Given $$E^{\circ}_{Cu^{2+}/Cu} = 0.34$$ V, $$E^{\circ}_{Cu^{2+}/Cu^{+}} = 0.15$$ V. Standard electrode potential for the half cell $$Cu^{+}/Cu$$ is

Standard electrode potentials are intensive properties and cannot be added or subtracted directly. Instead, we must convert them into their corresponding Gibbs Free Energy changes ($$\Delta G^\circ$$), which are extensive properties and can be algebraically combined:

$$\Delta G^\circ = -nFE^\circ$$

Where:

• $$n$$ is the number of moles of electrons transferred in the half-reaction.
• $$F$$ is the Faraday constant ($$96485 \text{ C mol}^{-1}$$).
• $$E^\circ$$ is the standard reduction potential.

• Reaction 1: Reduction of $$\text{Cu}^{2+}$$ to $$\text{Cu}$$

  $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad \left(n_1 = 2, \ E^\circ_1 = 0.34\text{ V}\right)$$ $$\Delta G^\circ_1 = -2 \cdot F \cdot (0.34) = -0.68F$$



• Reaction 2: Reduction of $$\text{Cu}^{2+}$$ to $$\text{Cu}^+$$

  $$\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \quad \left(n_2 = 1, \ E^\circ_2 = 0.15\text{ V}\right)$$ $$\Delta G^\circ_2 = -1 \cdot F \cdot (0.15) = -0.15F$$



• Reaction 3: Target Reduction of $$\text{Cu}^+$$ to $$\text{Cu}$$

  $$\text{Cu}^+ + e^- \rightarrow \text{Cu} \quad \left(n_3 = 1, \ E^\circ_3 = ?\right)$$ $$\Delta G^\circ_3 = -1 \cdot F \cdot E^\circ_3$$

By subtracting Reaction 2 from Reaction 1, we obtain our target Reaction 3:

$$\text{Reaction 3} = \text{Reaction 1} - \text{Reaction 2}$$

Therefore, their respective energy changes follow the same relationship:

$$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$$

Substitute the computed free energy values into the equation:

$$-1 \cdot F \cdot E^\circ_3 = -0.68F - (-0.15F)$$

$$-E^\circ_3 = -0.68 + 0.15$$

$$-E^\circ_3 = -0.53\text{ V}$$

$$E^\circ_3 = 0.53\text{ V}$$

Conclusion:

The standard electrode potential for the half-cell $$\text{Cu}^+/\text{Cu}$$ is calculated to be $$0.53\text{ V}$$.

Correct Option: B — 0.53 V