1.

Ag// AgBr(s) , underset(0.1M)(KBr)// underset(0.1M)(KCl), underset((3))(AgCl)//Ag the cell potential is 0.1x volts. What is x ? (Ksp AgBr = 10^(-16), AgCl= 10^(-11))

Answer»


Solution :`E = -(0.0591)/(1) log""(K_(sp) AgBr//0.1)/(K_(sp AgCl)//0.1) = -0.0591 XX log""((10^(-16))/(10^(-11)))`
`=- 0.0591 xx log(10^(-1)) = -0.0591 xx (-5) = + 0.2955 V = + 0.3V = 0.1 x ,x=3 `


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