`E_(Zn^(2+)//Zn)^(0)=-0.76V` The EMF of the cell `Zn//Zn_((1M))^(2+)"||"HCl(pH=2)|H_2(1 atm)"),Pt` is

For Zn^(2+)|Zn, E^(@)=-0.76V then EMF of the cell Zn|Zn^(2+)(1M)||2H^(+)(1M)|H_(2)(1atm) will be -

E_(Cu^(2+)|Cu)^@ = +0.337 V, E_(Zn^(2+)| Zn)^@ = - 0.762 V. The EMF of the cell , Zn|Zn^(2+) (0.01M)||Cu^(2+) (0.01M)|Cu is

E^(0) for the half cell Zn^(2+) // Zn is - 0.76 V . Emf of the cell Zn// Zn^(2+)(1M) // //H^(+)(1M) // H_2 at 1 atm is

E^@ for the half cell Zn^(2+)|Zn is -0.76V emf of the cell Zn|Zn^(2+) (1M) |2H^+ (1M) |H_2 (1atm) is

Calculate emf of the cell Zn_((s))|Zn^(2o+) (0.2M)|Ho+ (1.6M)H_2(g.1.8atm)|Pt^(a)t 25^@C .

E^@ values of Zn^(2+), Zn and Cu^(2+), Cu are respectively -0.76 V and + 0.34 . Calculate the EMF of the cell Zn//Zn^(2+) (0.1M)"//"Cu^(2+)(0.1 M)//Cu .

At 25^(@)C tempertaure, E_("cell")^ of the galvanic cell Zn|Zn^(2+)(aq, 1M)||H^(+)(aq)|H_(2)("g, 1 atm")|Pt is +0.583V . If E_(Zn^(2+)|Zn)^(@)=-0.76V , then pH of the aqueous solution present in the reduction half - cell is -

Calculate, emf of the cell at 25^@C Zn(s)|Zn^(2o+)(0.2M)|| H^(o+)(1.6M)|H_2(g, 1.8 atm)|Pt