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The emf of the cell Zn|underset((0.01 M)...

The `emf` of the cell `Zn|underset((0.01 M))(Zn^(2+))||underset((0.001 M))(Fe^(2+))|Fe`
at `298 K` is `0.2905`, then the value of equilibrium constant for the cell reaction is:

A

`e^(0.32//0.0295)`

B

`10^(0.32//0.0295)`

C

`10^(0.26//0.0295)`

D

`10^(0.32//0.0591)`

Text Solution

Verified by Experts

The correct Answer is:
B
`Zn + Fe^(2+) rarr Fe + Zn^(2+)`
`E_(cell) = E_(cell)^(@) + (0.059)/(2)log.([Fe^(2+)])/([Zn^(2+)])`
`0.2905 = E_(cell)^(@) + (0.059)/(2)log.((0.001))/(0.01)`
`:. E_(cell)^(@) = 0.2905 + 0.0295 = 0.32 V`
Now `E_(cell)^(@) = (0.059)/(2)log_(10)K_(c)`
`0.32 = (0.059)/(2)log_(10)K_(c)`
`:. K_(c) = 10^(.32//0.0295)`
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