1.

Derive expression to calculate emf (reduction) of the following half cells at 25^@C Cl_2 (g)|2Cl^(-) (Pt) (1 atm)

Answer»

SOLUTION :`1/2Cl_2(G)+e leftrightarrow CL^-` (reduction)
(1 atm)
`E_(Cl_2,Cl^-)=E_(cl_2,Cl^-)^@ -0.0591/1 LOG""([Cl^-])/([Cl]^(1/2))`
`=E_(Cl_2,Cl^-)^@-0.0591 log [Cl^-]`


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