Problem Solving Strategies - 2019-09-20

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3:22 AM

@JohnRennie morning sir

i was reading your last chat

with no body

when i say dipole is at origin ,i didnt get what does that mean ,i understand the dipole moment is from negative charge towards positive charge ,means the two charges lie on x axis

and same thing i didn,t understand for P

@JohnRennie sirtoday i have long long class of 8 hours so i have one more question ,i might no seen your answer till night but if you give hint of the question i will be greatful

@JohnRennie one more question two point charges are placed at distance R from each other in vacuum, the force between them is f1, one if charges is spread uniformly over the surface of a hemispherical bowl of radius R while the other is spread r and (r<R/2) and both have common Centre o, force on sphere is f1/n then n

4:20 AM

@JohnRennie Good morning sir :-)

@user8718165 morning :-)

@JohnRennie hi

@yuvrajsingh morning :-)

@JohnRennie have a look sir what i have posted ,i have a class at 10.30 plese sir

@yuvrajsingh I'm busy at the moment but I'll have a look in a few minutes.

4:32 AM

@JohnRennie ok,but sir you post you answer ,whenever you get free ,

@yuvrajsingh will do!

2 hours later…

6:06 AM

@yuvrajsingh I think n = 2

1 hour later…

7:35 AM

@JohnRennie hi you know electrochemistry?

@Aladdin kind of ...

8:00 AM

Calmel slectrodes and glass electrodes

Calomel you mean? The mercury electrode?

yes

@Aladdin what's the question?

why is kcl used In it

8:16 AM

@Aladdin You need Cl- because the reaction at the calomel cell produces or consumes Hg2Cl2

Why KCl is preferred over NaCl I don't know.

1 hour later…

9:37 AM

@JohnRennie hi

@Aladdin hi

imgur.com/a/NpqwEfo

Why can it act as both cathode and anode

@Aladdin In a cell the cathode is reduced and the anode is oxidised (or it might be the other way round, I always struggle to remember which way round it is).

Ok. For galvanic cell u are right

If you connect a calomel electrode to another electrode that is highly oxidising the calomel electrode will be oxidised so it acts as an anode.

If you connect it to an electrode that is highly reducing the calomel electrode will be reduced so it acts as the cathode.

This is true for all electrodes.

9:43 AM

Yep

So it just depends on the potential of whatever half cell you connect it to.

I needed help in a question

@Aladdin yes?

imgur.com/a/WkIkRGU

In Q21

My reaction is :

2Fe+3+2Hg+2cl- ------->2fe+2+Hg2cl2

Now when I put values in nerst equation, it is not giving value for Ecell

Isn't it $Fe^{3+} + Hg \to Fe^{2+} + Hg^+$ ?

9:54 AM

Is that hg+2 ion in ur last reactant

Ah wait, I think I misinterpreted what you wrote.

Calomel for anode

@Aladdin No, it's the monovalent mercury atom in Hg2Cl2

$FeCl_3 + Hg \to FeCl2 + \tfrac{1}{2} Hg_2CL_2$

At cathode :

Fe+3+e----->fe+2

Is this right

Yes

9:58 AM

At anode:

2Hg+2Cl- ------>Hg2cl2+2e

Is this right

Yes

So my net reaction is right?

Yes. I misread your equation because you've used the + signs in a rather confusing way.

Ah my mistake

@JohnRennie ohkk. I got the answer now

Looks like missed a factor of 2 somewhere..

@Aladdin cool :-)

10:05 AM

@JohnRennie imgur.com/a/zXl31Y2

How did they get Ej?

Doesn't that come from the basic equation for the free energy $G = RT \ln C$ ?

And $G = EnF$ ?

Not derivation for that in my book

Let me dig out my physical chemistry textbook ...

Can u post me link?

@JohnRennie if it has calomel, concentration and ion selective electrodes numerical, can u tell me it's name

It's just the Nernst equation. The cell is the same on both sides because the only difference is the concentration of you $M^{n+}$ - the membrane keeps the concentrations different.

So on side 1 you have $E_1 = E^o + RT/nF \ln C_1$ and on the other side $E_2 = E^o + RT/nF \ln C_2$

So the cell potential is $E_1 - E_2 = RT/nF \ln C_1 - RT/nF \ln C_2 = RT/nF \ln(C_1/C_2)$

The 2.303 is just the conversion of natural logarithms to base 10 logs.

10:30 AM

@JohnRennie book name?

Physical Chemistry by Walter J. Moore, but I think it's out of print.

Ahh so no pdf?

There's nothing special about the book. It was just the recommended course book in 1980 when I did first year physical chemistry.

Ok

Physical Chemistry by David W Ball is floating around the Internet as a pdf ...

