Westminster 6th form entrance exam chemistry sample paper solutions?

I was just looking at the solutions for the compulsory question in the Westminster chemistry sample paper, and the solutions don’t make sense to me…?

https://www.westminster.org.uk/academic-life/departments/sciences/chemistry/

For question one, there seems to be 8 oxygens on the left side but only 6 on the right according to the solution, so surely it is not balanced?

And for question 3, surely it would be YZX rather than XZY as the question asked for increasing order of reactivity.

Am I missing something or is it just a typo? I’m asking for peace of mind haha

Reply 1

anonymous56754

Original post by KellyC321

I was just looking at the solutions for the compulsory question in the Westminster chemistry sample paper, and the solutions don’t make sense to me…?
https://www.westminster.org.uk/academic-life/departments/sciences/chemistry/
For question one, there seems to be 8 oxygens on the left side but only 6 on the right according to the solution, so surely it is not balanced?
And for question 3, surely it would be YZX rather than XZY as the question asked for increasing order of reactivity.
Am I missing something or is it just a typo? I’m asking for peace of mind haha

I’m pretty sure you’re right for question 3 being YZX, the reactivity decreases down the group for group 7 elements and x is a gras, it’s either fluorine or chlorine so most reactive so at the end and then y before z because it says it’s less reactive.

Reply 2

anonymous56754

Original post by KellyC321

I was just looking at the solutions for the compulsory question in the Westminster chemistry sample paper, and the solutions don’t make sense to me…?
https://www.westminster.org.uk/academic-life/departments/sciences/chemistry/
For question one, there seems to be 8 oxygens on the left side but only 6 on the right according to the solution, so surely it is not balanced?
And for question 3, surely it would be YZX rather than XZY as the question asked for increasing order of reactivity.
Am I missing something or is it just a typo? I’m asking for peace of mind haha

Sorry for question 1 did you work out the 8 oxygens?

Reply 3

Nitrotoluene

Original post by KellyC321

I was just looking at the solutions for the compulsory question in the Westminster chemistry sample paper, and the solutions don’t make sense to me…?
https://www.westminster.org.uk/academic-life/departments/sciences/chemistry/
For question one, there seems to be 8 oxygens on the left side but only 6 on the right according to the solution, so surely it is not balanced?
And for question 3, surely it would be YZX rather than XZY as the question asked for increasing order of reactivity.
Am I missing something or is it just a typo? I’m asking for peace of mind haha

Hello Kelly!
Question one
You are right about the 8 and 6 oxygen atoms in the redox.

If you put the 8 oxygens in the reactants and products, you have an electronic charge balance issue; so the redox is still not well balanced.
I have balanced the redox and put it in the drawer for now. Let's see if you can come up with your own attempt at a balanced redox, or at least an approximation of the correct balance.

Bye,
Sandro

(edited 1 month ago)

Reply 4

KellyC321

Original post by anonymous56754

Sorry for question 1 did you work out the 8 oxygens?

My answer would be
3 C2H4O + Cr2O72- + 8 H+ = 3 C2H4O2 + 2 Cr3+ + 4 H2O

But the question doesn’t give a letter in front of the C2H4O2 so I wasn’t sure if that meant there should only be one of them or that I can put whatever number in front… idk haha

Reply 5

KellyC321

Original post by Nitrotoluene

Hello Kelly!
Question one
You are right about the 8 and 6 oxygen atoms in the redox.
If you put the 8 oxygens in the reactants and products, you have an electronic charge balance issue; so the redox is still not well balanced.
I have balanced the redox and put it in the drawer for now. Let's see if you can come up with your own attempt at a balanced redox, or at least an approximation of the correct balance.
Bye,
Sandro

Hello Sandro,
Thank you for your reply.
The only way I could balance the equation was like so:
3 C2H4O + Cr2O72- + 8 H+ = 3 C2H4O2 + 2 Cr3+ + 4 H2O
But the question doesn’t give a letter in front of C2H4O2. I wasn’t sure if that meant there was only supposed to be one?

Also, may I ask what putting it in the drawer means? Sorry for the trouble!

Reply 6

anonymous56754

Original post by KellyC321

My answer would be
3 C2H4O + Cr2O72- + 8 H+ = 3 C2H4O2 + 2 Cr3+ + 4 H2O
But the question doesn’t give a letter in front of the C2H4O2 so I wasn’t sure if that meant there should only be one of them or that I can put whatever number in front… idk haha

Yes, I just tried it out and I got the same equation as you. I’m not sure why there is no letter in front, just ignore that probably but I’m pretty sure you’re right!

Reply 7

Nitrotoluene

Original post by KellyC321

Hello Sandro,
Thank you for your reply.
The only way I could balance the equation was like so:
3 C2H4O + Cr2O72- + 8 H+ = 3 C2H4O2 + 2 Cr3+ + 4 H2O
But the question doesn’t give a letter in front of C2H4O2. I wasn’t sure if that meant there was only supposed to be one?
Also, may I ask what putting it in the drawer means? Sorry for the trouble!

Hello Kelly!

I came to the same conclusion as you yesterday.
Since both the atom count and charges are balanced, your redox reaction is indeed well balanced:

3 C2H4O + Cr2O7^2- + 8 H^+ = 3 C2H4O2 + 2 Cr^3+ + 4 H2O

Note: As you can see, the 8 initial oxygen atoms have nothing to do with the correct redox balance.

Quote
Also, may I ask what putting it in the drawer means?
Unquote
Sorry, but I slipped in an Italian idiom which means that when someone have finished a task, he/she is waiting and ready to do something or help (in other words, "to be in stand by"). I hope I explained myself well.

Kind regards,
Sandro

(edited 1 month ago)