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