Homework questions thread

[Surgeon]

I don't have a specific question but more of a concept based enquiry. I've got a SAC tomorrow and I've come to realize that I have a very superficial understanding of Redox equations. I would really appreciate it if someone could please post up the worked solutions to a redox question in which you need to write two half equations then join them!

Thanks so much in advance.

[Mr. Study]

Are you referring to the 'O-Na+' bit? If so, the reason that there isn't simply a straight line between them is because sodium doesn't form a covalent bond with oxygen there.

Hmm, So how exacctly would this type of bond come about? OR is it specific to the O-Na?

@Surgeon. These are from my own notes.

EDIT: There is a mistake on the first sheet! I did not balance electrons out! Argh, I type these notes on the summer holidays, My bad! IT should be 6e- on the LHS and NOT 10e- on the right!!

NVM, I update it just then. Got too insecure about that. >.>

EDIT 2: I never got around to finishing these notes but from the last 'sheet'. You just cancel down H₂O and cancel electrons out.

Hopefully the notes are clear enough!

[DisaFear]

I don't have a specific question but more of a concept based enquiry. I've got a SAC tomorrow and I've come to realize that I have a very superficial understanding of Redox equations. I would really appreciate it if someone could please post up the worked solutions to a redox question in which you need to write two half equations then join them!

Thanks so much in advance.

In case you want some more, in addition to Mr. Study's excellent notes

http://library.kcc.hawaii.edu/external/chemistry/bal_equations_rules.html

[charmanderp]

It's an ionic bond, rather than a covalent one. It's definitely not specific to 'O-Na+'. Covalent bonds are formed between non-metals only, when there is an equal sharing of electrons. I think sodium too readily donates its valence electron to ever be involved in a covalent bond, also.

[Mr. Study]

Ah! Ionic, Thanks for that charmander.

[thushan]

Are you referring to the 'O-Na+' bit? If so, the reason that there isn't simply a straight line between them is because sodium doesn't form a covalent bond with oxygen there.

Hmm, So how exacctly would this type of bond come about? OR is it specific to the O-Na?

@Surgeon. These are from my own notes.

EDIT: There is a mistake on the first sheet! I did not balance electrons out! Argh, I type these notes on the summer holidays, My bad! IT should be 6e- on the LHS and NOT 10e- on the right!!

NVM, I update it just then. Got too insecure about that. >.>

EDIT 2: I never got around to finishing these notes but from the last 'sheet'. You just cancel down H₂O and cancel electrons out.

Hopefully the notes are clear enough!

Hey Mr Study, just wanted to let you know that your notes are amazing - really well explained, very clear and concise - keep it up!

[stephanieteddy]

Hello there,

I am not quite sure if this has been asked previously,

but when we are drawing our hydrocarbon chains on the exam, do we have to do it in that zig zaggy pattern, or will the straight chain suffice?

Thank you!

[Surgeon]

Thanks so much guys. I really appreciate it. Can someone please explain the following to me:

Where does the bold 2e- come from?:


Q20.
In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed.

a)   Write half equations, and hence an overall equation, for the reaction.

a   Half equations:
Zn(s) --> Zn2+(aq) + 2e–
2MnO2(s) + 2H+(aq) + 2e– --> Mn2O3(s) + H2O(l)

[Phy124]

Thanks so much guys. I really appreciate it. Can someone please explain the following to me:

Where does the bold 2e- come from?:


Q20.
In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed.

a)   Write half equations, and hence an overall equation, for the reaction.

a   Half equations:
Zn(s) --> Zn2+(aq) + 2e–
2MnO2(s) + 2H+(aq) + 2e– --> Mn2O3(s) + H2O(l)

It balances the charge for the overall reaction. As there is two Hydrogen ions with a charge of +1, two electrons with charge -1 are needed. i.e. +2 + -2 = 0

edit: After re-reading, I think you may be asking how they get there, rather than why they are there.

As you can see in the first equation; two electrons are lost from that cell, these electrons then flow through the circuit  to the other half cell (anode -> cathode) and are gained. So pretty much the reductant transfers electrons to the oxidant - IIRC

[Surgeon]

Thanks so much guys. I really appreciate it. Can someone please explain the following to me:

Where does the bold 2e- come from?:


Q20.
In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed.

a)   Write half equations, and hence an overall equation, for the reaction.

a   Half equations:
Zn(s) --> Zn2+(aq) + 2e–
2MnO2(s) + 2H+(aq) + 2e– --> Mn2O3(s) + H2O(l)

It balances the charge for the overall reaction. As there is two Hydrogen ions with a charge of +1, two electrons with charge -1 are needed. i.e. +2 + -2 = 0

edit: After re-reading, I think you may be asking how they get there, rather than why they are there.

As you can see in the first equation; two electrons are produced (they are lost from that cell), these electrons then flow through the circuit  to the other half cell (anode -> cathode) and act as reactants (they are gained) - IIRC

ARGHH! I feel like a complete idiot now I completely overlooked the 2H+...

Thanks so much.

[soccerboi]

How much water must be added to 1.0L of a solution of a strong base of pH 13.0 in order to decrease the pH to 12.0?

[charmanderp]

A pH of 13 would mean a concentration of 0.1M. We need to get a pH of 12, which would require a concentration of 0.01M.

Using the equation C1V1=C2V2:

0.1x1=0.01xV2
>>V2=0.1/0.01
=10L

So therefore we would need to add 9 litres of water.

More simply though, every change in pH by one requires a dilution by a factor of 10.

[soccerboi]

How do you know pH of 13 requires concentration of 0.1M and pH 12 requires 0.01M?

Thanks 

[charmanderp]

Well a pH of 13 requires a pOH of 1, and a pH of 12 requires a pOH of 2. I really just kind of used memory to figure those out, it's hard for me to explain anything mathematic :p

Pay more attention to the point I made at the end though. A dilation by factor 10 will get you from 1 to 2 and from 14 to 13, etc.

[soccerboi]

Just need this clarified: On the exam, in regards to infra-red, we are only interested in/ considering bond stretching and not rocking,twisting or bending right?