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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

`-0.42` V

B

`0.42 V `

C

`-0.19 V `

D

`0.19 V `

Text Solution

Verified by Experts

The correct Answer is:
A
It is a concentration cell , therefore `E_("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
`therefore E = -0.059 (pH_(c)- pH_(a))`
`= -0.059 (11 - 3.87) = -0.42 V `
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