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How would you determine the standard electrode potential of the system Mg^(2+) | Mg ? |
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Answer» Solution :* 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. `Mg|Mg_((aq))^(2+)(1M)||H_((aq))^(+)(1M)|H_(2(g))(1" bar")|Pt_((S))` * Then, the emf of the cell is measured and this measured emf is the standard hydrogen electrode as the cathode is set up. `Mg|Mg_((aq))^(2+)(1M)||H_((aq))^(+)(1M)|H_(2(g))(1" bar")|Pt_((S))` * Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode. `E^(Theta)=E_(R)^(Theta)-E_(L)^(Theta)` * Here, `E^(Theta)` for the standard hydrogen electrode is ZERO. THEREFORE, `E^(Theta)=0-E_(L=E_(L)^(Theta))^(Theta)` |
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