# How do you calculate the standard cell potential of a cell constructed from Mg^(2+)/Mg andNi^(2+)/Ni. Which half-cell is the anode and which half-cell is the cathode?

Jun 15, 2018

${E}_{\text{Cell"^theta=+2.115 color(white)(l) "V}}$

• Cathode $\text{Mg"^(2+) "/" "Mg}$
• Anode $\text{Ni"^(2+) "/" "Ni}$

#### Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

• $\text{Mg"^(2+) (aq) + 2color(white)(l) "e"^(-) to "Mg"(s) color(white)(-)E^theta = -2.372 color(white)(l) "V}$
• $\text{Ni"^(2+) (aq) + 2color(white)(l) "e"^(-) to "Ni"(s) color(white)(-)E^theta = -0.257 color(white)(l) "V}$

The standard reduction potential ${E}^{\theta}$ resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions ($298 \textcolor{w h i t e}{l} \text{K}$, $1.00 \textcolor{w h i t e}{l} \text{kPa}$) is defined as $0 \textcolor{w h i t e}{l} \text{V}$ for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher ${E}^{\theta}$ and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower ${E}^{\theta}$ will experience oxidation and act the anode.

${E}^{\theta} \left(\text{Ni"^(2+) "/" "Ni") > E^theta("Mg"^(2+) "/" "Mg}\right)$

Therefore in this galvanic cell, the $\text{Ni"^(2+) "/" "Ni}$ half-cell will experience reduction and act as the cathode and the $\text{Mg"^(2+) "/" "Mg}$ the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E_"cell"^theta = E^theta ("Cathode") - E^theta ("Anode")
$\textcolor{w h i t e}{{E}_{\text{cell}}^{\theta}} = - 0.257 - \left(- 2.372\right)$
$\textcolor{w h i t e}{{E}_{\text{cell}}^{\theta}} = + 2.115$

Indicating that connecting the two cells will generate a potential difference of $+ 2.115 \textcolor{w h i t e}{l} \text{V}$ across the two cells.

References
[1] "1.2: Standard Electrode Potentials", Chemistry LibreTexts,
https://chem.libretexts.org/LibreTexts/University_of_California_Davis/UCD_Chem_002C/UCD_Chem_2C%3A_Larsen/Chapters/Unit_1%3A_Electrochemistry/1.2%3A_Standard_Electrode_Potentials

[2] "17.3: Standard Reduction Potentials", Chemistry LibreTexts, https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Book%3A_Chemistry_(OpenSTAX)/17%3A_Electrochemistry/17.3%3A_Standard_Reduction_Potentials