Chemistry Questions (1 Viewer)

[Aysce]

Re: Salt bridge and migration of ions

Bump.

 

[Aysce]

Potential difference in context of EMFs

I've googled what potential difference is in this context but all it is giving me are definitions that I don't understand and analogies. I know that it is synonymous for voltage, but I don't understand what it actually is and how it applies to the electrodes in the Galvanic cell. Can someone please explain this to me?

EMF definition in textbook:

The electromotive force or EMF of a galvanic cell is the potential difference (voltage) across the electrodes of the cell when a negligibly small current is being drawn. It is the maximum voltage that the cell can deliver.

Thank you!

 

[Dan895]

Re: Potential difference in context of EMFs

To my knowledge at least... Its just how much force the electrons coming out of the cell possess, therefore the greater the voltage the more work can be done by the system. Correct me if i'm wrong though...

 

[bleakarcher]

Re: Potential difference in context of EMFs

I've googled what potential difference is in this context but all it is giving me are definitions that I don't understand and analogies. I know that it is synonymous for voltage, but I don't understand what it actually is and how it applies to the electrodes in the Galvanic cell. Can someone please explain this to me?

EMF definition in textbook:

The electromotive force or EMF of a galvanic cell is the potential difference (voltage) across the electrodes of the cell when a negligibly small current is being drawn. It is the maximum voltage that the cell can deliver.

Thank you!

Voltage or potential difference is a measure of the amount of work required to move one coulomb of charge (a coulomb is the SI unit for charge) from one point in the circuit to another. So say the voltage (in the case of a galvanic cell) is 1.5 V, then it takes 1.5 J of energy to move 1 coulomb of charge between the electrodes.

 

[Aysce]

Re: Potential difference in context of EMFs

Voltage or potential difference is a measure of the amount of work required to move one coulomb of charge (a coulomb is the SI unit for charge) from one point in the circuit to another. So say the voltage (in the case of a galvanic cell) is 1.5 V, then it takes 1.5 J of energy to move 1 coulomb of charge between the electrodes.

Ah alright, thanks a lot bleakarcher.

 

[Aysce]

Re: Potential difference in context of EMFs

And bleakarcher, may I ask, what is potential? I'm also struggling to understand it via google :/

 

[bleakarcher]

Re: Potential difference in context of EMFs

And bleakarcher, may I ask, what is potential? I'm also struggling to understand it via google :/

Sorry for the late reply lol, was watching the federer match.

Electric potential is a measure of electrostatic potential energy per coulomb of charge. All points within an electric field have a certain electric potential and the work required to move a point charge between this two points is the voltage between those two points in the electric field or the potential difference (this is where the term comes from).

 

[Aysce]

Re: Potential difference in context of EMFs

Sorry for the late reply lol, was watching the federer match.

Electric potential is a measure of electrostatic potential energy per coulomb of charge. All points within an electric field have a certain electric potential and the work required to move a point charge between this two points is the voltage between those two points in the electric field or the potential difference (this is where the term comes from).

Haha all good man and thanks once again. Need to start learning physics again some time soon

 

[bleakarcher]

Re: Potential difference in context of EMFs

Haha all good man and thanks once again. Need to start learning physics again some time soon

Glad to help man.

 

[Aysce]

EMF help



I don't understand what I'm doing wrong for iii and iv :/

Here's my working for each:

iii. Splitting into half reactions:

Pb ---> Pb^(2+) + 2e E = - (-0.13) = +0.13V

Mg^2+ + 2e ----> Mg E = -2.37V

To gain total EMF of the reaction, add each together. Though I'm wrong and the answer is -2.50V

iv. I received two half reactions, each a reduction half reaction. The one with a higher standard electrode potential I kept it as a reduction reaction whilst the other one I reversed it into an oxidation reaction. I still do not yield the correct answer :/ The answer is -2.75V.



23.ii.a and b.

I got the answer for ii.a but I have no idea how I got it so an explanation would be nice (Answer is -0.40V). For b, I don't understand why the half reaction involves cadmium rather than hydrogen? Answer: Cd^2+ + 2e ---> Cd. An explanation of why so would also be great.

I have several more questions regarding EMFs but I'll wait for these to hopefully be answered first. Thanks guys!

 

[DamTameNaken]

Re: EMF help

Q25 iii. I get the same answer as you, -2.24 V. The answers might be wrong in this case.

Q25 iv. Half reactions are:

2Br- ---> Br^2+ + 2e- -1.09V

Al^3+ + 3e- ---> Al -1.66 V

Total: -2.75 V.

