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CalculateE_("cell")^(0),Delta G^(@) and equilibriumconstantfor thereaction2Cu^(+) to Cu^(2+)+Cu E_(Cu)^(0)| Cu = 0.52 V " and" E_(Cu) ^(0),Cu^(+) = 0.16V . |
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Answer» `E_(Cu^(+))^(0) | Cu =0.52 V , E_(Ca^(2+)) , Cu^(+)= 0.16 V` ` 1 F =96500 C` `E_("cell")^(0)= ?Delta G^(@) = ? ` (i) Theformulationof thecell : `Pt| Cu_((aq))^(+)Cu_((aq))^(2+) "||" Cu_((aq))^(+) | Cu_((s))` `((LHE .Cu_((aq))^(+) to Cu_((aq))^(+)+ e^(-))/(RHE. Cu_((aq))^(+)toCu_((s))))/(2Cu_((aq))^(+) to Cu_((aq))^(2+) + Cu_((s)))""underset"(Overall cell reaction )"underset"(Reduction atcathode)"("(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 Kis theequilibriumconstantfor theelectrochemicalredox 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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