The cell potential of the given cell is $0.34 \, V$. Write the cell reaction and calculate $E^o_{Cu^{2+}|Cu}$.
$Pt_{(s)} \mid H_{2(g)} (1 \, bar) \mid H^+_{(1 \, M)} \parallel Cu^{2+}_{(1 \, M)} \mid Cu_{(s)}$

(0.34 V) The cell reaction is as follows:
Anode (Oxidation): $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^-$
Cathode (Reduction): $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
Overall reaction: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2H^+_{(aq)} + Cu_{(s)}$
The cell potential is given by: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$
Given $E^o_{cell} = 0.34 \, V$ and $E^o_{H^+|H_2} = 0.00 \, V$ (Standard Hydrogen Electrode).
$0.34 \, V = E^o_{Cu^{2+}|Cu} - 0.00 \, V$
Therefore,$E^o_{Cu^{2+}|Cu} = 0.34 \, V$.