Calculate the potential of a `Zn -An^(2+)` electrode in which the molarity of `Zn ^(2+)` is `0.001M.` Given that `E_(Zn^(2+)//Zn)^(0)=-0.76V` `R=0.314JK^(-1) mol ^(-1), F= 96500C mol ^(-1).`

Calculate the potential of a `Zn -An^(2+)` electrode in which the mola

1.	Calculate the potential of a `Zn -An^(2+)` electrode in which the molarity of `Zn ^(2+)` is `0.001M.` Given that `E_(Zn^(2+)//Zn)^(0)=-0.76V` `R=0.314JK^(-1) mol ^(-1), F= 96500C mol ^(-1).`
Answer» Given electrode `Zn\|Zn^(+2)(0.001m)E_(Zn^(+2)//Zn)^(0) =-0.76V` Nernst equation is `E=E^(0) +(2.303RT)/(nF) log [M^(n+)]` Given `R=8.314J//K.`mole `F = 96500c//`mole `E= E^(0) +(0.059)/(2)log C` `=-0.76+(0.059)/(2)log 0.001` `=-0.76 -(0.059)/(2)xx3` `= -0.76-0.0295xx3` `=-0.76-0.0885 =-0.8485V`

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Calculate the potential of a `Zn -An^(2+)` electrode in which the molarity of `Zn ^(2+)` is `0.001M.` Given that `E_(Zn^(2+)//Zn)^(0)=-0.76V` `R=0.314JK^(-1) mol ^(-1), F= 96500C mol ^(-1).`

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