Discuss the method to determine cell potential of any cell when standard hydrogen electrode is considered as anode with suitable example.

Explanation

(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.