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`

Options

A. `x lt 0.01 M`
B. `x=0.01M`
C. `x gt 0.01M`
D. No prediction regarding x

Answer

B

Explanation

`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+)]`