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The emf of the cell, Zn|Zn^(2+)(0.01M)...

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

A

`(0.32)/(e^(0.0295))`

B

`(0.32)/(10^(0.0295))`

C

`(0.26)/(10^(0.0295))`

D

`(0.32)/(10^(0.0591))`

Text Solution

Verified by Experts

The correct Answer is:
B
Cell reaction is:
`Zn+Fe^(2+)rarrZn^(2+)+Fe` ltbr gt Here n=2
`E=E_(0)-(0.0591)/(2)"log"([Zn^(2+)])/(pFe^(2+)]`
`0.2957=E_(0)-0.02955"log"(0.01)/(0.001)`
`E_(0)=0.2957+0.02955=0.32525V`
`E_("cell")^(@)=(0.0591)/(2)V"log" k_(c)`
`therefore "log" k_(c)=(0.32525xx2)/(0.0591)=(0.6506)/(0.05591)`
`"log"_(10) k_(c)=(0.325)/(0.0295)`
`therefore k_(c)=(0.325)/(0.0295)`
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