Question

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 1

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.