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