Helppp

emf values
Cr2O72-(aq) + 14H+(aq) + 6e à 2Cr3+(aq) + 7H2O(l)
+1.33

Cu2+(aq) + e à Cu+(aq)
+0.15
Cu2+(aq) + 2e à Cu(s)
+0.34
Cu+(aq) + e à Cu(s)
+0.52

Using this, I need to prove that CuSO4 and CrCl3 don't react

I know copper is a better reducing agent and so it will itself be oxidised, this means it will be on LHS of the cell.

So doing 1.33 - any one of the Cu equations gives me a +ve result???

Just by speculating I know this reaction can't happen, but I can't seem to prove it... the only way I could get a negative result is if the dichromate equation showed oxidation, but this can't be possible because of its high electrode potential?

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Reply 1

Pigster

20

CuSO4 doesn't exist?!?

I expect you mean Cu2SO4(aq) doesn't exist. In which case, it because it disporportionates into Cu(s) + Cu2+(aq).

Also I suspect you're missing the data for Cl2 + 2e- -> 2Cl-, which I think will explain about CrCl3.

Reply 2

ps1265A

OP

14

Original post by Pigster

CuSO4 doesn't exist?!?

I expect you mean Cu2SO4(aq) doesn't exist. In which case, it because it disporportionates into Cu(s) + Cu2+(aq).

Also I suspect you're missing the data for Cl2 + 2e- -> 2Cl-, which I think will explain about CrCl3.

Nope, the question says CuSO4 :/

And I've only been given data that I've stated above, nothing for chlorine

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Reply 3

Pigster

20

It was a bit early when I read the OP...

Cr2O72-(aq) + 14H+(aq) + 6e à 2Cr3+(aq) + 7H2O(l) +1.33
AND
Cu2+(aq) + e à Cu+(aq) +0.15
OR
Cu2+(aq) + 2e à Cu(s) +0.34

The more +ve half equation will go to the RHS, the more -ve will go to the LHS.
The feasible reaction happens between Cr2O72- and either Cu+ or Cu forming Cr3+ and Cu2+.

The OP asks why they don't react. The reason is that the reverse reaction (that suggested above) is the feasible reaction, the reverse can't be feasible.

Reply 4

ps1265A

OP

14

Original post by Pigster

It was a bit early when I read the OP...

Cr2O72-(aq) + 14H+(aq) + 6e à 2Cr3+(aq) + 7H2O(l) +1.33
AND
Cu2+(aq) + e à Cu+(aq) +0.15
OR
Cu2+(aq) + 2e à Cu(s) +0.34

The more +ve half equation will go to the RHS, the more -ve will go to the LHS.
The feasible reaction happens between Cr2O72- and either Cu+ or Cu forming Cr3+ and Cu2+.

The OP asks why they don't react. The reason is that the reverse reaction (that suggested above) is the feasible reaction, the reverse can't be feasible.