That said, taking both it and Analytical Chemistry sounds like a good idea, provided that you don't decide to also take something like Differential Equations with the two of them. It might seem like a good idea at the time. It isn't.

Peelee
2014-04-06, 11:42 AM
In my experience, most chemists loathe organic chemistry with a passion reserved for child molesters and people who talk in theatres.

O-Chem is divisive. There's the chemists that decide to specialize in organic chemistry and adore it, and then there's the ones that absolutely detest the subject. I can't say I've seen all that much in between the two.

I'm also on the side of adoring it - Organic Chemistry I and II are, hands down, by far, the favorite classes I have ever taken in any subject. That said, taking both it and Analytical Chemistry sounds like a good idea, provided that you don't decide to also take something like Differential Equations with the two of them. It might seem like a good idea at the time. It isn't.

Knowing little to nothing about it, this is my impression of Organic:
https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/t1.0-9/1554397_10152103948944023_612079542_n.jpg
I fully expect to be in the camp of people who hate Organic Chem, but I would absolutely love to be pleasantly surprised. My tentative schedule was to have University Physics, Organic, and Analytical, until my wife pointed out that despite the fact that the labs all lined up on a single day, if the tests fell on the same day I would pretty much die. As of now, I'm thinking about finishing up Calc II and putting Physics off to another semester (which makes me really sad. If Kerbal Space Program has taught me anything, it's that I really like physics and I my NASA job application is clearly being held up by bureaucratic red tape and I should expect to hear back any day now), but math is also pretty fun. I'm still thinking Basically, as of the time of this writing, I don't have tentative schedule.

Also, assume I become a competent chemist. Would a minor in Math or a minor in Physics be better? Immediate goal is to get to grad school. End goal is to win a Nobel Prize, have at least a principle named after me, and be exactly filthy rich. Actual end goal is to go into space. I realize I'm far behind the curve and it's a ridiculous long shot, but it's friggin' space, and I wanna go there. Ideally the moon, but so long as it's space, I'll be happy. Seriously, you don't even know how happy

Knaight
2014-04-06, 12:21 PM
Knowing little to nothing about it, this is my impression of Organic:
https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/t1.0-9/1554397_10152103948944023_612079542_n.jpg
I fully expect to be in the camp of people who hate Organic Chem, but I would absolutely love to be pleasantly surprised.

It's pretty much all electrons moving. Sometimes it's lots of electrons moving around, and some mechanisms are fairly complex, but most of them are known. There's no magic involved, and only the occasional reaction where the mechanism is totally unknown.

Brother Oni
2014-04-07, 06:38 AM
O-Chem is divisive. There's the chemists that decide to specialize in organic chemistry and adore it, and then there's the ones that absolutely detest the subject. I can't say I've seen all that much in between the two.

I'm inclined to agree. The majority of the complaints I heard were from biologists grumbling that there's too much chemisty in it, while the chemists grumbled there's too much biology. :smalltongue:

Peelee
2014-04-07, 07:46 PM
I'm inclined to agree. The majority of the complaints I heard were from biologists grumbling that there's too much chemisty in it, while the chemists grumbled there's too much biology. :smalltongue:

Anything with biology has too much biology. Thanks all. Going for broke.

Now, back to the path to get me TO Organic.

Other than memorization of a given table, is there a way to identify relative strength of an acid/base for a given compound? Ex: which pair has the stronger conjugate base - H2CO3 or H2SO4? HCl or HF?

Livius
2014-04-07, 08:14 PM
Other than memorization of a given table, is there a way to identify relative strength of an acid/base for a given compound?

For acids, the equilibrium looks like this: HB + H2O <-> H3+ + B-. Ka = [H+]/[HB] determines how strong the acid is. The higher it is, the stronger the acid.

For bases, the equilibrium looks like this instead: B- + H2O <-> HB + OH-. Kb = [HB][OH-]/[B-] determines how strong the base is. Higher Kb are stronger.

For an acid and its conjugate base, Ka * Kb = 1. This means that stronger acids have weaker conjugate bases and vice versa.

The list of strong acids is HNO3, H2SO4, HClO4, HCl, HBr, and HI. They have Ka values >> 1.
Strong bases are generally alkali or alkaline earth metal hydroxides. Some examples are NaOH, Ca(OH)2, CsOH.

Therefore:

Ex: which pair has the stronger conjugate base - [b]H2CO3 or H2SO4? HCl or HF?

Peelee
2014-04-07, 08:32 PM
For an acid and its conjugate base, Ka * Kb = 1. This means that stronger acids have weaker conjugate bases and vice versa.

The list of strong acids is HNO3, H2SO4, HClO4, HCl, HBr, and HI. They have Ka values >> 1.
Strong bases are generally alkali or alkaline earth metal hydroxides. Some examples are NaOH, Ca(OH)2, CsOH.

So for the given examples, where no concentrations are given, you must know the strength of the given acids?

Livius
2014-04-07, 08:54 PM
So for the given examples, where no concentrations are given, you must know the strength of the given acids?

