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Ag// AgBr(s) , underset(0.1M)(KBr)// und...

`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))`

Text Solution

Verified by Experts

The correct Answer is:
C
`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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