Calculate the cathode electrode potential in this redox reaction

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](https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/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)




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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](https://i.stack.imgur.com/VlwUh.png)](https://i.stack.imgur.com/VlwUh.png)




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!


 2

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.


 2