An application of the iodine-thiosulphate titration- pos rep for whoever helpsWatch

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Thread starter 13 years ago
#1
Bound to be some amazing scientists here nobody in academic help. pos rep up for grabs!!!

Theory

the first high temperature superconductors were a group of oxides of yttrium, barium and copper formula YBA2Cu30x. The value of x depends on the oxidation state of the copper. To determine the oxidation state of the copper (and hence x), the copper from the superconductor is reacted with iodide ions. The copper is reduced to Cu 1+ and the iodide is oxidised to iodine. The iodide thus produced is titrated against sodium thiosulphate. n.b. Yttrium normally exists as a 3+ ion. Neither Yttrium nor barium reac with the iodide ions.

Practical

0.0100 moles of the superconductorwere used. The copper in it was reacted with excess iodide ions. The iodine produced was made upto 100cm cubed in a volumetric flask, and 10cm cubed portions were extracted with a pipette and titrated against 0.500 mol/dmcubed sodium thiosulphate. Concordant readngs of 8.00cmcubed were obtained.

Question

Calculate the oxidation state/s of the copper in the superconductor and hence the value of x.
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13 years ago
#2
whoa this is beyond me
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13 years ago
#3
I did the iodine-thiosulphate titration before, but I have NO idea what this is about!
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13 years ago
#4
x=2?
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13 years ago
#5
I am not used to dealing with these things but correct me if wrong but I make Cu in the IV oxidation state, therefore making x = 6.

Therefore, formula; YBa2Cu3O6

Although im confused, as surely the overall oxidation state of the compound should equal zero should it not, and in this case it dosent.
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13 years ago
#6
There was total of 0.02 mole of iodine liberated in the reaction with copper. 2 moles of iodides give 2 moles of electrons when oxidized to 1 mole of iodine. Oxidized iodine liberated 2*0.02 moles of electrons.

There was 0.03 mole of copper in the SC sample. 1 mole of copper at nth oxidation state accepts n-1 moles of electrons when reduced to +1. Copper accepted 0.03*(n-1) moles of electrons.

Number of electrons lost and gained is identical, thus 0.03*(n-1)=2*0.02 and n=7/3.

That gives x=7 (3 + 2*2 + 3*7/3 = 2x).

Funny.

Best,
Borek
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13 years ago
#7
This has been asked before:

http://www.thestudentroom.co.uk/showthread.php?t=214664

The answer came to x=7. Or 6.6 in my case...
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13 years ago
#8
0.0100 moles of the superconductor were used.
The copper in it was reacted with excess iodide ions.
The iodine produced was made upto 100cm cubed in a volumetric flask,
and 10cm cubed portions were extracted with a pipette and titrated against 0.500 mol/dm cubed
sodium thiosulphate. Concordant readings of 8.00cm cubed were obtained.

moles of thiosulphate = 0.008 x 0.5 = 0.004 moles

2S2O3(2-) + I2 --> S4O6(2-) + 2I-

2 moles thiosulphate equivalent to 1 mole of iodine...
therefore there is 0.004/2 = 0.002 moles of iodine in the 10cm3 aliquot.

therefore in 100cm3 there is 0.02 moles of iodine

the iodine was produced by the reaction with copper ions of unknown oxidation state:
0.02 moles of iodine release 0.04 moles of electrons by 2I- -2e --> I2

from the formula of the superconductor YBa2Cu3Ox there are 0.03 moles of copper
and these absorb 0.04 moles of electrons to arrive at Cu+

therefore each mole of copper ions seems to absorb 4/3 moles of electrons!!!

Possible a mixed oxidation state of Cu3+ Cu2+ Cu2+ ??

This makes the formula YBa2Cu3Ox needing
3+ from the Y3+
4+ from the 2Ba2+
4+ from the 3Cu1.33+
total = 11+

this would make the oxygen 5.5

so where am I going wrong?
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13 years ago
#9
Yeah, I now make the oxidation state 7/3 too, and hence x =7.

Not sure what I did yesterday...
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13 years ago
#10
(Original post by charco)
therefore each mole of copper ions seems to absorb 4/3 moles of electrons!!!
Exactly - plus 1 as it is left as Cu+.

4/3 + 1 = 7/3

Best,
Borek
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13 years ago
#11
(Original post by Borek)
Exactly - plus 1 as it is left as Cu+.

4/3 + 1 = 7/3

Best,
Borek

yup - that's the one - thanks Borek
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