Question

How would you determine the standard electrode potential of the system Mg²⁺ | Mg?

Answer 1

Set up an electrochemical cell consisting of Mg | MgSO₄ (1 M) as one electrode by dipping a magnesium wire in (1 M) MgSO₄ solution and the standard hydrogen electrode Pt, 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:

Mg | Mg²⁺ (1 M) || H⁺ (1 M) | H₂ (1 atm), Pt

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

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

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

It is observed that the EMF of the cell is 2.36 V.

Hence, the standard electrode potential (E°) for the Mg²⁺ | Mg system will be −2.36 V.