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Discuss the method to determine cell potential of any cell when standard hydrogen electrode is considered as anode with suitable example. |
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Answer» Solution :(a) Reduction potential : A 298K the emf of the cell, standard hydrogen electrode and second half-cell constructed by taking standard hydrogen electrode as anode (reference half-cell) and the other half-cell as cathode gives the reduction potential of the other half-cell. Standard: `UNDERSET("anode half-cell")("hydrogen electrode half-cell")||underset("cathode half-cell")("another half-cell")` Cell potential=(Reduction potential of other cell) =emf value of another half-cell. (b) Explanation with general example: (i) If the concenctrations of the oxidised and the reduced forms of the SPECIES in the right-hald half-cell are unity, then the cell potential is equal to the standard electrode potential. General: cell `Pt|H_(2(g))("1 bar")|H_((aq))^(+)(1M)|``|underset("metal ION")(Mn_((aq))^(+)(1M))|underset("metal")(M_((S)))` (ii) `E^(Theta)=E_(R)^(Theta)-E_(L)^(Theta)`: Where `E^(Theta)`=standard reduction potential of cell `E_(R)^(Theta)`=standard reduction potential of cell present on right side. `E_(L)^(Theta)=`standard reduction potential of cell present on left side. As `E_(L)^(Theta)`=for standard hydrogen electrode is zero `therefore E^(Theta)=(E_(R)^(Theta)-0.0)` `E^(Theta)=E_(R)^(Theta)`. . (i) (iii) Hydrogen half-cell || copper half-cell: `Pt|H_(2(g))("1 bar")|H_((aq))^(+)(1M)||Cu_((aq))^(2+)(1M)|Cu_((S))` The measured emf of the cell is +0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction: Oxidation left hand side : `H_(2(g)) to 2H_((aq))^(+)+2e^(-)""therefore E_(L)^(Theta)=0.0V` Reduction right hand side: `Cu_((aq))^(2+)+2e^(-)toCu_((S))` `therefore E_(cell)^(Theta)=(E_(R)^(Theta)-E_(L)^(Theta))=0.34` `therefore E_(R)^(Theta)=E_(cell)^(Theta)=0.34V` `therefore E_(cell)^(Theta)=E_(R)^(Theta)` `Cu_((aq))^(2+)(1M)+2e^(-)toCu_((S))` and its standard reduction potential `Cu^(2+)|Cu=+0.34V` So, as `Pt_((S))|H_(2("g, 1 bar"))|H_((aq,1M))^(+)||Zn_((aq,1M))^(2+)|Zn=-0.76V` Note : When standard hydrogen electrode is on left side and working as anode then other half-cell has POSITIVE reduction potential value and such cell possess positive emf value. this shows that `Cu^(2+)` ion does not get reduced easilywith respect to `H^(+)` ions and hence reaction does not occur in BACKWARD direction. |
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