1.

Cu^(2+)+2e^(-)toCu,E^(@)=+0.34V,Ag^(+)+1e^(-)toAg,E^(@)=+0.80V (i) Construct a galvanic cell using the above data. (ii) For what concentration of Ag^(+) ions will the emf of the cell be zero at 25^(@)C, if the concentration of Cu^(2+) is 0.01 M? (log 3.919=0.593).

Answer»


Solution :(i) For `E_(cell)^(@)` to be +ve, oxidation will occur at copper ELECTRODE and reduction at SILVER electrode. Hence, the cell will be represented as : `Cu|Cu^(2+)||Ag^(+)|Ag`
(ii) `Cu+2Ag^(+)toCu^(2+)+2Ag,n=2`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))`. hence, `0.46V-(0.0591)/(2)"log"(0.1)/([Ag^(+)]^(2))`. CALCULATE `[Ag^(+)]`


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