Yes. You'd have to know the strength of the given acids even if the acid concentration was given. "Strong" != "concentrated". Strong acids dissociate almost entirely, so 1M HCl solution has 1M H+ and 1M Cl- and a pH of 0. Weak acids don't dissociate very much, so 1M HCH3CH2O2 (acetic acid, ka = 1.8 x 10-5) has ~ 1M HCH3CH2O2, sqrt(ka) = 0.004 M H+, sqrt(ka) = 0.004 M CH3CH2O2-, and a pH of 2.7. They are equally concentrated, but HCl is much stronger.

Luckily, since ka only depends on the acid and solvent, it can be easily looked up.

Peelee
2014-04-07, 09:29 PM
Yes. You'd have to know the strength of the given acids even if the acid concentration was given. "Strong" != "concentrated". Strong acids dissociate almost entirely, so 1M HCl solution has 1M H+ and 1M Cl- and a pH of 0. Weak acids don't dissociate very much, so 1M HCH3CH2O2 (acetic acid, ka = 1.8 x 10-5) has ~ 1M HCH3CH2O2, sqrt(ka) = 0.004 M H+, sqrt(ka) = 0.004 M CH3CH2O2-, and a pH of 2.7. They are equally concentrated, but HCl is much stronger.

Luckily, since ka only depends on the acid and solvent, it can be easily looked up.

And that answered my next question as well. Awesome!

So "strong" and "weak" have no bearing on, say, how much it will hurt it you stick your finger in a big ol' jar of a basic solution, and the concentrations do. Correct?

Livius
2014-04-08, 01:37 AM
So "strong" and "weak" have no bearing on, say, how much it will hurt it you stick your finger in a big ol' jar of a basic solution, and the concentrations do. Correct?

Not exactly. The stronger acid/base will hurt more unless the concentrations are extremely unbalanced (say, 1M HCH3CH2O2 vs 0.004 M HCl). The reason for this comes from reaction kinetics.

Let's say you have a strip of zinc that you want to dissolve in acid by the following reaction: Zn + 2H+ -> Zn+2 + H2. (The rate at which this would happen) = (some constant) * (the surface area of the copper in the acid) * (the H+ concentration of the acid).

Using the solutions from my last post, strong acid 1M HCl and weak acid 1M HAc (Ac- = acetate anion, HAc = acetic acid), the [H+] is ~250 times greater (1 M / 0.004 M) for the HCl. This means that the reaction will happen 250 times more quickly and thus be much more vigorous. Assuming (probably dangerously) that the released heat can be dumped effectively to the environment and there is sufficient acid, the end result will be the same: a completely dissolved strip of zinc. However, if the HCl reaction took 1 minute to complete, the HAc reaction would take over 4 hours.

Knaight
2014-04-08, 01:43 AM
So "strong" and "weak" have no bearing on, say, how much it will hurt it you stick your finger in a big ol' jar of a basic solution, and the concentrations do. Correct?

How much it will hurt if you stick your finger in a big ol' jar of basic or acidic solution is probably best measured by PH. Strong acids and bases have concentrations much further from 7 than weak acids and bases do at the same concentration, and further from 7 corresponds to more painful. So, both of these have a bearing on the situation.

Concentrations are a big thing here. For example, NaOH is a very strong base, but if you put a drop of it in a gallon of water and stick your finger in it, you'll be totally fine. Similarly, acetic acid is on the weak end, but glacial acetic acid (maximum concentration) will produce a burning sensation.

With that said - at the same concentration, "strong" vs. "weak" makes a very big difference. I've worked with 6 M HCl quite a bit, and it's something I'd be worried about spilling on myself. Even 1 M HCl isn't something I'd like to come in contact with. 1 M acetic acid? I wouldn't want that getting in a cut, but skin contact isn't really of that much concern. 6 M acetic acid? That could actually hurt, but it really wouldn't be that bad.

If you want to look at one number, look at PH. Generally speaking, things with a PH of about 2-12 produce levels of hydroxide and free protons (which actually get attached to water molecules in a number of ways) that aren't dangerous for skin. That doesn't mean that you should bathe in them, or that repeated application won't cause problem, but they won't feel like burning. 0-2 and 12-14 are sketchier, but also comparatively rare, though they do show up in household items. Anything outside of the 0-14 range is all sorts of fun. Things that break the pH scale entirely are even more fun.

Brother Oni
2014-04-08, 02:20 AM
How much it will hurt if you stick your finger in a big ol' jar of basic or acidic solution is probably best measured by PH.

Outside of special cases, bases tend to do more damage to organic tissues than acids though, so sticking your finger into a big jar of a basic solution would hurt more given an equivalent ion concentration.

As you've said later though, there are things that break the pH scale - I've spilled 2M H2SO4 and 1M HCl on myself with minimal damage (dried my skin out for a couple of days) and even trifluoroacetic acid (affected skin peeled off after a couple of days) and I'm more concerned with dropping things than any actual damage when handling 10M NaOH (it makes soap out of the triglycerides in your skin).

I could take every safety precaution known and still be sweating buckets when handling HF, despite it being the weakest halogen acid.

1 M acetic acid? I wouldn't want that getting in a cut, but skin contact isn't really of that much concern. 6 M acetic acid? That could actually hurt, but it really wouldn't be that bad.

