1.

Calcualte emf of the following at 25^(@) C Fe | Fe^(2+) (0.001) || H^(+) (0.01 M) | H_(2) (g) (1 bar) | Pt(s) E^(@) (Fe^(2+)[Fe])^(@) = -0.44 V, E(H^(+)|H_(2))^(@) = 0.00 V

Answer»

Solution :The CELL reaction is Fe (s) `+2H^(+)(AQ) to Fe^(2+)(aq) + H_(2)`
`E_("cell")^(@) = 0.00 -(0.44) = 0.44` volt
`E_("cell") = E_("cell")^(@) - 0.059/2 LOG ([Fe^(2+)])/([H^(+)]^(2)`
`E_("cell") = 0.44 - 0.059/2 log ([0.001])/([0.01])^(2)`
`=0.44 - 0.0295 XX 1 = 0.44 - 0.0295 = 0.4105` volt


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