The EMF of the cell `Ni|Ni^(2+)` (0.01M)
`Cl^(-)(0.01M)//Cl_(2)`, pt is__ V if the SRP of nickel and chlorine electrodes are `-0.25` and `+1.36 V` respectively

The EMF of the cell Ni//Ni^(2+)(0.01)// // Cl^(-)(0.01M)//Cl_(2) , pt is -----V if the SRP of nickel and chlorine electrodes are -0.25V and +1.36V respectively

The potential of the cell Cu, Cu^(2+) (0.1M)"//"HCl (xM), Cl_(2), Pt is 1.07 V. If the standard potential of copper and chlorine electrodes are 0.34 V and 1.36 V, calculate the concentration of HCI.

For the cell , Hg,Hg_2 Cl_2//Cl (0.1M)//Cl^(-)(0.01M)//Cl_2, Pt E_("cell")^@ is 1.10V , Hence , E_("cell") is

E^(@) values of Ni^(2+) , Ni and Cl_(2), Cl^(-) are respectively -0.25V and +1.37V . Calculate the EMF of the cell Ni, Ni^(2+) (0.01M)//Cl^(-)(0.1M), Cl_(2) , pt. Potential of nickel electrode is given as,

Calculate E of the cell with Cu^(2+) (0.1)//Cu electrode and Cl^(-)(0.1M)// 1/2Cl_(2) , Pt electrode. E^(@) Cu^(2+)//Cu= +0.33V and E^(@) 1/2 Cl_(2)//Cl^(-) = 1.36V

The cell Pt|H_(2)(g,01 bar) |H^(+)(aq),pH=x||Cl^(-)(1M)|Hg_(2)Cl_(2)|Hg|Pt has emf of 0.5755 V at 25^(@)C the SOP of calomel electrode is -0.28V then pH of the solution will be

The cell Pt|H_(2)(g) (1bar) |H^(+)(aq),pH=x||Cl^(-)(1M)|Hg_(2)Cl_(2)|Hg|Pt has emf of 0.5755 V at 25^(@)C the SOP of calomel electrode is -0.28V then pH of the solution will be

If the SRP of nickel and chlorine electrodes are -0.25 V and + 1.36 V respectively . The EMF of the cell Ni//Ni^(2+) (0.01M)// // Cl^(-) (0.01M) // Cl_2 , Pt is - V

Calculate the EMF of cell Ni//Ni^(2+) (0.01M)"//"Cl^(-)0.1M//Cl_(2) , Pt E^(@)Ni^(2+)//Ni= -0.250V : E^(@)Cl_(2)//Cl^(-)= +1.360V

The e.m.f of the cell, Ni//Ni^(2+)(1M)////Cl^(-)(1M)//Cl_(2),Pt is E^(@)Ni^(2+)//Ni= -0.25V,E^(@)1//2Cl_(2)//Cl^(-)=1.36V