VCE Chemistry Question Thread

[odeaa]

You do not have to remember any reactions in any cells for the exam including Leclanche. If the half equation isnt in the data book it will be given to you.
If youre teacher says you need to know it for a SAC then you should probably learn it then

Swoite, that makes it heaps easier

[warya]

How do you find the mole of Metal deposited given only the amount of Faradays?

The question says an electric charge of 0.03F is passed through a cell containing AgNO3 (1M), where the metal is deposited onto the negative electrode. The amount in mole of Ag deposited is... then there is mutli choice where the answer is 0.03mol of Ag

Im confused as I thought you have to use Q=nF to find n(e) therefore at least 2 bits of info given 

[paper-back]

How do you find the mole of Metal deposited given only the amount of Faradays?

The question says an electric charge of 0.03F is passed through a cell containing AgNO3 (1M), where the metal is deposited onto the negative electrode. The amount in mole of Ag deposited is... then there is mutli choice where the answer is 0.03mol of Ag

Im confused as I thought you have to use Q=nF to find n(e) therefore at least 2 bits of info given 

A Faraday is the amount of charge 1 mol of electrons have
0.03F being passed through the circuit is equivalent to 0.03 x [1 mol of electrons] passed through circuit (Therefore n(e-)=0.03mol)
As the reduction of Ag is: Ag⁺+e^--> Ag
n(e-)=n(Ag)
Hence number of mol of Ag deposited = 0.03mol

[warya]

A Faraday is the amount of charge 1 mol of electrons have
0.03F being passed through the circuit is equivalent to 0.03 x [1 mol of electrons] passed through circuit (Therefore n(e-)=0.03mol)
As the reduction of Ag is: Ag⁺+e^--> Ag
n(e-)=n(Ag)
Hence number of mol of Ag deposited = 0.03mol

So how does Q relate to this then? I thought charge is measured in C  not F, so confused 

[Orb]

quick question (easy) one b/c i'm terrible at chem:

is the permanganate ion a strong or weak oxidant and why?

P.S answers say strong oxidant but i thought it's weak from the electrochemical series O.o

[warya]

quick question (easy) one b/c i'm terrible at chem:

is the permanganate ion a strong or weak oxidant and why?

P.S answers say strong oxidant but i thought it's weak from the electrochemical series O.o

So the higher it is on the electrochemical series (on the left), the stronger it is as an oxidant, F2 is the strongest oxidant as it is the highest on the left, and MnO4- is not far below it, meaning its a strong oxidant

[odeaa]

Im struggling to understand a question from Thushan's notes (keep in mind i am chemically retarded here lol)
A mixture of copper (II) nitrate and potassium chloride was electrolysed in acidid solution (HCl) using inert graphite electrodes. Which species are produced at which electrode, assuming standard conditions?

The notes split everything up into ions, and then says that both Cl and H2O can be oxidised, and that H2O is more likely to be oxidised. How do you know that from the electrochemical series? Isnt the strongest reductant (more likely to be oxidised) at the bottom of the table? Or have I got my definition wrong?

Regrets of playing snowboard party in chem for the last few months...

[BakedDwarf]

So how does Q relate to this then? I thought charge is measured in C  not F, so confused 

1 faraday = 96500 C
0.03 faraday = 96500 x 0.03
                     = 2895 C

Q = n(e-) x F
    = n(e-) x 96500

2895 = n(e-) x 96500
n(e-) = 0.03 mol

[paper-back]

So how does Q relate to this then? I thought charge is measured in C  not F, so confused 

Q is charge(Q)
1 Faraday(F) is 96500C(Q)
0.03F = 0.03 x 96500C = 2895C (Q)

Using this information as BakedDwarf has; you can get 0.03mol

[BakedDwarf]

Im struggling to understand a question from Thushan's notes (keep in mind i am chemically retarded here lol)
A mixture of copper (II) nitrate and potassium chloride was electrolysed in acidid solution (HCl) using inert graphite electrodes. Which species are produced at which electrode, assuming standard conditions?

The notes split everything up into ions, and then says that both Cl and H2O can be oxidised, and that H2O is more likely to be oxidised. How do you know that from the electrochemical series? Isnt the strongest reductant (more likely to be oxidised) at the bottom of the table? Or have I got my definition wrong?

Regrets of playing snowboard party in chem for the last few months...

