1.

Calculate the emf of the following cell at 25^(@) C Zn| Zn^(2+) (0.001 M) || H^(+) (0.01 M) | H_(2) (g) (1 bar) | Pt (s) E_(Zn^(2+)//Zn)^(@) = -0.76 V, E_(H^(+)//H_(2))^(@) = 0.00 V

Answer»

Solution :The cell reaction is:
`Zn(s) + 2H^(+) (aq) to Zn^(2+) (aq) + H_(2)`, Then n=2
`E = E^(@) - 0.059/2 LOG ([Zn^(2+)])/([H^(+)]^(2)`
`=0.76 - 0.0295 log ([0.001])/([0.01]^(2)) = 0.76 - 0.0295 log 10`
`=(0.76 - 0.0295) "VOLT" = 0.7305` volt


Discussion

No Comment Found

Related InterviewSolutions