1.

E_(cell)=0.78 volt for the following cell. underset(Fe_((s))|Fe_((aq))^(2+))||underset((0.01M)(Cu_((aq))^(2+)|Cu_((s))) E_(Fe//Fe^(2+)(aq))^(@)=0.44V,E_(Cu//Cu^(2+)(aq))^(@)=-0.34V

Answer»

x cannot be predicted
x=0.01M
`xgt0.01M`
`xlt0.01M`

SOLUTION :If a solution contains two or more catios, the cation to be deposited on the cathode is the one which has HIGHER reduction potential. Here the REACTION will be
`Fe+Cu^(2+)TOFE^(2+)+Cu`
`E_(cell)=E_(cell)^(o)-(0.0591)/(n)"log"([Fe^(2+)])/([Cu^(2+)])`
`0.78=0.78-(0.591)/(n)"log"[(Fe^(2+))/(Cu^(2+))]`
or `0="log"(x)/(0.01)implies(x)/(0.01)=1impliesx=0.01`.


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