`E^0``(Zn^(2+)`/Zn)`=-0.76V ``E^0`(`Cu^(2+)`/Cu)`=0.34V ``E^0``(``Fe^(3+)``/Fe^(2+)`)=0.77V `E^0`(`H^(+)``/``H_(2)``=0V` The correct order of reducing nature of Zn,Cu,`Fe^(2+)` and `H_(2)` is

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@) =0.43V E_(Cu^(+)//Cu)^(@)=0.55V E_(Cu^(2+)//Cu^(+))^(@) =

Given E_(Ag^(+)//Ag)^(@)=+0.80 V, E_(Cu^(2+)//Cu)^(@)=+0.34 V, E_(Fe^(3+)//Fe^(2+))^(@)=+0.76 V, E_(Ce^(4+)//Ce^(3+))^(@)=+1.60 V Which of the following statements is not correct ?

Given that E_(Fe^(2+)//Fe)^(.)=-0.44V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V if Fe^(2+),Fe^(3+) and Fe solid are kept together then

If E_(Fe^(2+))^(@)//Fe = -0.441 V and E_(Fe^(3+))^(@)//Fe^(2+) = 0.771 V The standard EMF of the reaction Fe+2Fe^(3+) rarr 3Fe^(2+) will be:

E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

Given E^(c-)._(Ag^(o+)|Ag)=+0.80V, E^(c-)._(Co^(2)|Co)=-0.28V, E^(c-)._ (Cu^(2+)|Cu)=+0.34V E^(c-)._(Zn^(2+)|Zn)=-0.76V Which metal will corrode fastest ?

Which of the following will act as cathode when connected to standard hydrogen electrode which has E^(@) value given as zero ? (i) Zn^(2+)//Zn, E^(@)=-0.76V (ii) Cu^(2+)//Cu, E^(@)=+0.34 V (iii) Al^(3+)//Al, E^(@)=-1.66 V (iv) Hg^(2+)//Hg, E^(@)=+0.885 V (A)(i) and (ii) (B)(ii) and (iv) (C)(i) and (iii) (D)(i), (ii), (iii) and (iv)

Given: E_(Zn^(+2)//Zn)^(@) =- 0.76V E_(Cu^(+2)//Cu)^(@) = +0.34V K_(f)[Cu(NH_(3))_(4)]^(2+) = 4 xx 10^(11) (2.303R)/(F) = 2 xx 10^(-4) At what concentration of Cu^(+2) emf of the cell will be zero (at 298K) and concentration of Zn^(+2) is remains same: