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

(N/A) At $298 \ K$,the $EMF$ of a cell constructed by taking the standard hydrogen electrode $(SHE)$ as the cathode (reference half-cell) and the other half-cell as the anode,gives the reduction potential of the other half-cell.
General cell representation:
$M_{(s)} | M_{(aq)}^{n+} (1 \ M) || H_{(aq)}^{+} (1 \ M) | \frac{1}{2} H_{2_{(g)}} (1 \ bar) | Pt_{(s)}$
Using the formula $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$,since $E^{\circ}_{cathode} = E^{\circ}_{H^{+}/H_2} = 0.0 \ V$,we get $E^{\circ}_{cell} = -E^{\circ}_{anode}$.
Example: Zinc half-cell as anode and $SHE$ as cathode:
$Zn_{(s)} | Zn_{(aq)}^{2+} (1 \ M) || H_{(aq)}^{+} (1 \ M) | H_{2_{(g)}} (1 \ bar) | Pt_{(s)}$
The measured $E^{\circ}_{cell}$ is $-0.76 \ V$. Therefore,the standard reduction potential of $Zn^{2+}/Zn$ is $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$.
When the $SHE$ is on the right side (cathode),the other half-cell acts as the anode,and the cell potential is negative if the metal is more reactive than hydrogen.
Cathode reaction: $H^{+}_{(aq, 1 \ M)} + e^{-} \rightarrow \frac{1}{2} H_{2_{(g)}} (1 \ bar)$
Overall reaction: $Zn_{(s)} + 2H^{+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + H_{2_{(g)}}$