# A2 Chem - disportionation!!!!!!!!!

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#1
A disproportionation reaction occurs when a species M+ spontaneously undergoes simultaneous
oxidation and reduction.
2M+(aq) → M2+(aq) + M(s)
The table below contains E data for copper and mercury species.
Using these data, which one of the following can be predicted?
A Both Cu(I) and Hg(I) undergo disproportionation.
B Only Cu(I) undergoes disproportionation.
C Only Hg(I) undergoes disproportionation.
D Neither Cu(I) nor Hg(I) undergoes disproportionation.

the answer is B, but does anyone understand how to get it?

Any help appreciated
Thanks
0
12 years ago
#2
(Original post by amo17)

The table below contains E data for copper and mercury species.
Using these data, which one of the following can be predicted?
I think we need this table first.
0
#3
sorry, didn't notice that this didn't copy too.

Cu2+(aq) + e– → Cu+(aq) + 0.15
Cu+(aq) + e– → Cu(s) + 0.52
Hg2+(aq) + e– → Hg+(aq) + 0.91
Hg+(aq) + e– → Hg(l) + 0.80

Thanks
0
12 years ago
#4
Well the first equation (2M+(aq) → M2+(aq) + M(s)) tells you that during disproportionation, an aqueous 2+ ion and and a solid is formed. From the E values you can see that C+ forms a solid, where as Hg+ forms a liquid; so the only possible answer can be B.

Not sure if there's any other way of doing this question, but that's how I'd work it out.
0
12 years ago
#5
To do the copper disproportionation the hypothetical equation would involve reacting the reverse of equation 1 with equation2 (from the data list)

The Cu+ in equation 1 is the species getting oxidised and the Cu+ in equation 2 is the species getting reduced.

E = E(red) - E(ox) = +0.52 - +0.15 = +0.37
This is a positive value and larger than 0.3 suggesting that the system is both feasible and spontaneous and will go all the way.

Repeat the same process for the mercury...
1
12 years ago
#6
Its:

Cu+ + Cu+ -------> Cu + Cu^2+

The copper (i) has oxidised the other and it has been reduced itself.
0
3 months ago
#7
(Original post by charco)
To do the copper disproportionation the hypothetical equation would involve reacting the reverse of equation 1 with equation2 (from the data list)

The Cu+ in equation 1 is the species getting oxidised and the Cu+ in equation 2 is the species getting reduced.

E = E(red) - E(ox) = +0.52 - +0.15 = +0.37
This is a positive value and larger than 0.3 suggesting that the system is both feasible and spontaneous and will go all the way.

Repeat the same process for the mercury...
Thank you for this from 12 years in the future.
1
3 months ago
#8
(Original post by Da14a)
Thank you for this from 12 years in the future.
0
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