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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=0.01M`

B

`Xlt0.01M`

C

`Xgt 0.01M`

D

X can not be predicted

Text Solution

Verified by Experts

The correct Answer is:
A
`E= (2.37-0.45) - (0.06)/(2) log ((X)/(0.01))`
1.92 - 1.92 = `(0.06)/(2) log (X)/(0.01)Rightarrow log 100X = 0 Rightarrow X= 0.01`
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