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?

  • 2
    $\begingroup$ 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? ;) $\endgroup$
    – Jan
    Jun 12 '15 at 18:32
  • $\begingroup$ @Jan Water! Right. Didn't think about that. Thanks. $\endgroup$
    – 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 $\ce{Mg^2+}$.

Thus, hydrogen gas is ‘deposited’ on the cathode first and bubbles away while the solution gets removed.

It is not possible to deposit considerably less noble metals on the cathode in normal, aqueous solutions.


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.

  • 1
    $\begingroup$ It doesn't answer this. $\endgroup$
    – Mithoron
    Sep 17 '15 at 14:45

Since the reduction potential is negative, the process becomes non spontaneous. Moreover, there are hydrogen ions already present having zero assumed reduction potential. Therefore, Mg won't get deposited


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