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Calculate the potential of the cell at 2...

Calculate the potential of the cell at 298 K :
`Cd//Cd^(2+)(0.1M)||H^(+)(0.2M)//Pt, H_(2)(0.5atm)`
`"Given "E^(@)" for "Cd^(2+)//Cd=-0.403V, R=8.314J^(-1)" mol"^(-1), F="96500 C mol"^(-1)`.

Text Solution

Verified by Experts

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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