VCE Chemistry Question Thread

[Willba99]

Hi guys! I'm working through the 2013 VCAA exam and I have a few questions:

For Q29 (see attached image) I have no clue how to even begin to look at this question. Secondary galvanic cells have always made little sense to me.

For c)i) I'm not sure how to do acidity constants. can someone help here? I'm not sure of they're on the study design anymore so I think its safest to learn how to do them!!

Cheers

[Eric11267]

Hi guys! I'm working through the 2013 VCAA exam and I have a few questions:

For Q29 (see attached image) I have no clue how to even begin to look at this question. Secondary galvanic cells have always made little sense to me.

For c)i) I'm not sure how to do acidity constants. can someone help here? I'm not sure of they're on the study design anymore so I think its safest to learn how to do them!!

Cheers

To my knowledge acidity constants is not on the study design but if you want to know just do a quick google search

For Q29 I suggest you first look at the E^o values and decide which reactions would occur in the primary galvanic cells. Then for the secondary galvanic cells it would just be the reverse of these reactions, and you would look through the options and see which in consistent with the equations.

[Willba99]

To my knowledge acidity constants is not on the study design but if you want to know just do a quick google search

For Q29 I suggest you first look at the E^o values and decide which reactions would occur in the primary galvanic cells. Then for the secondary galvanic cells it would just be the reverse of these reactions, and you would look through the options and see which in consistent with the equations.

cheers

so. this is my thought process about Q29:

The second reaction has an E0 of 1.69 which is higher than the first reaction which has an E0 of -.36. Therefore, in discharge, Reaction 2 undergoes reduction, meaning Reaction 1 is the oxidation half reaction. That means, in discharge, Reaction 1 goes to the left and Reaction 2 to the right. In recharge, they go opposite ways i.e. Reaction 1 goes to the right and Reaction 2 to the left.

Therefore B is the answer, as Pb is a product in both the reduction and oxidation reactions.

How does this sound??

[Eric11267]

cheers

so. this is my thought process about Q29:

The second reaction has an E0 of 1.69 which is higher than the first reaction which has an E0 of -.36. Therefore, in discharge, Reaction 2 undergoes reduction, meaning Reaction 1 is the oxidation half reaction. That means, in discharge, Reaction 1 goes to the left and Reaction 2 to the right. In recharge, they go opposite ways i.e. Reaction 1 goes to the right and Reaction 2 to the left.

Therefore B is the answer, as Pb is a product in both the reduction and oxidation reactions.

How does this sound??

Spot on

[Willba99]

Spot on

Thanks Eric

Does anyone happen to know if we need to be able to use mmHg (as in, do we have to know how to convert it to kPa)?

[Syndicate]

Does anyone happen to know if we need to be able to use mmHg (as in, do we have to know how to convert it to kPa)?

Yes, but you don't have to memorise it. It's on page 6 in the data book.

[usernameincorrect]

I don't think you have to know the acid-base stuff (can't say 100% though). If I am correct the only fermentation reaction you have to know is the fermentation of glucose (by bacteria in an digester) to produce bioethanol and CO2.

Thank you!

[Willba99]

Yes, but you don't have to memorise it. It's on page 6 in the data book.

phew

[usernameincorrect]

Can someone please explain simply how to do part c of this galvanic cells question. Ive seen the answers but don't really understand why it is 0.55 + 0.62. Thanks!!

[Natasha.97]

Can someone please explain simply how to do part c of this galvanic cells question. Ive seen the answers but don't really understand why it is 0.55 + 0.62. Thanks!!

Ive attached the image as a link because when I tried to upload, file was 1.3mb and it said max file size is 1024kb. After compressing it and reuploading, again it said file was too large, but the maximum now is only 512kb. Strange?

http://compressjpeg.com/download/0o7vfamuuevn058i/o_1br5umd2gmd5fe1t8jlirlkia/IMG_6806-min.JPG?rnd=0.871545898985

Hi!

- As Iron produces a higher EMF with Tin compared to Cadmium, Iron is the stronger oxidant (most easily reduced)
- This results in Cadmium being oxidised at the anode (reduction half-cell) by Iron at the cathode (oxidation half-cell)
- The standard EMF of a galvanic cell = EMF of oxidation half-cell + EMF of reduction half-cell = 0.55 + 0.62

Hope this helps

[usernameincorrect]

Hi!

- As Iron produces a higher EMF with Tin compared to Cadmium, Iron is the stronger oxidant (most easily reduced)
- This results in Iron being oxidised at the anode (reduction half-cell) by Tin at the cathode (oxidation half-cell)
- The standard EMF of a galvanic cell = EMF of oxidation half-cell + EMF of reduction half-cell = 0.55 + 0.62

Hope this helps

Hello! Doesn't that mean iron is reduced, and cadmium is oxidase and not the other way around? Also, I still don't really understand why we use, for example the value 0.62V for the Fe3+/Fe2+ only half cell if the 0.62 value was obtained from a cell that used Fe3+/Fe2+ and Sn4+/Sn2+? Confused af hahaha

[Rieko Ioane]

Hello,

For this https://imgur.com/a/H4MIg

Is the correct name 3,4-dimethylpent-2-ene. The answer names it starting from the other side of the molecule, but alkenes take pref. over alkyl groups?

Thanks

[Syndicate]

Hello,

For this https://imgur.com/a/H4MIg

Is the correct name 3,4-dimethylpent-2-ene. The answer names it starting from the other side of the molecule, but alkenes take pref. over alkyl groups?

Thanks

You should be right (I am assuming that you have written 3,4-dimethylpent-2-ene).

[Rieko Ioane]

You should be right (I am assuming that you have written 3,4-dimethylpent-2-ene).

Yup, thanks brother.

[imactuallyalegend]

Can someone explain the difference between carbon and hydrogen environments?
Also, what's the easiest way to find the number of carbon and hydrogen environments?

Thanks!