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;
<br /> \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}<br />

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 :(