Saved Bookmarks
| 1. |
Expalin SHE as a reference electrode. |
|
Answer» Solution :STANDARD Hydrogen Electrode (SHE) is used as the reference electrode. It has been assigned an ARBITRARILY emf of zero volt. It consists of a platinum electrode in contact with 1M HCl solution and 1 atm hydrogen gas. The hydrogen gas is bubbled through the solution at `25^(@)C`. SHE can act as a cathode as well as an anode. The Half cell reactions are given below. If SHE is used as a cathode, the reduction reactions is `2H^(+)(AQ,1M)+2e^(-) rarr H_(2)(g,1 atm)""E^(@)=0" volt"` If SHE is used as an anode, the oxidation reaction is `H_(2)(g,1atm) rarr 2H^(+)(aq, 1M)+2e^(-)""E^(@)="volt"` Illustration : Let us calculate the reduction potential of zinc electrode dipped in zinc sulphate solution using SHE. Step 1 : The following galvanic cell is constructed using SHE `Zn_((s))abs(Zn^(2+)(aq,1M))abs(H^(+)(aq,1M)|H_(2)(g,1atm))pt_((s))` Step 2 : The emf of the above galvanic cell is measured using a volt meter. In this case, the measured emf of the above galvanic cell is 0.76V. Calculation We know that, `E_(cell)^(@)=(E_("ox")^(@))_(Zn|Zn^(2+))+(E_(RED)^(@))_(SHE)` `E_(Cell)^(@)=0.76 " and " (E_(red)^(@))_(SHE)=0V`. Substitute these values in the above equation `rArr 0.76V=(E_("ox")^(@))_(Zn|Zn^(2+))+0V` `rArr (E_("ox")^(@))_(Zn|Zn^(2+))=0.76V` This oxidation potential corresponds to the below mentioned half cell reaction which takes PLACE at the cathode. `Zn rarr Zn^(2+)+2e^(-)` (Oxidation) The emf for the reverse reaction will give the reduction potential `Zn^(2+)+2e^(-) rarr Zn, E^(@)=-0.76V` `""therefore (E_(red)^(@))_(Zn^(2+)|Zn)=-0.76V` 
|
|