Question

How would you determine the standard electrode potential of the system \[\ce{Mg^{2+} | Mg}\]?

Answer 1

Set up an electrochemical cell consisting of \[\ce{Mg | MgSO4 (1 M)}\] as one electrode by dipping a magnesium wire in \[\ce{(1 M) MgSO4}\] solution and the standard hydrogen electrode \[\ce{Pt}\], \[\ce{H2 (1 atm) | H+ (1 M)}\] as the second electrode.

Measure the cell's EMF and note the direction of deflection in the voltmeter. The direction of deflection indicates that electrons flow from the magnesium electrode to the hydrogen electrode, implying that oxidation occurs on the magnesium electrode and reduction on the hydrogen electrode.

Thus, the cell may be represented as follows:

\[\ce{Mg | Mg^{2+} (1 M) || H+ (1 M) | H2 (1 atm), Pt}\]

\[\ce{E^Θ_{cell} = E^Θ_{{H^{+}}/{{H_2}}} - E^Θ_{{Mg^{2+}/Mg}}}\]

Put \[\ce{E^Θ_{{H^{+}}/{{H_2}}}}\] = 0

Hence \[\ce{E^Θ_{{Mg^{2+}/Mg}}}\] = \[\ce{- E^Θ_{cell}}\]

It is observed that the EMF of the cell comes out to be 2.36 V.

Hence, the standard electrode potential for the \[\ce{Mg^{2+} | Mg}\] system will be E^Θ = −2.36 V.