Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s)` Given that `E_(Ag^(+)//Ag)^(@)=0.80V and E_(Cu^(2+)//Cu)^(@)=0.34V`

Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)

1.	Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s)` Given that `E_(Ag^(+)//Ag)^(@)=0.80V and E_(Cu^(2+)//Cu)^(@)=0.34V`
Answer» 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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Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s)` Given that `E_(Ag^(+)//Ag)^(@)=0.80V and E_(Cu^(2+)//Cu)^(@)=0.34V`

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