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298K पर सेल Zn|Zn^(2+) (0.01M)||Fe^(2+...

298K पर सेल
`Zn|Zn^(2+) (0.01M)||Fe^(2+) (0.001M)` Fe का विधुत वाहक बल (emf ) ०.2905 वाल्ट है तब सेल अभिक्रिया के लिए साम्य स्थिरांक का मान है

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+) to Zn^(2+) +Fe(n=2)`
`E=E^@-0.0591/2 log Q`
`0.0295 =E^@-0.0591/2 log""0.1/0.001`
`E^@= 0.2905+0.0295=0.32` वोल्ट
`E^@=0.0591/2 log K_(समय )`
`0.32=0.0591/2 log K_(साम्य )`
`=0.02945 log K_(साम्य )`
`K_(साम्य )`= `10^(0.32//0.0295)`
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The emf of the cell, Zn|Zn^(2+) (0.01 M)||Fe^(2+) (0.001 M)|Fe at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is :