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

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

Reply 1

Wenzel

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.

Reply 2

amo17

OP

2

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

Reply 3

Wenzel

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.

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...

Reply 5

Carlo08

Its:

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

The copper (i) has oxidised the other and it has been reduced itself.

Reply 6

Da14a

16

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...