Not that I'm encouraging piracy of course :-)

@Aladdin I might even have a copy if the PDF somewhere ...

10:36 AM

Can u send if u have

Yes

Thanks a lot!

If you can find it Physical Chemistry by Peter Atkins is also well recommended. I don't a have a copy of that.

The Atkins book has a solutions manual as well.

I have atkins

But there is no mention of calomel or ion selective electrode there

Not In 8th and 10th edition

Ball does describe the calomel electrode.

@JohnRennie page no forthe derivation

@Aladdin do you mean the derivation for that membrane cell?

If so I didn't find that in Ball. I worked it out.

10:45 AM

Ah ok6

@JohnRennie one more doubt

@Aladdin yes?

imgur.com/a/2TQvGG0

For Eanode shouldn't it be 1/C1 inside log?

Zn+2 ions ate on left side nof the reaction

What it's doing is writing the half cell reactions for the two sides, then subtracting them to get the potential difference between the two sides.

Yeah

So it's writing the forward reaction in both cases.

If you wrote the forward reaction and the reverse reaction you can do this but then you'd have to add them not subtract them.

Both ways will end up with $\ln(C_2/C_1)$

i.e. $\ln(C_2) - \ln(C_1) = \ln(C_2) + \ln(1/C_1)$

11:24 AM

It's still not making sense...

Maybe nerst equation for oxidation reaction is different ?

4 hours later…

3:11 PM

@JohnRennie yes you are. Correct but how you arrive at this

@yuvrajsingh hi

The ball will behave as a point charge, so we have a point charge at the centre of a charged hemisphere.

So if we can calculate the field at the centre of a charged hemisphere the force on the ball will just be $F = EQ_b$, where $Q_b$ is the charge on the ball.

Calculating the field at the centre of a charged hemisphere is a standard problem and you'll find lots of web pages and YouTube videos explaining how to do it. You have to do it by integration and it is a somewhat messy calculation. The end result is that the field at the centre is $k Q_h /(2R^2)$, where $Q_h$ is the charge on the hemisphere and $R$ is the radius of the hemisphere.

So the force ends up as $F = kQ_b Q_h/(2R^2)$, which is half the original force between the charges.

3:37 PM

@JohnRennie thank you sir

@JohnRennie what I do was

To solve it by gauss law

Where E=Sigma /4epsilon

@yuvrajsingh yes

Of she'll on it

F=qb. E

But then I got n=1

F=kq/4pier^2

@JohnRennie I do not know where I am wrong

Start with E = sigma / 4e, which is correct.

OK

The charge is Q_h and the area of a hemisphere is 2 pi R^2, so sigma = Q_h/(2 pi R^2)

Substitute for sigma and you get E = Q_h (2 * 4pi e * R^2)

Which is E = k Q_h / (2 R^2)

3:46 PM

@JohnRennie when we solve it reverse

OK sir thank you

@JohnRennie one more short question

@yuvrajsingh yes?

4:05 PM

@JohnRennie have a very fundamental doubt. A black body can be assumed as a cavity with a small hole with radiation leaking out of it. As the temperature of the black body is increased we can assume the the charge particles, electrons, on the metal surface will behave as harmonic oscillators and the energy of the harmonic oscillation will be equal to the energy density of the radiation inside the cavity at thermal equilibrium.

My doubt is that at thermal equilibrium, the cavity(assuming it to be uniform) will have a uniform temperature. Since the oscillations of the charged particles is du…

The temperature of a system is an average property. For example in a gas the average velocity of the molecules is related to the temperature by 1/2 m v^2 = 3/2 kT.

But this is only an average and some molecules move more slowly while others move faster.

In any thermal system there are random fluctuations in energy distribution such that at any instant some parts of the system are hotter and some are colder.

It is because of these random fluctuations that the black body radiation is spread over a range of wavelengths rather than being at a single wavelength.

@JohnRennie the black body is emitting entire frequency range, each electron much oscillate at different frequency in order to radiate different frequencies.

@yuvrajsingh black body radiation doesn't come from the oscilaltions of single isolated electrons. It comes from collective oscillations in the total electron density.

I think there are questions on the main site about the origin of black body radiation ...

@JohnRennie You misunderstand me. Yes, the distributions are the same for the same temperature, but the individual kinetice energies and frequencies are all over the spectrum. The questioner seems to think that for temperature = 300K there is only one frequency allowed

@yuvrajsingh then they are wrong :-)

4:15 PM

@JohnRennie then distribution is I think not the correct word

@yuvrajsingh I'd need to see the book, or whatever it is, to comment on how the term distribution is being used.

@JohnRennie I do not have it right now, it our discussion today among friends

I need to go now. I'll be back tomorrow as usual.

Good day sir @JohnRennie

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