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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

`e^(0.32//0.0295)`

B

`10^(0.320//0.0295)`

C

`10^(0.26//0.0295)`

D

`10^(0.32//0.0591)`

Text Solution

Verified by Experts

(A)Eq. of Ni`(NO_3)_2` = `5xx0.4xx2` =4,
Eq. of `Ni^(2+)` lost = `("i"."t")/96500` = `(2.4125xx10xx60xx60)/96500` =0.9
`:.` Eq. of Nii`(NO_3)_2` left = 4-0.9=3.1,
`:.` Molarity of i`(NO_3)_2` = `3.1/(5xx2)` =0.31
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