Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed) Watch

Poll: How pumped up are you for this exam?-(warning)-(bad jokes arene this poll!)
"Titanium-I'm not going to corrode (even at high temperatures)" (A*) (22)
16.67%
"Benzene's my middle name, give me the paper in a week and I'll ace it!" (A) (27)
20.45%
"Yeah, I'm fairly electrophillic (positively charged) about the exam" (B) (27)
20.45%
"I'm in the middle of the salt bridge, but I will pass-eventually" (C) (21)
15.91%
"I'm feeling rather electroNegative about this exam" (D) (18)
13.64%
"Benzene, what's that?" (E) (6)
4.55%
"Chemistry, what's that?" (F) (11)
8.33%
This discussion is closed.
posthumus
Badges: 20
Rep:
?
#481
Report 6 years ago
#481
(Original post by LeaX)
Attachment 206207
Attachment 206208

Could someone possibly help me on this question please? I understand how you find the moles of MnO4- initially but after that first step I'm confused. Thank you.
Are these question from electrochemistry/redox chapter or the d block chapter?

Could you post the question too if possible, thanks
0
posthumus
Badges: 20
Rep:
?
#482
Report 6 years ago
#482
(Original post by jojo1995)
omg - lol - how awks - sorry i did not know that thank you
I have no idea tbh
but if the rod and the solution are separate, the rod is negative and the positive solution. idk - im scared lol this stuff is confusing
ahaha its fine ! Its confusing the hell out of me too.

I only just started electrochemistry today... still have 2 more chapters to do (out of 5). Then Unit 1,2 & 4 revision
0
shumen9523
Badges: 6
Rep:
?
#483
Report 6 years ago
#483
(Original post by posthumus)
ahaha its fine ! Its confusing the hell out of me too.

I only just started electrochemistry today... still have 2 more chapters to do (out of 5). Then Unit 1,2 & 4 revision
Hi, check out this link:
http://bouman.chem.georgetown.edu/S02/lect25/lect25.htm
1
David Tennant
Badges: 0
Rep:
?
#484
Report Thread starter 6 years ago
#484
(Original post by posthumus)
Okay thanks for the response, so just to confirm:
The zinc rod is the cathode & the copper one is the anode.

In the book it also say the anions are attracted to the anodes ? Is this incorrect then... because where the positive anode is the solution is negative? And those electrolytes in the salt bridge are attracted to the solution.

Also the zinc rod is negative because the electrons stay on the rod when the zinc is oxidized ? and the zinc ions go into the solution.

The copper rod is positive but 2e- & Cu^2+ are forming to create Cu which is neutral. The confusing bit is - shouldn't the copper rod then be neutral ?
But electrons move from the zinc cathode towards the copper anode to make it more negative so although we are creating Cu, the electrons make it more negative.
But if this is true then why is the copper electrode positive :confused:!

I will ask my teacher when the holiday ends!
0
LeaX
Badges: 20
Rep:
?
#485
Report 6 years ago
#485
(Original post by posthumus)
Are these question from electrochemistry/redox chapter or the d block chapter?

Could you post the question too if possible, thanks
The d block chapter. The question is at the top, there's two pictures on that post although they look like they've merged to form one haha.
0
posthumus
Badges: 20
Rep:
?
#486
Report 6 years ago
#486
Thanks for the link, it looks really useful! I may just print that out


(Original post by David Tennant)
But electrons move from the zinc cathode towards the copper anode to make it more negative so although we are creating Cu, the electrons make it more negative.
But if this is true then why is the copper electrode positive :confused:!

I will ask my teacher when the holiday ends!
From my understanding the Zn gets oxidized - passes the electrons over to Copper where the Cu2+ is converted into copper. And that about all I understand

But yh for example I don't get why the anode is positive exactly... or if that is just relative to the rest of the circuit. If the salt bridge were removed the anode would be neutral?

Do you know why in Zn <-----> Zn^2+ + 2e- the equilibrium is to the right ? :confused:

Electrochemistry is just too damn confusing & yes please do !
0
posthumus
Badges: 20
Rep:
?
#487
Report 6 years ago
#487
(Original post by LeaX)
The d block chapter. The question is at the top, there's two pictures on that post although they look like they've merged to form one haha.
Okay I actually haven't started this chapter yet but...

write down what is going on:

MnO4- + 8H+ + 5e- -------> Mn^2+ + 4H20

Fe^2+ -----> Fe^3 + e-

to substitute in the second equation you have to multiply out by 5 (since there are 5e- in the first one). That would explain why you multiply the moles of MnO4- by 5

so you do the same for the 2nd one

& I'm sure you can carry on from there !!
0
JoshL123
Badges: 2
Rep:
?
#488
Report 6 years ago
#488
Hi everyone. Are we given the redox potentials in our exam?
0
mkhan51
Badges: 0
Rep:
?
#489
Report 6 years ago
#489
Hi, we have the data booklet and they have the redox potentials in them (I think)


Posted from TSR Mobile
0
Gnome :)
Badges: 2
Rep:
?
#490
Report 6 years ago
#490
(Original post by JoshL123)
Hi everyone. Are we given the redox potentials in our exam?
Yep This is what we get in our exam
0
David Tennant
Badges: 0
Rep:
?
#491
Report Thread starter 6 years ago
#491
(Original post by posthumus)
Thanks for the link, it looks really useful! I may just print that out




From my understanding the Zn gets oxidized - passes the electrons over to Copper where the Cu2+ is converted into copper. And that about all I understand

But yh for example I don't get why the anode is positive exactly... or if that is just relative to the rest of the circuit. If the salt bridge were removed the anode would be neutral?

