# Why isn't Mg deposited on the cathode after electrolysis?

Well here's a question from my book:

A solution containing one mole per liter of each $\ce{Cu(NO3)2}$, $\ce{AgNO3}$, $\ce{Hg2(NO3)2}$ and $\ce{Mg(NO3)2}$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are:

• $\ce{Ag+/Ag} = +0.80$,
• $\ce{Hg2^2+/Hg} = +0.79$,
• $\ce{Cu^2+/Cu} = +0.34$,
• $\ce{Mg^2+/Mg} = -2.37$

The sequence of deposition of the metals on the cathode will be?

The answer to the above question is: Ag, Hg, Cu.
Why isn't Mg deposited on the cathode? Is it because the reduction potential is negative?
If yes, what exactly is the reason?

• I have improved the formatting with $\LaTeX$ (edit pending approval). For more information on how to do so yourself, see here and here. – Jan Jun 12 '15 at 18:30
• Your solution is aquaeous. There is another cation present. It very much has something to do with magnesium's potential being positive. Do you maybe see it? ;) – Jan Jun 12 '15 at 18:32
• @Jan Water! Right. Didn't think about that. Thanks. – Tabish Mir Jun 13 '15 at 15:11

You are in aquaeous solution. Everywhere where there is an aquaeous solution, you have a certain concentration of a further cation: $\ce{H+}$. It has a potential of $\pm 0\,\mathrm{V}$ by definition, thereby being nobler than $\ce{Mg^2+}$.
Reduction takes place on cathode. Thus the element having most reduction potential will get reduced and get deposited. In your question the reduction potential of $\ce{Mg}$ is the least of all the ones that you have mentioned hence it will not get deposited.