How do you calculate the standard cell potential of a cell constructed from $M g^{2 +}$ $Mg^(2+)$ / $M g$ $Mg$ and $N i^{2 +}$ $Ni^(2+)$ / $N i$ $Ni$ . Which half-cell is the anode and which half-cell is the cathode?

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How do you calculate the standard cell potential of a cell constructed from $M g^{2 +}$ $Mg^(2+)$ / $M g$ $Mg$ and $N i^{2 +}$ $Ni^(2+)$ / $N i$ $Ni$ . Which half-cell is the anode and which half-cell is the cathode?

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Answer 1 · 2018-06-15T10:40:44

$E_{Cell}^{θ} = + 2.115 l V$ $E_"Cell"^theta=+2.115 color(white)(l) "V"$

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

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]

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

The standard reduction potential $E^{θ}$ $E^theta$ 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 ( $298 l K$ $298 color(white)(l) "K"$ , $1.00 l kPa$ $1.00 color(white)(l) "kPa"$ ) is defined as $0 l V$ $0 color(white)(l) "V"$ 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 $E^{θ}$ $E^theta$ and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower $E^{θ}$ $E^theta$ will experience oxidation and act the anode.

$E^{θ} ({Ni}^{2 +} / Ni) > E^{θ} ({Mg}^{2 +} / Mg)$ $E^theta ("Ni"^(2+) "/" "Ni") > E^theta("Mg"^(2+) "/" "Mg")$

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

$E_{cell}^{θ} = E^{θ} (Cathode) - E^{θ} (Anode)$ $E_"cell"^theta = E^theta ("Cathode") - E^theta ("Anode")$
$E_{cell}^{θ} = - 0.257 - (- 2.372)$ $color(white)(E_"cell"^theta) = -0.257 - (-2.372)$
$E_{cell}^{θ} = + 2.115$ $color(white)(E_"cell"^theta) = +2.115$

Indicating that connecting the two cells will generate a potential difference of $+ 2.115 l V$ $+2.115 color(white)(l) "V"$ 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

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How do you calculate the standard cell potential of a cell constructed from $M g^{2 +}$ $Mg^(2+)$ / $M g$ $Mg$ and $N i^{2 +}$ $Ni^(2+)$ / $N i$ $Ni$ . Which half-cell is the anode and which half-cell is the cathode?

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How do you calculate the standard cell potential of a cell constructed from $M g^{2 +}$ $Mg^(2+)$ / $M g$ $Mg$ and $N i^{2 +}$ $Ni^(2+)$ / $N i$ $Ni$ . Which half-cell is the anode and which half-cell is the cathode?