# Will magnesium undergo a redox reaction with zinc(II) or copper(II)?

Magnesium can do a redox reaction with $$\ce{Cu^2+}$$ as well as with $$\ce{Zn^2+}$$, since Mg has the lowest reduction potential $$E_0$$. (I am aware that a reaction may or may not be feasible as redox, and this is according the reduction potential of the two half-equations)

$$\begin{array}{l|l|r} \ce{Cu^2+ & Cu} &\pu{0.34 V} \\ \ce{Zn^2+ & Zn} &\pu{-0.76 V} \\ \ce{Mg^2+ & Mg} &\pu{-2.37V} \\ \end{array}$$

I'm supposing we have in one beaker, a solution of $$\ce{Cu^2+}$$ ions and $$\ce{Zn^2+}$$ ions and a piece of magnesium. So the possible reactants are: $$\ce{Cu^2+}$$, $$\ce{Zn^2+}$$, and $$\ce{Mg}$$.

The potential difference between $$\ce{Mg}$$ and $$\ce{Cu^2+}$$ is greater than with $$\ce{Zn^2+}$$. So what's the answer?

My guess is "with $$\ce{Zn^2+}$$", since by comparing with another example, ethanol ($$\ce{CH3CH2OH}$$) reduces $$\ce{MnO4-}$$ to give $$\ce{CH3CHO}$$ instead of $$\ce{CH3COOH}$$, if the quantity of $$\ce{MnO4-}$$ was limited. $$\begin{array}{l|l|r} \ce{MnO4- & Mn^2+} & \pu{1.51 V} \\ \ce{CH3CHO & CH3CH2OH} & \pu{0.19 V} \\ \ce{CH3COOH & CH3CH2OH}& \pu{0.05 V} \end{array}$$

So $$\ce{CH3CH2OH}$$ reacted with $$\ce{MnO4-}$$ to give $$\ce{CH3CHO}$$, that leads me to deduce that in the case of multiple possibilities (i.e. feasible redox reactions), the reaction with the least potential difference takes place (since $$1.51 − 0.19 < 1.51 − 0.05$$).

From another hand, I guess it's familiar that the other option of having $$\ce{Mn^2+}$$ reacting with $$\ce{Mg}$$ is true as being held between the most oxidant which is $$\ce{Mn^2+}$$ and the most reductant which is solely $$\ce{Mg}$$ in our case.

• I wonder if an answer could be in antioxidant in the body. – YoussefDir Dec 24 '15 at 10:36

Let’s assume for a second that magnesium reacts only with the zinc. The reaction would occur according the following equation:

$$\ce{Mg + Zn^2+ -> Mg^2+ + Zn}$$

You have already identified, if I read your question correctly that the redox potentials allow this and that the reaction is spontaneous etc. Assume that you have equimolar amounts of magnesium, zinc and copper for the rest of this thought experiment.

Now our new solution would contain magnesium(II), copper(II) and zinc(0) (okay, the zinc would deposit itself somewhere, but you get the picture). Notice anything? Yes, zinc and copper can react according to the well-known and often-displayed electrochemical reaction (depositing copper on a blank piece of zinc metal). The equation for that one is:

$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$$

Again, check out the standard potentials to see that this reaction is possible and spontaneous.

If we add the two reactions together, we get an effective reaction:

$$\ce{Mg + Zn^2+ + Zn + Cu^2+ -> Mg^2+ + Zn + Zn^2+ + Cu}$$

$$\ce{Mg + Cu^2+ -> Mg^2+ + Cu}$$

You notice that the zinc is little more than a spectator ion.

Now how does this differ from the ethanol case? Well, permanganate can (and will) react with ethanol to form ethanal, and it can (and will) react with ethanal to form ethanoic acid — more rapidly. However, if permanganate (and therefore: electrons) is the limiting reagent, we end up with a mixture of ethanol, maybe ethanal and ethanoic acid. By a reaction analogous to the reverse Cannizarro reaction, these can comproportionate:

$$\ce{CH3CH2OH + CH3COOH <=>> 2 CH3CHO}$$

Therefore, the overall reaction seems to be one where ethanol only reacted with permanganate to only form ethanal, while in fact ethanoic acid was formed, but in the end the equilibrium favoured ethanal.

$$\ce{Mg^2+}$$ will undergo a redox reaction with both the species.

Have a look at the electrode potential values at this link: http://www.physchem.co.za/data/electrode_potentials.htm via the Internet Archive

What you'll find is that the half reaction of $$\ce{Mg^2+}$$ has a highly negative electrode potential $$-2.37$$ while the $$\ce{Zn^2+}$$ and $$\ce{Cu^2+}$$ half reactions have less negative $$E_0$$ values [$$-0.76$$ and $$0.34$$ respectively]. A highly negative electrode potential indicates a strong tendency to reduce other species, while it gets oxidized itself. The $$E_0$$ values for the net reactions in both cases is positive. [Try it out: $$E(\text{cathode})-E(\text{anode})$$].

Consider the reverse reaction, i.e.
\begin{align} \ce{Cu + Mg^2+ &-> Mg + Cu^2+}\\ \tag{anode} \ce{Cu &-> Cu^2+ + 2e-} & (E_0 &= +0.34)\\ \tag{cathode} \ce{Mg^2+ + 2e- &-> Mg} &(E_0 &= -2.37) \end{align}
The $$E_0$$ of the net reaction is $$E(\text{cathode})-E(\text{anode})$$ which turns out to be negative. The change in Gibbs free energy- given by the expression $$-nFE_0$$, is thus positive for this reaction, implying that it is not feasible - i.e. $$\ce{Mg^2+}$$ will not be reduced by $$\ce{Cu^2+}$$. If you try this out with $$\ce{Zn^2+}$$, you'll end up with the same conclusion.

So, in response to your question - $$\ce{Mg^2+}$$ will participate in a redox reaction with both the species in which $$\ce{Mg}$$ gets oxidized while simultaneously reducing the other species.

• Yash, I'm aware that E(cathode)−E(anode) must be positive so the reaction is feasible. I have edited my question to clarify it. What I actually meant is that in one container we have Mg, and both the $\ce{Zn^2+}$ and $\ce{Cu^2+}$. – YoussefDir Dec 10 '15 at 21:02