1.

Calculate the emf of the cell, Cd|Cd^(2+)(0.001M)||Fe^(2+)(0.6M)|Fe at 25^(@)C. The standard reduction potential of Cd//Cd^(2+) and Fe//Fe^(2+) electrodes are -0.403 and -0.441 volt respectively.

Answer»


Solution :For the cell as REPRESENTED, `E_(cell)^(@)=E_(Fe^(2+)//Fe)^(@)-E_(Cd^(2+)//Cd)^(@)=-0.441-(0.403)=-0.038V`
Cell reaction: `Cd+Fe^(2+)(0.6M)toCd^(2+)(0.001M)+Fe`
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cd^(2+)])/([Fe^(2+)])=-0.038-(0.0591)/(2)"log"(10^(-3))/(0.6)=-0.038+0.0821=0.0441V`
ALTERNATIVELY, CALCULATE electrode potentials of L.H.S. & R.H.S. ELECTRODES separately. then
`E_(cell)=E_(RHS)-E_(LHS)`.


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