A galvanic cell is made by dipping a Zn - rod in 0.1 (M) `Zn(NO_(3))_(2)` and a Pb - rod in `0.25 (M) Pb(NO_(3))_(2)` solution. Calculate EMF of the cell, write the cell reaction and represent the cell. Given : `E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Zn^(2+)|Zn)^(@)=-0.77V.`

A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn (NO_(3))_(2) solution and metallic plate of lead in 0.02M Pb(NO_(3))_(2) solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. ("Given" : E^(@) Zn^(2+)ZN=0.76V, " " E^(@)Pb^(2+)//Pb = -0.13V)

A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn (NO_(3))_(2) solution and metallic plate of lead in 0.02M Pb(NO_(3))_(2) solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. ("Given" : E^(@) Zn//Zn^(2+)=0.76V, " " E^(@)Pb^(2+)//Pb = -0.13V)

Find the E_(cell)^(@) for the following cell reaction Fe^(+2)+Zn rarr Zn^(+2)+Fe Given E_(Zn//Zn^(+2))^(@)=0.76V,E_(Fe//Fe^(+2))^(@)=+0.41 V

A half - cell is constructed by dipping a Zn- rod into a solution of ZnSO_(4) at 25^(@)C . If the concentration of Zn^(2+) ion in the solution be 0.01 (M), calculate the oxidation potential of the half- cell, Given : E_(Zn^(2+)|Zn)^(@)=-0.76V .