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At 25^(@)C temperature the cell potent...

At `25^(@)C` temperature the cell potential of a given electrochemical cell is 1.92 V
`Mg(s)|Mg^(2+)(Aq)x M||Fe^(2+)(aq)0.01 M |Fe(s)`
Given `E_(Mg//mg^(2+))(Aq)=2.37 v`
`E_(Fe//Fe^(2+))^(@)(Aq)=0.45 V`
Find the value of x

A

`x=0.01 M`

B

`x lt 0.01 M`

C

`x gt 0.01 M`

D

x cannot be predicted

Text Solution

Verified by Experts

The correct Answer is:
A
`E_("cell")^(@)=E_("cathode")-E_("anode")`
`=-0.45 -(-2.37)=1.92 V`
`Mg(s)|Mg^(2+)(aq)xxM||Fe^(2+)(Aq)0.01 Mfe(s)`
The cell reaction is
`Mg+Fe^(2+) rarr Mg^(2+) rarr Mg^(2+) +Fe`
`E_(cell) =E_(cell)^(@)-(0.059)/(2) log [(Mg^(2+))]/[(Fe^(2+))]`
`1.92 =1.92 -(0.59)/(2) log (x)/(0.01)`
`therefore x=0.01 M`
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