1.

When copper plate is kept in 0.1 M solution of CuSO_(4) at 298K temperature and if 70% dissociation is occurred then calculate the potential of copper electrode.

Answer»

Solution :REACTION : `Cu^(2+)+2E^(-) to Cu_((S))`
CONCENTRATION of `Cu^(2+)`=molarity`xx`percentage of dissociation
`therefore [Cu^(2+)]=0.1xx(70)/(100)=0.07M`
As per Nernst EQUATION, where, `E_(Cu^(2+)|Cu)^(Theta)=+0.34V`
`E_(cell)=E_(cell)^(Theta)-(0.059)/(n)"log"([Cu_((S))])/([Cu^(2+)])""n=2`
`=0.34-(0.059)/(2)"log"(1)/(0.07)`
`=0.34-(0.059)/(2)"log"(1)/(0.07)`
`0.34-(0.0295)log(14.286)`
`=0.34-(0.03407)=0.3059V`
If dissociation is less than 100% then `[Cu^(2+)]` and potential both decreases.


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