1.

The standard reduction potential of Ag^(+)|Ag electrode is 0.80 volt. Calculate the standard electrode potential of Cl^(-)|AgCl|Ag at 25^(@)C. Given solubility product, K_(sp)(AgCl)=1.8xx10^(-10).

Answer»

Solution :APPLY
`E_(cell)^(@)=0.0591xxlog [Ag^(+)][Cl^(-)]=0.591 log K_(sp) (AGCL)`
`E_(cell)^(@)=-0.576` volt
`E_(cell)^(@)`= OXID. pot. anode + RED. pot. CATHODE
Red. Pot. Cathode `[Cl^(-)|AgCl|Ag]=-0.576-(-0.80)`
`=-0.576+0.80=0.224` volt


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