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Find the equilibrium constant for the re...

Find the equilibrium constant for the reaction `:`
`Cu^(2+)+In^(2+)hArrCu^(o+)+In^(3+)`
Given that `E^(c-).._(Cu^(2+)|Cu^(o+))=0.15V,E^(c-)._(ln^(2+)|ln^(o+))=-0.4V,`
`E^(c-)._(ln^(3+)|ln^(o+))=-0.42V`

Text Solution

Verified by Experts

The correct Answer is:
`K_(c)=10^(10)`

`-nFE^(c-)=-0.59F`
`-E^(c-)._(cell)F=-0.59F`
` E^(c-)._(cell)=0.59`
`E_(cell)=E^(c-)-(0.0591)/(n)logK_(c)`
`0.59=(0.0591)/(1)log K_(c)`
Hence, `logK_(c)=(0.59)/(0.059)=10^(10)`
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