`E^@` values of `N^(2+), Ni and Cl_(2), Cl^(-)` are respectively `-0.25 V and +1.37 V`. Calculate the EMF of the cell, `Ni , Ni^(2+) (0.01M)"//"Cl^(-)(0.1M), Cl_(2),Pt`.

Explanation

Potential of nickel electrode is given as,
`E = E^@ + 0.0295 log [Ni^(2+)] = -0.25 + 0.0295 log 10^(-2) = -0.25 + 0.0295(-2) = -0.305V`.
Potential of chlorine electode is given as,
`E = E^@ - 0.059 log [Cl^(-)] = +1.37 - 0.059 log 10^(-1) = 1.43 V`.
EMF of cell l= E (chlorine electrode) - E(nickel electrode) = `1.43 - (-0.305) = 1.735 V`.