?Given the following half-reactions and associated standard reduction potentials: \(\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarro
Chapter 20, Problem 20.4(choose chapter or problem)
Given the following half-reactions and associated standard reduction potentials:
\(\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) \\ E_{\text {red }}^{\circ}=-0.86 \mathrm{~V}\)
\(\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) \\ E_{\mathrm{red}}^{\circ} &=-0.43 \mathrm{~V}\)
\(\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}\longrightarrow=+0.49 \mathrm{~V}\)
(a) Write the equation for the combination of these half cell 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.
Text Transcription:
AuBr4 -(aq) + 3 e- \longrightarrow Au(s) + 4 Br-(aq) E_red^{\circ} = -0.86 V
Eu3+(aq) + e- \longrightarrow Eu2+(aq) E° red = -0.43 V
IO-(aq) + H2O(l) + 2 e- \longrightarrow I -(aq) + 2 OH-(aq) E_red^{\circ} = +0.49 V
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