Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array} $$

First, determine which half-reaction is taking place at the cathode (reduction) and which is taking place at the anode (oxidation). The substance with the higher standard reduction potential value will act as the cathode, as it has a stronger tendency to gain electrons and undergo reduction. In this case, gold (Au) has a higher standard reduction potential (\(\mathscr{E}^\circ = 1.50 V\)) than thallium (Tl), which has a standard reduction potential of \(\mathscr{E}^\circ = -0.34 V\). Thus, the cathode half-reaction is: $$ \mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}; \quad \mathscr{E}^{\circ} = 1.50 \mathrm{~V} $$ And the anode half-reaction is: $$ \mathrm{Tl}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{Tl}; \quad \mathscr{E}^{\circ} = -0.34 \mathrm{~V} $$

Now that we have identified the cathode and the anode, we can use their respective standard reduction potentials to calculate the overall cell potential: $$ E_{cell}^° = E_{cathode}^° - E_{anode}^° = E_{Au}^° - E_{Tl}^° $$ Plug in the given values for the standard reduction potentials and calculate the cell potential: $$ E_{cell}^° = 1.50\,\text{V} - (-0.34\,\text{V}) = 1.50\,\text{V} + 0.34\,\text{V} = 1.84\,\text{V} $$ Therefore, the overall cell potential for this galvanic cell is \(1.84\,\text{V}\).