The potential of the cell at `25^(@)C ` is

Given `pK_(a)` of `CH_(3)COOH` and `pK_(b)` of `NH_(4)OH=4.74`

Options

A. `-0.04V`
B. `0.04V`
C. `-0.189V`
D. `0.189V`

Answer

c

Explanation

It is concentration cell, therefore, `E^(c-)._(cell)=0.`
Acidic buffer at anode `:`
`:. pH=pK_(a)+log[(Sal t)/(Aci d)]`
`=4.74+log .(0.1)/(0.01)=4.74+1=5.74`
Basic buffer at cathode `:`
`:. pOH=pK_(b)+log[(Sal t)/(Base)]`
`=4.74+log.(0.2)/(0.1)`
`=4.74+0.03=5.04`
`:. pH=14-5.04=8.96`
`E_(cell)=-0.059(pH_(c)-pH_(a))`
`=-0.050(8.96-5.74)`
`=-0.059xx3.22=-0.189V`