Q23. ii.
It's important to note that we treat the hydrogen electrode as if it has a voltage of zero, (this is just so scientists can more easily use it as a reference for other electrode potentials. You can use their results or just look at the chart for standard electrode potentials and copy down their equation for cadmium

b.
We can ignore the hydrogen half equation because it always has a voltage of zero. Therefore any voltage that comes out of the system is entirely due to the cadmium. The answer is Cd^2+ + 2e- ---> Cd and not the hydrogen half reaction

 

[Aysce]

Re: EMF help

Thanks a lot, mate

 

[Aysce]

Quick question

For Iodine-120, the nucleus is unstable but how come it is not expected to be a beta emitter? Can someone please explain?

Also when trying to prove it is unstable, I attained the neutronroton ratio to be 1.26:1 - Isn't this close enough to be considered stable? Since the atomic number (Z) is 53 and that the N ratio for a stable isotope with Z of approximately 50 is 1.3:1? I've rounded to two D.P for another isotope (e.g. Barium-56) and it was stable rather than unstable in this case (N ratio was 1.46:1 where when Z=50, the N ratio for isotope stability is approximately 1.5:1).

Thanks for the help!

 

[Peter-PF7]

Re: Quick question

Coz check the periodic table, the stable atomic mass is approx. 127. Therefore I-120 is neutron deficient as the mass is lower. Neutron deficitent particles undergo beta-PLUS decay to convert proton to neutron and positron, NOT beta decay....

At least, I think this is the explanation haha...

Out of curiosity, didnt you complete HSC last year lol?

 

[Aysce]

Re: Quick question

Thanks for the information, Peter! Would love it if someone were to offer a second opinion

Where did you get this info from btw? Just researching on the net?

And yes I did finish the HSC last year.

 

[Madridismo]

Re: Quick question

Isn't it 1.5:1 for a large isotope and 1:1 for a small isotope?

Damnit I forgot all this chem >.<

 

[Aysce]

Re: Quick question

Isn't it 1.5:1 for a large isotope and 1:1 for a small isotope?

Damnit I forgot all this chem >.<

In the textbook they gave the approximate N ratios for Z<20, Z=50 and Z=80 which correlate to 1:1, 1.3:1 and 1.5:1 respectively. So yeah you're right - good memory

 

[Madridismo]

Re: Quick question

In the textbook they gave the approximate N ratios for Z<20, Z=50 and Z=80 which correlate to 1:1, 1.3:1 and 1.5:1 respectively. So yeah you're right - good memory

There was a 1.3:1?

I think I might need to refresh my chem before uni starts!

 

[Aysce]

Re: Quick question

There was a 1.3:1?

I think I might need to refresh my chem before uni starts!

Maybe a light review wouldn't hurt.

Anyways, can you confirm whether Peter's answer is correct? (It makes sense and I've searched up the answer but then again, I wanna know if it's defs legit lol)

 

[Omed62]

Re: Quick question

Maybe a light review wouldn't hurt.

Anyways, can you confirm whether Peter's answer is correct? (It makes sense and I've searched up the answer but then again, I wanna know if it's defs legit lol)

Hey Aysce,

How's uni going cuz,


the question is pretty simply, I took this from my notes today, see if it helps your.

See from the Periodic Table that Iodine has atomic number 53, so the proton: neatron ratio is 53: 67. so you can see that atomic no. is less than less than 83, so it is considered as a stable isotope.

• A stable isotope’s nucleus doesn’t change
• A radioactive isotope has an unstable nucleus which releases particles/energy to become more stable.
o Any isotope with atomic number > 83, is unstable
o Stability of nucleus also depends on neutron-to-proton ratio.
o Most stable nuclei are found in an area of graph known as the ‘zone of stability’
• Radioactivity is the spontaneous change in composition of an unstable nucleus leading to the emission of radiation.

• Alpha-decay
o Occurs when there are too many protons AND neutrons (nucleus too heavy).


• Beta-decay
o Occurs when there are too many neutrons for the # of protons present. A neutron turns into a proton and electron
o Mass number remains same, atomic number +1, electron emitted.


• Positron emission
o Occurs when there are too many protons for the # of neutrons present. A proton becomes a neutron and positron
o Mass number remains same, atomic number -1, positron emitted.


• Electron capture
Occurs when there are too many protons for the no. of neutrons present. An inner-orbital electron is captured by the nucleus, converting a proton into a neutron.

• Gamma radiation
o Emission of high energy photons enables nucleus to lose excess energy and become more stable

yes peters is absolutely right, also check my above notes!

I am currently in year 12, aiming for medicine too lol.