For a given data : E_(Mg^(2+) // Mg)^(@) = -2.37 V , E_(Cu^(2+)// Cu)^(@) = + 0.34 V Calculate the emf of the cell in which the following reaction takes place . underset((0.0001 M)) ( Mg_((s))) + Cu_((aq))^(2+)underset((0.001M)) to Mg_((aq))^(2+) + Cu_((s))

For a given data : E_(Mg^(2+) // Mg)^(@) = -2.37 V , E_(Cu^(2+)// Cu)^

1.	For a given data : E_(Mg^(2+) // Mg)^(@) = -2.37 V , E_(Cu^(2+)// Cu)^(@) = + 0.34 V Calculate the emf of the cell in which the following reaction takes place . underset((0.0001 M)) ( Mg_((s))) + Cu_((aq))^(2+)underset((0.001M)) to Mg_((aq))^(2+) + Cu_((s))
Answer» Solution :Limiting molar CONDUCTIVITY of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.If the molar conductivity of cation is denoted by and that of anion `X^@`. Where `V_(+)` and `V_(-)` are the number of cations and anions per formula unit of an electrolyte. (ii)Mercury cell is a primary cell. SUITABLE for low CURRENT devices. Anode: Zinc - Mercury amalgam Cathode:Paste of HgO and Carbon Electrolyte: Paste of KOH and ZnO. Cell reactions:Anode:`Zn(HG) 4- 2OH to ZnO(s) + H_(2)O + 2e^(-)` Cathode:`HgO + H_(2)O_(4)- 2e^(-) to Hg(Z) 4- 2OH^(-)` Over all reaction:`Zn(Hg) + HgO(s) to ZnO(s) + Hg(Z)` Cell potential1.35 V.