InterviewSolution
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InterviewSolution
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How would you determine the standard electrode potential of `Mg^(2+)|Mg`? |
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Answer» We will set up a cell consisting of `Mg|MgSO_(4)(1M)` as one electrode (by dipping a magnesium wire in 1M `MgSO_(4)` solution) and standard hydrogen electrode Pt, `H_(1)` (1 atm) `H^(+)|(1M)` as the second electrode and measure the EMF of the cell and also note the direction of deflection in the voltmeter. the direction of deflection shows that electronsf low from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode. Hence, the cell may be represented as: `Mg|Mg^(2+)(1M)||H^(+)(1M)|H_(2),(1atm),Pt` `E_(cell)^(@)=E_(H^(+),1//2H_(2))^(@)-E_(Mg^(2+),Mg)^(@)` put `E_(H^(+),1//2H_(2))^(@)=0` Hence, `E_(Mg^(2+),Mg)^(@)=-E_(cell)^(@)`. |
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