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# ocr a f325 revision thread watch

1. (Original post by Y_123456)
Could someone tell me why Ni have 2 unpaired d electrons?
Attachment 225790
Think about it - nickel has 8 3d electrons to go into five orbitals. That means 3 filled orbitals and 2 half-filled ones
2. (Original post by _HabibaH_)
I don't know :/ The MS explained it like this.

Oh gawd. Do you happen to know the year of the paper?
3. (Original post by Dslash)
Just divide by 3.
ARE YOU SURE? i thought the I2 and Cu2+ will have same ratio. So you are saying no of mol of I2 in one equation = no of mol of 3I2 in another ?
4. can anyone explain the answer to the question from jan 13 to do with entropy, where you had to work out temperatures and prove delta g = 0 or something? can't remember it now but looked at mark scheme earlier and didn't get it. it was the calculation after the bit about ice melting and entropy. would be really appreciated, thanks

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5. (Original post by Y_123456)
Could someone tell me why Ni have 2 unpaired d electrons?
Attachment 225790
The way the the sub shells full is, up-up-up-up-up-up-up-up-up-up-down-down-down-down-down-down-down-down-down-down
6. (Original post by eggfriedrice)
Oh gawd. Do you happen to know the year of the paper?
Yep, Jan 11.
7. (Original post by AyshaK)

Will Zn2+ ions in solution oxidise Fe metal to fe2+ ion?

First the half equations are
Zn 2+ + 2e- ---> Zn Eo = -0.76 V
Fe 2+ + 2e- ----> Fe E0 = -0.44 V

here we can see the most positive electrode potential is Fe 2+ so that will oxidise Zn to Zn2+ so Fe2+ acts as oxidising agent therefore Zn will not oxidise Fe metal to Fe2+

so the reaction stated in the question is not feasible if that makes sense?

Yes, I get it now, thank you , cause in the Jan 2011 paper, they asked us to predict which reactions will take place?
8. something tells me, we will see a shocking paper tomorrow
9. (Original post by georgiaaaxo)
can anyone explain the answer to the question from jan 13 to do with entropy, where you had to work out temperatures and prove delta g = 0 or something? can't remember it now but looked at mark scheme earlier and didn't get it. it was the calculation after the bit about ice melting and entropy. would be really appreciated, thanks

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deltaG = deltaH - TdeltaS
t = deltaH / delta S = (answer)
deltaG = DeltaH - (answer x delta S)
therefore was proved DeltaG=0
10. Can someone explain in bullet points the factors that affect enthalpy of hydration and lattice enthalpy?
11. Guys, someone explain electrode potentials and feasibility and how you know which has been reduced and oxidised and how the equation shows this and the movement on electrons?! It's all gone to mush in my brain thanks x

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12. By the way, you know the Zn and Cr2O7 redox, why are their eight electrons in the Cr2O7 redox?
13. (Original post by DudeBoy)
Yes well done
Do you think we'll need to know this as it came up in jan13?
14. (Original post by _HabibaH_)
Yep, Jan 11.
I'm looking at the paper now and cant find that question D:
15. (Original post by AyshaK)
deltaG = deltaH - TdeltaS
t = deltaH / delta S = (answer)
deltaG = DeltaH - (answer x delta S)
therefore was proved DeltaG=0
thanks! not sure why I didn't get that :/ will look at it tomorrow, cheers

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16. (Original post by v :))
Guys, someone explain electrode potentials and feasibility and how you know which has been reduced and oxidised and how the equation shows this and the movement on electrons?! It's all gone to mush in my brain thanks x

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The more positive value is reduced, and the less positive value is oxidised. If the cell potential difference is below 0.4V it's not feasible. Reduction half cell for reduction has electrons on the left hand side of the equation
17. (Original post by georgiaaaxo)
thanks! not sure why I didn't get that :/ will look at it tomorrow, cheers

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18. (Original post by v :))
Guys, someone explain electrode potentials and feasibility and how you know which has been reduced and oxidised and how the equation shows this and the movement on electrons?! It's all gone to mush in my brain thanks x

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The species with the lower electrode potential will become the negative electrode; the higher electrode potential becomes the positive one.

Think of a battery - electrons move from the negative terminal to the positive one. So the reaction at the negative electrode will release electrons (i.e. it will be oxidation - the equilibrium will go to the left). The positive electrode gains electrons - reduction.

The overall reaction in a cell is made by combining the half equations at each electrode. A reaction is feasible if a hypothetical cell between the two reactants would have said cell reaction and not the reverse.
19. can someone explain fuel cell and the equation in hydrogen-oxygen fuel cell
20. Say we are given the titration process and we find the no of moles of I2 is the last stage of the process, now we have to back track to the previous equation, however in this equation there is 3I2 and Cu2+ - so do we triple our no of I2 moles and then divide by three or do we only divide by three? ? ?

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