Represent the cell in which the following reaction takes place: Mg(s)+2Ag^(+)(0.0001M)toMg^(2+)(0.130M)+2Ag(s) Calculate its E_(cell). Given that E_(Mg^(2+)//Mg)^(@)=-2.37V and E_(Ag^(+)//Ag)^(@)=+0.80V

Represent the cell in which the following reaction takes place: Mg(s)

1.	Represent the cell in which the following reaction takes place: Mg(s)+2Ag^(+)(0.0001M)toMg^(2+)(0.130M)+2Ag(s) Calculate its E_(cell). Given that E_(Mg^(2+)//Mg)^(@)=-2.37V and E_(Ag^(+)//Ag)^(@)=+0.80V
Answer» Solution :Here, we are given reduction potential as `E_(Mg^(2+),Mg)=-2.37,E_(AG^(+),Ag)=+0.80V` As the emf of the cel must be positive, this can be so only if oxidation takes PLACE at the MAGNESIUM electrode, hence, the electrode reaction will be `MgtoMg^(2+)+2e^(-)` (At anode) `2Ag^(+)+2e^(-)to2Ag` (At cathode) Thus, the cell may be REPRESENTED as `Mg\|Mg^(2+)(0.130M)\|\|Ag^(+)(0.0001M)\|Ag` Standard emf of the cell will be: `E_(cell)^(@)=`Std. Red. Pot. Of R.H.S. electrode-Std. Red. Pot. Of L.H.S. electrode=0.80-(-2.37)=3.17V The overall reaction is: `Mg+2Ag^(+)hArrMg^(2+)+2Ag""(n=2)` Applying nernst eqn., we get `E_(cell)=E_(Cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)])/([Ag^(+)]^(2))` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"(0.130)/((10^(-4))^(2))=E_(Cell)^(@)-(0.0591)/(2)"log"(0.130)/(10^(-8))` `=3.17-0.02955log(1.30xx10^(7))=3.17-0.02955xx(7.1139)=3.17-0.21=2.96`volt. ALTERNATIVELY, this problem may be solved by first calculating the electrode potentials of the two electrodes separately and then calculating the emf from the electrode potentials.