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The emf of the cell, Zn|Zn^(2+) ( 0.05...

The emf of the cell,
`Zn|Zn^(2+) ( 0.05 M) ||Fe^(2+) ( 0.002 M ) Fe at 298 K ` is 0.2957 V then th value of equilibrium constant for the cell reaction is

A

`e^((0.34 )/( 0.0295 ))`

B

`10^((0.34 )/( 0.0295 ))`

C

`10^((0.25)/( 0.0295))`

D

`10^((0.25 )/( 0.0591))`

Text Solution

Verified by Experts

The correct Answer is:
B
`F_("cell") = E_("cell")^@ - ( 0.0591 )/(n) log ([Zn^(2+)])/( [Fe^(2+)])`
`0.2957 = E_("cell")^@ - ( 0.0591 )/(2) log ( 0.05)/(0.002)`
` E_("cell")^@ = 0.2957 + 0.0413 = 0.3370 V `
Now we know `E_("cell")^2 = ( 0.0591 )/(n) log K`
` log k=( 0.3370 xx 2 )/( 0.0591 )= ( 0.34 )/( 0.0295 )`
`k=10^((0.34 )/( 0.0295 ))`
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