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In this problem I'm asked to calculate the electrode potential of the magnesium electrode ($E_{Mg^{2+}/Mg}$) given the red-ox reaction:$$\text{Mg(s)}+2\text{Ag}^+(10^{-2}\text{M})\leftrightarrow 2\text{Ag(s)}+\text{Mg}^{2+}(10^{-2}\text{M})$$

The information I'm given is:

  1. Temperature equal to $278\,K$ (the standard).
  2. The molar concentration of the $\text{Mg}^{2+}$ ions is of $0.01\,\text{M}$.
  3. The standard reduction potential of the $Mg$ electrode is: $E^{º}_{Mg^{2+}/Mg}=-2.34\,V$.

I know I have to use the Nernst Equation, but my solution and the book's answer are very different.

My solution:

$$E_{Mg^{2+}/Mg}=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#ff3300}{\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}}=\ldots=\boxed{-2.28\,V}$$

My book's solution: $$E_{Mg^{2+}/Mg}=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#00ff00}{\mathbf{\log\frac{1}{\left[\text{Mg}^{2+}\right]}}}=\ldots=\boxed{-2.40\,V}$$

($n=2$ because 2 moles of electrons where exchanged per 1 mole of reaction)


I've colored the part where I go wrong (basically the logarithm). I don't know exactly why the book put the concentration in the denominator, since $\text{Mg}^{2+}$ is a product (not a reactant), therefore following what I studied, its concentration should be in the numerator.

Any help? Sorry for asking such an elementary question, but I'm still learning the basic stuff in Chemistry!

Note I attach an image of the solution provided by the book enter image description here

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Zhe's answer is the best. Though, I will try to offer a different (mathematical) approach, which will hopefully enable you to see the answer yourself, or rather, verify that your book's solution is indeed correct.

According to you, the standard oxidation potential of magnesium electrode would be $E_{Mg/Mg^{2+}}=E^{º}_{Mg/Mg^{2+}}-\dfrac{0.059}{n}\,\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}$. Agreed?

Also, note that $E_{Mg/Mg^{2+}}=-E_{Mg^{2+}/Mg}$ always holds. Agreed?

Then, $E_{Mg^{2+}/Mg}=-E_{Mg/Mg^{2+}}$ $$=-(E^{º}_{Mg/Mg^{2+}}-\dfrac{0.059}{n}\,\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}})$$ $$=(-E^{º}_{Mg/Mg^{2+}})-\dfrac{0.059}{n}\,\cdot(-1)\cdot\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}$$. $$=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\mathbf{\log\frac{1}{\left[\text{Mg}^{2+}\right]}}$$.

Which is exactly the answer given by your book.

I hope it helps!

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The equation that corresponds to the standard reduction potential is:

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

Notice that this is not the actual half reaction in your total equation, where magnesium is oxidized. But the book is determining the reduction potential, so the magnesium ion is a reactant not a product. In the reaction quotient, the concentration/activity of magnesium ion should be in the denominator.

I'm not sure I agree that we should choose the standard reduction potential as the electrode potential for the electrode where oxidation occurs, but at least the book is consistent.

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  • $\begingroup$ According to my book, the equation that corresponds to the standard reduction potential is the same as you wrote, but with the arrow pointing to the other side. I'm a bit lost, could you clarify this in your answer please? I am beginning with the Chemistry of reactions $\endgroup$ – Jose Lopez Garcia Feb 1 '18 at 14:35
  • $\begingroup$ That would be incorrect. The reaction I have written is the reduction. If you reverse the reaction, that would be an oxidation. @JoseLopezGarcia $\endgroup$ – Zhe Feb 1 '18 at 14:41
  • $\begingroup$ But the $Mg$ is the anode in the galvanic cell. It's being oxidized by the $Ag$. Is this correct? $\endgroup$ – Jose Lopez Garcia Feb 1 '18 at 15:34
  • $\begingroup$ (In my first comment I said "reduction" but I meant "oxidation", apologies) $\endgroup$ – Jose Lopez Garcia Feb 1 '18 at 15:35
  • $\begingroup$ Yes, that's my whole point. I don't understand why the book is talking about the half reaction as if it were a reduction. It doesn't make sense to me. But if you are going to talk about it as a reduction, then the magnesium ion is a reactant not a product, which means it is in the denominator of the reaction quotient. $\endgroup$ – Zhe Feb 1 '18 at 15:56

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