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The equilibrium constant for the cell Cu...

The equilibrium constant for the cell `Cu(s)+2Ag^+(aq)rarrCu^(2+)(aq)+2Ag(s)`,at 298K is [Given,`E_((Ag)^+/(Ag))^o=0.8V` and `E_((Cu)^(2+)/(Cu))^o=0.34V`]

Text Solution

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The cell may be represented as: `Cu|Cu^(2+)(aq)||Ag^(+)(aq)|Ag`
`E_(cell)^(@)=E_(RHS)^(@)-E_(LHS)^(@)=0.80V-(0.34V)=046V`
`E_(cell)^(@)=(0.0591)/(n)" log "K_(c)`
For the given reaction, `n=2,E_(cell)^(@)=0.46V`
`therefore0.46=(0.0591)/(2)logK_(c)" or "logK_(c)=(0.46xx2)/(0.0591)=15.5668" "thereforeK_(c)"Antilog "15.5668=3.6xx10^(15)`.
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