If  $E^{\circ }\left({\mathrm{Cu}}^{2+}\mid \mathrm{Cu}\right)=0.34\mathrm{V}$ and $E^{\circ }\left({\mathrm{Cu}}^{2+},{\mathrm{Cu}}^{+}\mid \mathrm{Pt}\right)=0.15\mathrm{V}$ then $E^{\circ }\left({\mathrm{Cu}}^{+}\mid \mathrm{Cu}\right)$ is about 

From the given data, we have

$\text{ (i) }{\mathrm{Cu}}^{2+}+2\mathrm{e}=\mathrm{Cu}$     $\mathrm{\Delta }{G}_1^{\circ }=-2F{E}^{\circ }{\mathrm{Cu}}^{2+}\mid \mathrm{Cu}$

$\text{ (ii) }{\mathrm{Cu}}^{2+}+\mathrm{e}\to {\mathrm{Cu}}^{+}$   $\mathrm{\Delta }{G}_2^{\circ }=-F{E}^{\circ }{\mathrm{Cu}}^{2+},{\mathrm{Cu}}^{+}\mid \mathrm{Pt}$

Subtracting Eq. (ii) from Eq. (i), we have

${\mathrm{Cu}}^{+}+\mathrm{e}=\mathrm{Cu}\mathrm{\Delta }{G}_3^{\circ }=-F{E}_{{\mathrm{Cu}}^{+}\mid \mathrm{Cu}}^{\circ }$

Hence $\mathrm{\Delta }{G}_3^{\circ }=\mathrm{\Delta }{G}_1^{\circ }-\mathrm{\Delta }{G}_2^{\circ }$

$-F{E}_{{\mathrm{Cu}}^{+}\mid \mathrm{Cu}}^{\circ }=-2F{E}_{{\mathrm{Cu}}^{2+}}^{\mathrm{Cu}}+F{E}_{{\mathrm{Cu}}^{2+}}^{\circ },{\mathrm{Cu}}^{+}\mid {\mathrm{Pt}}^{\circ }$

${{E^{\circ }}_{{\mathrm{Cu}}^{+}}}^{\mathrm{Cu}}=2\times 0.34\mathrm{V}-0.15\mathrm{V}=0.53\mathrm{V}$