Calculate equilibrium constant for tha disproportionation reaction :
`2Cu^(2+)(aq)rarrCu(s)+Cu^(2+)(aq)" at " 25^(@)C`
`("Given " E_((Cu^(+)//Cu))^(@)=0.52 V, E_((Cu^(2+)//Cu^(+)))^(@)=0.16 V)`.

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1 Answers

`2Cu^(+)(aq)rarrCu(s)+Cu^(2+)(aq)`
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cu^(2+)(aq)])/([Cu^(+)(aq)]^(2))=E_(cell)^(@)(0.0591)/(1)"log"([Cu^(2+)(aq)])/([Cu^(+)(aq)]^(2))`
`([Cu^(2+)(aq)])/([Cu^(+)(aq)]^(2))=K_(c)" and "E_(cell)=0,E_(cell)^(@)=E_(Cu^(++)//Cu)^(@)-E_(C^(2+)//Cu)^(@)=0.52-0.16=0.36 V`.
`:. " "log K_(c)=(E_(cell)^(@))/((0.0591 V))=(0.36 V)/((0.0591 V))=6.09`
`K_(c)="Antilog "6.09=1.2xx10^(6)`

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