# 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: $$\ce{Ag}$$, $$\ce{Hg}$$, $$\ce{Cu}$$.

Why isn't Mg deposited on the cathode? Is it because the reduction potential is negative?
If yes, what exactly is the reason?

• 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 aqueous solution. Everywhere where there is an aqueous solution, you have a certain concentration of a further cation: $$\ce{H+}$$. It has a potential of $$\pu{\pm0 V}$$ 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.