1.

E^(@) values of Ni^(2+), Niand 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,

Answer»

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


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