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Calculate the potential (emf) of the cell Cd|Cd^(2+)(0.10M)||H^(+)(0.20M)|Pt,H_(2)(0.5atm) (given E^(@) for Cd^(2+)//Cd=-0.403V,R=8.314JK^(-1)mol^(-1),F=96,500" C "mol^(-1)) |
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Answer» Solution :The cell REACTION is: `Cd+2H^(+)(0.20M)toCd^(2+)(0.10M)+H_(2)(0.5atm)` `E_(Cell)^(@)=E_(H^(+),1//2H_(2))^(@)-E_(Cd^(2+),Cd)^(@)=0-(-0.403)=0.403V` ltBrgt Applying NERNST EQUATION to the cell reaction, ltbr `E_(cell)=E_(cell)^(@)-(2.303nRT)/(nF)"LOG"([Cd^(2+)]xxP_(H_(2))^(**))/([H^(+)]^(2))=0.403-(2.303xx8.314xx298)/(2xx96500)"log"(0.1xx0.5)/((0.2)^(2))` `=0.403-0.003=0.400V` |
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