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निम्नलिखित सेल का विद्युत वाहक बल 0.6 वो...

निम्नलिखित सेल का विद्युत वाहक बल 0.6 वोल्ट है। Ag (s) |AgBr (s) KBr (0.01M)| `|AgNO_(3) (0.001 M)|`Ag(s)AgBr के `K_(sp)` को `1xx10^(-x)`के रूप में व्यक्त किया जाता है, x है: [`(2.303RT)/F`= 0.06 V लीजिए

Text Solution

Verified by Experts

The correct Answer is:
A
`"Ag"toAg^+(L)+e^-`
`Ag^+(0.001M)+e^-"toAg`
`Ag^+(0.001M)toAg^+(L)`
`0.6=0-0.06/1log(K_(sp)/(10^-2xx10^-3))`
`10^-10=K_(sp)/10^-5impliesK_(sp)=10^-15`
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