Home
Class 12
CHEMISTRY
For a cell involving two electrons chang...

For a cell involving two electrons changes, `E_(cell)^(@)=0.3" V"` at `25^(@)C`. The equilibrium constant for the reaction is :

A

`10^(10)`

B

`3xx10^(-2)`

C

10

D

`10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
D

`logK_(C )=(nE_(cell)^(@))/(0.0591)=((2xx0.3" V"))/((0.0591" V"))~~=10`.
`K_(C )="Antilog "10=10^(10)`
Promotional Banner

Topper's Solved these Questions

  • ALCOHOLS AND PHENOLS

    DINESH PUBLICATION|Exercise COMPLETION QUESTION|2 Videos
  • ALCOHOLS AND PHENOLS

    DINESH PUBLICATION|Exercise NCERT In-Text Questions|9 Videos
  • ALDEHYDES AND KETONES

    DINESH PUBLICATION|Exercise Interger|5 Videos

Similar Questions

Explore conceptually related problems

For a cell involving two electron change , E_(cell)^(@) = 0.3 V at 25^(@) C . The equilibrium constant of the reaction is

The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25^(@) C . The equilibrium constant of the reaction is : (F=96,500C mol^(-1) : R=8.314 Jk^(-1)mol^(-1)

For a cell involving one electron E_(cell)^(0)=0.59V and 298K, the equilibrium constant for the cell reaction is: [Given that (2.303RT)/(F)=0.059V at T=298K ]