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

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

Given `pK_(b)` of `NH_(4)OH=4.74`

A

`0.05V`

B

`-0.05V`

C

`-0.28V`

D

`0.28V`

Text Solution

Verified by Experts

The correct Answer is:
b
It is a concentration cell, therefore, `E^(c-)cell=0`.
For weak base `NH_(4)OH`,
`pOH=(1)/(2)(4.74-log10^(-3))=3.87`
`:. pH=14-3.87=10.13`
For strong base `NaOH`,
`Poh=3,Ph=14-3=11`
`:. E_(cell)=-0.059(pH_(c)-pH_(a))`
`=-0.059(11-10.13)`
`=-0.059xx0.87`
`=-0.05V`
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