Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=&+0.49 \mathrm{~V} \end{aligned} $$ (a) Write the equation for the combination of these halfcell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

For the largest positive emf, we want to combine the strongest reducing agent with the strongest oxidizing agent. From the given half-reactions, it is clear that \(\mathrm{AuBr}_{4}^-(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)\) has the lowest value of standard reduction potential, which indicates it is the strongest reducing agent. The half-reaction with the highest value of the standard reduction potential is \(\mathrm{IO}^-(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)\), making it the strongest oxidizing agent. Now, let's balance the electrons in the two half-reactions and then combine them to form the overall reaction.

Since we have 3 electrons in the reduction half-reaction and 2 electrons in the oxidation half-reaction, the least common multiple is 6. Therefore, we balance the two half-reactions by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 3: $$ \begin{aligned} 2[\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)] \\ 3[ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)] \\ \end{aligned} $$ Now, combining the balanced half-reactions: $$ \begin{aligned} 2\mathrm{AuBr}_{4}^{-}(a q) + 3\mathrm{IO}^{-}(a q)+3\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{Au}(s) + 8 \mathrm{Br}^{-}(a q) + 3\mathrm{I}^{-}(a q)+6 \mathrm{OH}^{-}(a q) \end{aligned} $$

To calculate the emf for this combination of half-cell reactions, we first sum the standard potentials of the oxidation and reduction half-reactions: $$ \begin{aligned} E_{cell}^{\circ}=E_{red}^{\circ} + E_{ox}^{\circ}=(-0.86 \, \mathrm{V})+(0.49 \, \mathrm{V}) =-0.37 \, \mathrm{V} \end{aligned} $$ Since the calculated emf is negative, we have to reverse the overall reaction, which makes the standard reduction potential positive.

The final reaction with the largest positive emf is: $$ \begin{aligned} 2\mathrm{Au}(s) + 8 \mathrm{Br}^{-}(a q) + 3\mathrm{I}^{-}(a q)+6 \mathrm{OH}^{-}(a q) \longrightarrow 2\mathrm{AuBr}_{4}^{-}(a q) + 3\mathrm{IO}^{-}(a q)+3\mathrm{H}_{2} \mathrm{O}(l) \\ E_{cell}^{\circ}=0.37 \, \mathrm{V} \end{aligned} $$

For the smallest positive emf, we want to combine the weakest reducing agent with the weakest oxidizing agent. From the given half-reactions, it is clear that \(\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)\) has the least negative value of standard reduction potential, which indicates it is the weakest reducing agent. However, since the half-cell reaction for the strongest oxidizing agent has already been used to find the largest positive EMF, we must find the next weakest oxidizing agent for the smallest positive EMF. In this case, we already know the weakest reducing agent we must reverse to act as the weakest oxidizing agent: $$ \begin{aligned} \mathrm{Eu}^{2+}(a q) \longrightarrow \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \end{aligned} $$ Now, let's combine this half-reaction with the weakest reducing agent to form the overall reaction with the smallest positive EMF.

Since there is only one electron involved in both the oxidation and reduction half-reactions, we can directly combine the two reactions: $$ \begin{aligned} \mathrm{Eu}^{3+}(a q)+\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) + \mathrm{Eu}^{2+}(a q)+1 \mathrm{e}^{-} \end{aligned} $$

To calculate the emf for this combination of half-cell reactions, we first sum the standard potentials of the oxidation and reduction half-reactions: $$ \begin{aligned} E_{cell}^{\circ}=E_{red}^{\circ} + E_{ox}^{\circ}=(-0.86 \, \mathrm{V})-(-0.43 \, \mathrm{V}) =-0.43 \, \mathrm{V} \end{aligned} $$ Similarly, since the calculated emf is negative, we have to reverse the overall reaction, which makes the standard reduction potential positive.

The final reaction with the smallest positive emf is: $$ \begin{aligned} \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) + \mathrm{Eu}^{2+}(a q)+1 \mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{3+}(a q)+\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \\ E_{cell}^{\circ}=0.43 \, \mathrm{V} \end{aligned} $$ So, the final answers are: (a) Largest positive emf: \(E_{\text{cell}}^{\circ} = 0.37 \, \mathrm{V}\) (b) Smallest positive emf: \(E_{\text{cell}}^{\circ} = 0.43 \, \mathrm{V}\)