`Mg(s) |Mg^(2+) (0.1M)||Cu^(2+) (1xx 10^(-1) M) |Cu(s)` सेल के EMF की गणना कीजिए। दिया है। `E_(Cu^(2+))^(0)|Cu=0.34 V`
`E_(Mg^(2+)^(0) |Mg = - 2.37 V`

Calculate the e.m.f of the cell Mg(s)//Mg^(2+)(0.1 M)||Cu^(2+)(1.0xx10^(-3) M)//Cu(s) Given E_(Cu^(2+)//Cu)^(@)=+0.34 V and E_(Mg^(2+)//Mg)^(@)=-2.37 V

Calculate the e.m.f. of the following cell at 298 K. Mg_((s))|Mg^(2+)(0.001M)||Cu^(2+)(0.0001M)|Cu_((s)) Given E^(@)(Cu^(2+)|Cu)=0.34V and E^(@)(Mg^(+)|Mg)=-2.37V .

Write the Nernst equation and emf of the following cell at 298 K. (i) Mg(s) |Mg^(2+) ( 0.001 M) || Cu^(2+) ( 0.0001M ) | Cu(s) (ii) Fe(s) | Fe^(2+) ( 0.001M) || H^(+) ( 1M) | H_(2)(g) (1 "bar") | Pt(s) Given that, E_(Mg^(2+) //Mg)^(@) = - 2.36V , E_(Cu^(2+) Cu)^(@) = 0.34V, E_(Fe^(2+)Fe)^(@) = -0.44V

Calculate equilibrium constant for the reaction : Mg(s) |Mg^(2+)(0.001 M) || Cu^(2+)(0.0001 M)|Cu(s) Given E_((Mg^(2+)//Mg))^(@)=-2.37" V " , E_((Cu^(2+)//Cu))^(@)=0.34" V "

Calculate the emf of the cell. Mg(s)|Mg^(2+) (0.2 M)||Ag^(+) (1xx10^(-3))|Ag E_(Ag^(+)//Ag)^(@)=+0.8 volt, E_(Mg^(2+)//Mg)^(@)=-2.37 volt What will be the effect on emf If concentration of Mg^(2+) ion is decreased to 0.1 M ?

Calculate the emf of the cell. Mg(s)|Mg^(2+) (0.2 M)||Ag^(+) (1xx10^(-3))|Ag E_(Ag^(+)//Ag)^(@)=+0.8 volt, E_(Mg^(2+)//Mg)^(@)=-2.37 volt What will be the effect on emf If concentration of Mg^(2+) ion is decreased to 0.1 M ?