1.

Following cell has EMF 0.7995 V . Pt| H_(2) (1) atm) |HNO_(3) (1M)|| AgNO_(3) (1 M)| Ag If we add enough KCl to Ag cell so that the final Cl^(-) is 1 M . Now the measured emf of the cell is 0.222 V . The K_(sp) of AgCl would be -

Answer»

`1 xx 10^(-9.8)`
`1 xx 10^(-19.6)`
`2 xx 10^(-10)`
`2.64 xx 10^(-14)`

Solution :`2 Ag^(+) + H_(2) to 2 H^(+) + 2 Ag`
`E = E^(@) - (0.0591)/(2) "log" ([H^+]^(2))/(P_(H_(2)) xx [Ag^(+)]^(2))`
`0.222 = 0.7995 - (0.0591)/(2) "log" (1)/([Ag^(+)]^(2))`
`[Ag^(+)] = 10^(-9.8)`
`K_(SP) = [Ag^(+)] [CL^(-)] = (10^(-9.8)) xx (1) = 10^(-9.8)`


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