Yes, the strongest reductant is the bottom right of the table. Hence, because H2O is lower than Cl- on the right hand side of the electrochemical series, it is  a stronger reductant and most likely going to be oxidised instead of Cl-

[Orb]

So the higher it is on the electrochemical series (on the left), the stronger it is as an oxidant, F2 is the strongest oxidant as it is the highest on the left, and MnO4- is not far below it, meaning its a strong oxidant

Thanks, but I can't quite see where MnO4- is? O.o

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chemdata-w.pdf

I can only see Mn2+ and that's a reading of -1.03.. :/

[odeaa]

Yes, the strongest reductant is the bottom right of the table. Hence, because H2O is lower than Cl- on the right hand side of the electrochemical series, it is  a stronger reductant and most likely going to be oxidised instead of Cl-

That makes sense, but why cant the K+ be oxidised? Its almost at the bottom

[warya]

Thanks, but I can't quite see where MnO4- is? O.o

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chemdata-w.pdf

I can only see Mn2+ and that's a reading of -1.03.. :/

Thats weird it is in the series in heinemann 

[warya]

1 faraday = 96500 C
0.03 faraday = 96500 x 0.043
                     = 2895 C

Q = n(e-) x F
    = n(e-) x 96500

2895 = n(e-) x 96500
n(e-) = 0.03 mol

Where did the 0.043 come from?

[paper-back]

Im struggling to understand a question from Thushan's notes (keep in mind i am chemically retarded here lol)
A mixture of copper (II) nitrate and potassium chloride was electrolysed in acidid solution (HCl) using inert graphite electrodes. Which species are produced at which electrode, assuming standard conditions?

The notes split everything up into ions, and then says that both Cl and H2O can be oxidised, and that H2O is more likely to be oxidised. How do you know that from the electrochemical series? Isnt the strongest reductant (more likely to be oxidised) at the bottom of the table? Or have I got my definition wrong?

Regrets of playing snowboard party in chem for the last few months...

So in this question we have the following elements in solution:
K+, Cl-, Cu2+, H+, H2O

Using the electrochemical we can derive a smaller electrochemical series with the elements presented
So in descending order beginning from the top we'd write(All the elements present in solution are underlined):
H₂O₂+2H⁺+2e^-<->2H₂O
Cl₂+2e^- <-> 2Cl^-
O₂+4H⁺+4e^-<->2H₂O
O₂+2H⁺+2e^-<->H₂O₂
O₂+2H₂O+4e^-<->4OH_-
Cu²⁺+2e^-<->Cu
2H⁺+2e^-<->H₂
2H₂O+2e^-<->H₂+2OH^-
K⁺+e^-<->K

Now we can cancel out two reactions as O2 is present on the 'reactant side' and it is is not present in solution:
H₂O₂+2H⁺+2e^-<->2H₂O
Cl₂+2e^- <-> 2Cl^-
O₂+4H⁺+4e^-<->2H₂O
O₂+2H⁺+2e^-<->H₂O₂
O₂+2H₂O+4e^-<->4OH_-
Cu²⁺+2e^-<->Cu
2H⁺+2e^-<->H₂
2H₂O+2e^-<->H₂+2OH^-
K⁺+e^-<->K

This leaves us with the remaining reactions. Now remember from galvanic cells that; as you move down the electrochemical series the more likely the species on the right hand side is to become oxidised (i.e. the greater it's ability to act as a reductant)
Now, moving down our little electrochemical series we see that the species with the greatest ability to be oxidised is the H2O as it is the lowest species underlined on the right hand side. Therefore we can note that reaction as occurring at the anode:
H₂O₂+2H⁺+2e^-<->2H₂O
Cl₂+2e^- <-> 2Cl^-
O₂+4H⁺+4e^-<->2H₂O
O₂+2H⁺+2e^-<->H₂O₂
O₂+2H₂O+4e^-<->4OH_-
Cu²⁺+2e^-<->Cu
2H⁺+2e^-<->H₂
2H₂O+2e^-<->H₂+2OH^-
K⁺+e^-<->K

By moving up the electrochemical series, we see that the Cu2+ is most likely to be reduced as it is the upper-most species of the left hand side (remember how the ability of a species (on the left hand side) to become reduced increases as we move up the electrochemical series). Therefore we can also highlight that reaction:
H₂O₂+2H⁺+2e^-<->2H₂O
Cl₂+2e^- <-> 2Cl^-
O₂+4H⁺+4e^-<->2H₂O
O₂+2H⁺+2e^-<->H₂O₂
O₂+2H₂O+4e^-<->4OH_-
Cu²⁺+2e^-<->Cu
2H⁺+2e^-<->H₂
2H₂O+2e^-<->H₂+2OH^-
K⁺+e^-<->K

These leaves us with the following reactions occurring at the cathode and anode respectively:
Cu²⁺+2e^-->Cu
2H₂O ->4e^-+4H⁺+O2