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Explain equilibrium state in Daniell cell and derive its equilibrium constant. |
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Answer» Solution :* Daniell cell reaction: `Zn_((S))+Cu_((aq))^(2+) to Zn_((aq))^(2+)+Cu_((S))` In anode half cell the oxidation of Zn into `Zn^(2+)` take place and so the concentration of `Zn^(2+)` keeps on INCREASING while the concentration of `Cu^(2+)` keeps on decreasing. At the same time voltage of the cell as read on the VOLTMETER keeps on decreasing. After some time, we shall note that there isno changein the concentration of `Cu^(2+) and Zn^(2+)` ions and at the same time, voltmeter gives zero reading. this indicates that equilibrium has been attained. `Zn_((S)) +Cu_((aq))^(2+) hArr Zn_((aq))^(2+)+Cu_((S))` * Equilibrium constant at equilibrium `=K_(C)` Concentration of `Zn^(2+)` at equilibrium=`[Zn^(2+)]` Concentration of `Cu^(2+)` at equilibrium=`[Cu^(2+)]` Then, `K_(C)=([Zn_((aq))^(2+)])/([Cu_((aq))^(2+)])` * So in this CONDITION `E_(cell)=0.0V and n=2`, then nernst equation for Daniell cell at equilibrium will be as follows: `0=E_(cell)=E_(cell)^(THETA)-(2.303RT)/(2F)"log"([Zn^(2+)])/([Cu^(2+)])` `THEREFORE E_(cell)^(Theta)=(2.303RT)/(2F)logK_(C)""therefore E_(cell)^(Theta)=(0.059)/(2)log" "K_(C)|"Where, "R=8.314JK^(-1)mol^(-1)""F=96487" C "mol^(-1)``T=298K` but `E_(cell)^(Theta)=1.1V` is in Daniell cell `therefore log" "K_(C)=(1.1xx2)/(0.059)=37.288` `therefore K_(C)="Antilog "37.288=2xx10^(37)` |
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