Calculate `E_("cell ")^(0) ,Delta G^(@)` and equilibrium constant for the reaction `2Cu^(+) to Cu^(2+) +Cu`
`E_(Cu)^(0) | Cu = 0.52 V " and " E_(Cu) ^(0) ,Cu^(+) = 0.16 V .`

Answer

`E_("cell ")^(0) =0.36 V ; Delta G^(@) =- 34.74 kJ ;` Equilibrium constant = `K = 1.2 xx 10^(6) mol^(-1) dm^(-3)`

Explanation

Given : Cell reaction `: 2Cu_((aq))^(+) to Cu_((aq))^(2+) + Cu_((s))`
`E_(Cu^(+))^(0) | Cu = 0.52 V , E_(Ca^(2+)) , Cu^(+)= 0.16 V`
` 1 F = 96500 C`
`E_("cell")^(0) = ? Delta G^(@) = ? `
(i) The formulation of the cell :
`Pt| Cu_((aq))^(+) Cu_((aq))^(2+) "||" Cu_((aq))^(+) | Cu_((s))`
`((LHE .Cu_((aq))^(+) to Cu_((aq))^(+) + e^(-))/(RHE. Cu_((aq))^(+) to Cu_((s))))/(2Cu_((aq))^(+) to Cu_((aq))^(2+) + Cu_((s)))" "underset"(Overall cell reaction )"underset"(Reduction at cathode)"("(Oxidation at anode)")`
`:. n=1`
`E_("cell ")^(0) = E_(cu^(+) |Cu)- E_(Cu^(+2))^(0) .Cu^(+)`
`=0.52- 0.16 =0.36 V`
(ii) `Delta G^(@) =- nFE_("Cell")^(0) =- 1xx 96500 xx 0.36`
`=- 34740 J`
`=- 34.74 kJ`
(ii) If K is the equilibrium constant for the electrochemical redox reaction then
`K= ([Cu^(2+)])/([Cu^(+)]^(2)) `
`E_("cell ")^(0) = (0.0592)/(n) log_(10) k`
`:. log_(10) K = (nxx E_("cell ")^(0))/(0.0592)`
`=(1xx 0.36)/(0.0592)`
`=6.081 `
`:. K = A16 .081 = 1x2 xx 10^(6) ("mol" dm^(-3))^(-1) (or "mol"^(-1) dm^(3))`