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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. |
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Answer» Solution :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`. |
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