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Assuming that hydrogen begaves as an ideal gas, calculate EMF of the cell at `25^(@)C` when `P(1)=640mm and P_(2)=425mm` `Pt|H_(2)(P_(1))|HCl|H_(2)(P_(2))|Pt`A. `0.005V`B. `0.004V`C. `0.003V`D. `0.002V` |
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Answer» Correct Answer - A As the cell is a concentration cell, `E_(cell)^(@)-0` Now, the cell reaction is Left electrode: `H_(2)(P_(1)=640mm)hArr2H^(+)+2e^(-)` Right electrode: `:2H^(+)+2e^(-)hArrH_(2)(P_(2)=425mm)` Net cell reaction: `H_(2)(P_(1)mm)hArrH_(2)(P_(2)mm)` Now, from Nernst equation `=0-(0.059)/(2)log.(425)/(640)=0.005V` |
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