My only complaint about working with glacial acetic acid is that it makes me hungry for fish and chips. :smalltongue:

Knaight
2014-04-08, 02:38 AM
I disagree with this. As you've said later, there are things that break the pH scale - I've spilled 2M H2SO4 and 1M HCl on myself with minimal damage (dried my skin out for a couple of days) and even trifluoroacetic acid (affected skin peeled off after a couple of days) and I'm only worried about handling 2M NaOH since it makes it more likely to drop stuff than any actual damage (it makes soap out of the triglycerides in your skin).

I could take every safety precaution known and still be sweating buckets when handling HF, despite it being the weakest halogen acid.

I figure none of that's likely to come up with Gen Chem II chemicals. Plus, it's far better than concentration or acid strength individually.

As for HF - it won't burn skin. When it comes to damage on account of it being an acid, from the free protons*, it's fairly minimal. Even when fairly concentrated (provided that FHF- isn't coming into this), it's weak enough that the PH is close to 7. It's just that it also happens to be a really, really nasty poison, and capable of seeping in well past the skin. A stronger acid applied directly well under the skin to something like muscle tissue or bone would also be quite bad.

*Not actually free.

My only complaint about working with glacial acetic acid is that it makes me hungry for fish and chips. :smalltongue:
I can see that. I haven't worked with huge amounts of it, but I've already inured myself to the smell even if it's not under a fume hood. It's still noticeable (unlike acetone), but certainly not strong.

Peelee
2014-04-14, 02:27 PM
Ok, so I've been enjoying electrochemistry so far, mostly because it's really cool and I'm understanding it way better than anything else my professor talked about this semester. Then she pulls this out as an example (I'm pulling it from memory, so if I have the second half of the shorthand backwards, lemme know):

Write the shorthand notation for the following equation.

Pb(s) + Br2(l) --> Pb2+(aq) + 2Br-(aq)
Pb2+(aq)+2e- -->Pb(s) E*= -0.13 V
Br2(l)+2e- --> 2Br-(aq) E*= +1.07 V

Solution:

Pb(s) | Pb2+(aq) || 2Br-(aq) | Br2(l) | Ti

Where the hell did the Titanium come from? We'd just done the reverse (pulling the equation from the shorthand), but she'd presented this as a separate problem; If this question was asked on a test, would the fact that Titanium is an electrode be given, or would I have to figure that out, and if so, how?

Also, why does the titanium not go into the standard chemical reaction notation? Is it excluded due to being a spectator?

Livius
2014-04-14, 02:43 PM
Solution:

Pb(s) | Pb2+(aq) || 2Br-(aq) | Br2(l) | Ti

Where the hell did the Titanium come from? We'd just done the reverse (pulling the equation from the shorthand), but she'd presented this as a separate problem; If this question was asked on a test, would the fact that Titanium is an electrode be given, or would I have to figure that out, and if so, how?

For electrochemical cells using liquids or gases as the cathode or anode, a metal electrode is generally used as the reaction surface since it is solid (also, non-polar molecules don't dissolve in aqueous solutions very well and you need somewhere for the reactions to take place at a reasonable rate or the cell won't be at equilibrium. In this case it is Ti because: Ti(s) -> Ti2+(aq) + 2e- has a electrochemical potential between Pb and Br2 and doesn't react with either of them very much.

Also, why does the titanium not go into the standard chemical reaction notation? Is it excluded due to being a spectator?

The titanium isn't even a spectator. It's just a surface for the bromine to adsorb onto and react from. Another example is H2(g) | Pt, where you have a platinum rod connecting an aqueous solution and hydrogen gas.

Peelee
2014-04-14, 02:58 PM
For electrochemical cells using liquids or gases as the cathode or anode, a metal electrode is generally used as the reaction surface since it is solid (also, non-polar molecules don't dissolve in aqueous solutions very well and you need somewhere for the reactions to take place at a reasonable rate or the cell won't be at equilibrium. In this case it is Ti because: Ti(s) -> Ti2+(aq) + 2e- has a electrochemical potential between Pb and Br2 and doesn't react with either of them very much.
So if given a chemical equation for Daniell cell with only one solid metal, I should be expected to be able to figure out what the second electrode is?

The titanium isn't even a spectator. It's just a surface for the bromine to adsorb onto and react from. Another example is H2(g) | Pt, where you have a platinum rod connecting an aqueous solution and hydrogen gas.
So as long as a given metal has an electrochemical potential between the solutions, it can be used as an electrode?

Livius
2014-04-20, 08:54 AM
So if given a chemical equation for Daniell cell with only one solid metal, I should be expected to be able to figure out what the second electrode is?

It's probably easier to memorize the gas/electrode pairs that generally go together (like remembering H2 as H2|Pt). If you forget what the electrode is for a particular pair and can't look it up, Pt, Os, Ir are good guesses since they're less reactive.

So as long as a given metal has an electrochemical potential between the solutions, it can be used as an electrode?

Not necessarily; if it reacts with anything else in the system, it is a bad choice for an electrode. It just also has to have a potential between the solutions or it would become the anode/cathode itself.