Question
How would you determine the standard electrode potential of the system $Mg^{2+}|Mg?$
✓
Answer
The standard electrode potential of $Mg^{2+} | Mg$ can be measured with respect to the standard hydrogen electrode, represented by $Pt_{(s)}, H_{2( g )} (1atm) | H^{+(aq)} (1M).$
A cell, consisting of $Mg | MgSO_4$ $($aq $1M)$ as the anode and the standard hydrogen electrode as the cathode, is set up.
$\text{Mg}|\text{Mg}^{2+}(\text{aq, 1M})||\text{H}^+(\text{aq, 1M})|\text{H}_2(\text{g, 1 bar}),\text{Pt}_{(s)}$
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
$\text{E}^\ominus=\text{E}^\ominus_\text{R}-\text{E}^\ominus_\text{L}$
Here, $\text{E}^\ominus_\text{R}$ for the standard hydrogen electrode is zero.
$\therefore\ \text{E}^\ominus=0-\text{E}^\ominus_\text{L}$
$=-\text{E}^\ominus_\text{L}$
A cell, consisting of $Mg | MgSO_4$ $($aq $1M)$ as the anode and the standard hydrogen electrode as the cathode, is set up.
$\text{Mg}|\text{Mg}^{2+}(\text{aq, 1M})||\text{H}^+(\text{aq, 1M})|\text{H}_2(\text{g, 1 bar}),\text{Pt}_{(s)}$
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
$\text{E}^\ominus=\text{E}^\ominus_\text{R}-\text{E}^\ominus_\text{L}$
Here, $\text{E}^\ominus_\text{R}$ for the standard hydrogen electrode is zero.
$\therefore\ \text{E}^\ominus=0-\text{E}^\ominus_\text{L}$
$=-\text{E}^\ominus_\text{L}$
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