1.

If Fe|Fe^(2+)(xM)||Cu^(2+)(0.01M)|Cu has cell potential of 0.78V then x= . . . E_(Fe^(2+)//Fe)^(@)=-0.44V,E_(Cu^(2+)//Cu)^(@)=+0.34V).

Answer»

`xgt0.01M`
`X=0.01M`
`x lt 0.01M`
No prediction regarding x

Solution :(i) `E_(cell)^(@)=E_("reduction")^(@)("Cathode")-E_("reduction")^(@)("Anode)`
`=E_(Cu^(2+)//Cu)^(@)-E_(FE^(2+)/?Fe)^(@)`
`=+0.34-(-0.44)`
=0.78V
So `E_(cell)=E_(cell)^(@)` Hence x=0.01 M
(ii) Cell reaction :
`Fe_((S))+Cu^(2+)(0.01M) t Fe^(2+)(xM)+Cu_((S))`
So n=2
`E_(cell)=E_(cell)^(@)-(0.0592)/(n)"log"([Fe^(2+)])/([Cu^(2+)])`
`therefore 0.78=0.78-(0.0592)/(2)"log"(x)/(0.01)`
`therefore logx-log(0.01)=0`
`therefore logx+2=0`
`therefore logx=-2`
`therefore x=10^(-2)=0.01M`


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