`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.` The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction isA. `10^(0.32//0.0591)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `e^(0.32//0.295)`

`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.` The `EMF` of the above cell

1.	`Zn\|Zn^(2+)(a=0.1M)\|\|Fe^(2+)(a=0.01M)Fe.` The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction isA. `10^(0.32//0.0591)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `e^(0.32//0.295)`
Answer» Correct Answer - B `Zn\|Zn^(+2) (0.01M)\|\|Fe^(+2) (0.001M)\|Fe` `E_(cell)^(@) = E_(cell)^(@) - (RT)/(nF) In K` `E^(@) = E_(Cell)^(@) - (RT)/(2F) In K` `E^(@) = 0.2905 - (0.0591)/(2) log.(10^(-3))/(10^(-2))` `=0.2905 - (0.0591)/(2) log 10^(-1)` `= 0.2905 +(0.0591)/(2) = 0.2905 +0.02950` `= 0.32V` `E^(@) = (RT)/(2F) In K` `E^(@) = (0.0591)/(2) log K` `log K = (0.32)/(0.0295) rArr K = 10^(0.32//0.0295)`

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`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.` The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction isA. `10^(0.32//0.0591)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `e^(0.32//0.295)`

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