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

(N/A) Reduction potential: At $298 \ K$,the $emf$ of a cell constructed by taking the standard hydrogen electrode as the anode (reference half-cell) and the other half-cell as the cathode gives the reduction potential of the other half-cell.
Standard hydrogen electrode half-cell $\mid$ another half-cell
(anode half-cell) $\mid$ (cathode half-cell)
Cell potential $=$ (Reduction potential of other cell) $=$ $emf$ value of the other half-cell.
$(b)$ Explanation with a general example:
$(i)$ If the concentrations of the oxidized and reduced forms of the species in the right-hand half-cell are unity,the cell potential equals the standard electrode potential.
General cell: $Pt \mid H_{2(g)} (1 \ bar) \mid H^+_{(aq)} (1 \ M) \| M^{n+}_{(aq)} (1 \ M) \mid M_{(s)}$
$(ii)$ $E^{\ominus} = (E^{\ominus}_R - E^{\ominus}_L)$:
Where $E^{\ominus}$ is the standard reduction potential of the cell,$E^{\ominus}_R$ is the standard reduction potential of the right-hand electrode,and $E^{\ominus}_L$ is the standard reduction potential of the left-hand electrode.
Since $E^{\ominus}_L = 0.0 \ V$ for the standard hydrogen electrode,$\therefore E^{\ominus} = (E^{\ominus}_R - 0.0) = E^{\ominus}_R$.
$(iii)$ Hydrogen half-cell $\mid$ Copper half-cell:
$Pt \mid H_{2(g)} (1 \ bar) \mid H^+_{(aq)} (1 \ M) \| Cu^{2+}_{(aq)} (1 \ M) \mid Cu_{(s)}$
The measured $emf$ of the cell is $+0.34 \ V$,which is the standard electrode potential for the reaction:
Oxidation (left): $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^- \ (E^{\ominus}_L = 0.0 \ V)$
Reduction (right): $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
$\therefore E^{\ominus}_{cell} = (E^{\ominus}_R - E^{\ominus}_L) = 0.34 \ V \implies E^{\ominus}_R = 0.34 \ V$.
Thus,the standard reduction potential for $Cu^{2+} \mid Cu$ is $+0.34 \ V$.