1.

Electrochemical cell Mg_((S))|Mg_((aq))^(2+)(xM)||Fe^(2+)(0.01M)|Fe_((l)) has electric charge of 1.92 V then x= . . . .M E_(Mg||Mg^(2+))^(@)=2.37V,E_((Fe|Fe^(2+)))^(@)=0.45V

Answer»

`x LT 0.01 M`
`x=0.01M`
`x GT 0.01M`
No PREDICTION regarding x

Solution :`E_(CELL)^(@)=E_(Mg|Mg^(2+))^(@)-E_(Fe|Fe^(2+))^(@)`
`=2.37-0.45=1.92V`
And `E_(cell)^(@)=1.92V` is given
So, `E_(cell)^(@)=E_(cell)^(@)` and both half cell n=2,
so both half cell has equal concentration
`therefore x=0.01M=[Mg^(2+)]=[Fe^(2+)]`


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