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Given the following cell at 25^(@)C ...

Given the following cell at `25^(@)C`

What will be the potential of the cell ?
Given `pK_(a) ` of `CH_(3)COOH=4.74`

A

`-0.42V`

B

`0.42V`

C

`-0.19V`

D

`0.19V`

Text Solution

Verified by Experts

The correct Answer is:
a
It is a concentration cell, therefore, `E^(c-)._(cell)=0`
`pH of W_(A)=(1)/(2)(pK_(a)-log c)`
`=(1)/(2)(4.74-log 10^(-3))=3.87`
`pH of NaOH=14-3=11`
`:. E=-0.059(pH_(c)-pH_(a))`
`=-0.059(11-3.87)=-0.42V`
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