Do you know why in Zn <-----> Zn^2+ + 2e- the equilibrium is to the right ? :confused:

Electrochemica is just too damn confusing & yes please do !
Equillibrium is to the right because electrons are required for the reduction of copper...So electrons are drawn away and the right side of the equation gets smaller...Therefore more Zn will react to form Zn+2 ions to replace the electrons lost and maintain an equilibrium-This is what is imagine it is, and it causes a shift to the right.

Perhaps it is positive relative to the rest of the circuit, electrons are being lost from it to go to the copper and so it loses negative charge.
If the salt bridge was removed then nothing would happen, as the circuit would be incomplete, just like cutting the wire in an electric circuit...so yes it would be neutral!

The bit which gets me, even more than the half cells is calculating Ecell values, the equation doesn't seem to work in all situations and is really confusing. (I use Ecell=RHS-LHS)
0
JoshL123
Badges: 2
Rep:
?
#492
Report 6 years ago
#492
What is the difference between standard reduction potentials and standard electrode potentials ?
0
JoshL123
Badges: 2
Rep:
?
#493
Report 6 years ago
#493
(Original post by Gnome :))
Yep This is what we get in our exam

(Original post by mkhan51)
Hi, we have the data booklet and they have the redox potentials in them (I think)


Posted from TSR Mobile

Thanks!
0
David Tennant
Badges: 0
Rep:
?
#494
Report Thread starter 6 years ago
#494
(Original post by JoshL123)
What is the difference between standard reduction potentials and standard electrode potentials ?
Aren't they the same?
If you've got the Hodder book turn to page 143, the green box and look at the second definition:

The Standard electrode (reduction) potential, E, of a standard half-cell is the e.m.f of that half-cell relative to a standard hydrogen electrode under standard conditions.Standard electrode electrode potentials are sometimes called standard redox potentials.
0
JoshL123
Badges: 2
Rep:
?
#495
Report 6 years ago
#495
(Original post by David Tennant)
Aren't they the same?
If you've got the Hodder book turn to page 143, the green box and look at the second definition:
I just saw two tables on the Ed excel resources guide with both of them
0
GeorgeL3
Badges: 5
Rep:
?
#496
Report 6 years ago
#496
(Original post by LeaX)
Attachment 206207
Attachment 206208

Could someone possibly help me on this question please? I understand how you find the moles of MnO4- initially but after that first step I'm confused. Thank you.
Here are the first set of equations needed.
Half equation for Manganese being reduced:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Half equation for Iron being oxidised:
Fe2+ → Fe3+ + e-

Full equation for manganate oxidising iron:
MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

The First step now is to calculate the number of moles used to react with the Iron.
Moles of MnO4- in 1st titration: n=cv/1000, n=(0.02 x 18.2)/1000 = 3.64x10^-4 moles.

From the balanced equation we can see this is 5x less than the moles of Iron present, so the moles of iron are:
5 x 3.64x10^-4 = 1.82x10^-3 moles

Now before the second titration, zinc is added so it reduces the iron(III) to iron(II):
Zn + 2Fe3+ → Zn2+ + 2Fe2+
So now in the second equation, all of the Iron is Iron(II).

So now we do the same process with this second titration:
Moles of MnO4- = (0.02 x 25.3)/1000 = 5.06x10^-4 moles
Moles of Fe(II) = 5 x 5.06x10^-4 = 2.53x10^-3

Now this total amount of Fe(II) includes the moles of Fe(III) in it, so to find the moles of Fe(III) we need to take away the first answer for moles of iron:
Moles of Iron(III) = (2.53x10^-3) - (1.82x10^-3 moles) = 7.1x10^-4 moles

Now from the Iron and Zinc equation you can see that the number of moles of Zinc are half that of Iron(III):
Moles of zinc = (7.1x10^-4 moles) / 2 = 3.55x10^-4 moles

Now you can just use Mass = Moles x RMM (RMM of Zn is 65.4):
3.55x10^-4 x 65.4 = 0.023217g

So your answer for mass of Zinc is 0.0232g
1
Weaselmoose
Badges: 9
Rep:
?
#497
Report 6 years ago
#497
(Original post by senz72)


UMS is out of 60 but raw marks are out of 40 (14+14+12)

Yeah but didn't you write 48/54? Aha I'm lost
0
senz72
Badges: 16
Rep:
?
#498
Report 6 years ago
#498
(Original post by Weaselmoose)
Yeah but didn't you write 48/54? Aha I'm lost
Whoops!

I meant like if you get 37 out of 40, this will be 48 out of 60.
But if you get 38 out of 40, this will be 54 out of 60.

Sorry. :/
0
JRP95
Badges: 2
Rep:
?
#499
Report 6 years ago
#499
(Original post by senz72)
Whoops!

I meant like if you get 37 out of 40, this will be 48 out of 60.
But if you get 38 out of 40, this will be 54 out of 60.

Sorry. :/
Would 39/40 be 56 or 58 ums?
0
senz72
Badges: 16
Rep:
?
#500
Report 6 years ago
#500
(Original post by JRP95)
Would 39/40 be 56 or 58 ums?
Not sure But I'd go for 57.

The only reason why I know 38 and 37 is because that is the A* and A boundary respectively.
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 19 Oct '19
  • Coventry University
    Undergraduate Open Day Undergraduate
    Sat, 19 Oct '19
  • University of Birmingham
    Undergraduate Open Day Undergraduate
    Sat, 19 Oct '19

Why wouldn't you turn to teachers if you were being bullied?

They might tell my parents (11)
6.04%
They might tell the bully (18)
9.89%
I don't think they'd understand (32)
17.58%
It might lead to more bullying (69)
37.91%
There's nothing they could do (52)
28.57%

Watched Threads

View All