Electrolysis in Cu/Hg electrolytic cell

[steven10137]

Homework Statement

http://img143.imageshack.us/img143/7135/electrolysisrf2.jpg

What is the ratio of number of moles of copper deposited to the number of moles of mercury deposited? {n(Cu):n(Hg)}

2. The attempt at a solution

Well the species present are;
$$\begin{array}{l}<br /> Cu^{2 + } /SO_4 ^{2 - } \;ions \\ <br /> Hg_2 ^{2 + } /NO_3 ^ - \;ions \\ <br /> and\;H_2 O \\ <br /> \end{array}$$

so reading down the Eo table from left to right; the first species found is $$Cu^{2 + }$$ ions.
So at the cathode we have; $$Cu^{2 + } + 2e^ - \to Cu$$

Now reading up the Eo table from right to left; the first species found is $$Hg_2 ^{2 + }$$
So I would assume that at the anode we have; $$Hg_2 ^{2 + } \to 2Hg^{2 + } + 2e^ -$$

Now the question states that it is the ratio of copper to mercury produced. How can this be?
Do I just assume that the mercury ions are now further electrolysed to mercury by;
$$Hg^{2 + } + 2e^ - \to Hg_{(l)}$$

Could someone please explain how this works and what rules I am supposed to be following as I am a little confused ...

cheers
Steven

 

[steven10137]

Anybody

- sorry I have an exam Friday :(