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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` |
Answer» Correct Answer - 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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