Question

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`

Answer 1

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`