1.

Calculate the electrode potential of given electrode Pt, Cl_(2) (1.5 "bar") | 2Cl^(-) (0.01 M), Solu E_(Cl_(2)//2Cl^(-))^(@) = 1.36 V tion:

Answer»

Solution :The REACTION of electrode is
`{:(Cl_(2)(g) + 2E^(-), to, 2Cl^(-)),(1.5 "bar",,0.01 M):}`
`E = e^(@) -(0.0591)/n LOG ([Cl^(-)]^(2))/P_(Cl_(2))`
`=1.36 - (0.0591)/2 log (0.01)^(2)/1.5 = 1.483 V`


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