Utilizing the half-cell potential information provided in Table 10.1, what would be the most likely spontaneous outcome in a system composed of bromine, fluorine, bromine ions and fluoride ions?
Br_2(l) + 2e^- $\rightarrow$ 2 Br ^-_(aq)Potential = 1.09 V
F_2(g) + 2e^- $\rightarrow$ 2 F ^-_(aq)Potential = 2.87 V

Question

Utilizing the half-cell potential information provided in Table 10.1, what would be the most likely spontaneous outcome in a system composed of bromine, fluorine, bromine ions and fluoride ions?
Br_2(l) + 2e^-

\rightarrow

2 Br ^-_(aq)Potential = 1.09 V
F_2(g) + 2e^-

\rightarrow

2 F ^-_(aq)Potential = 2.87 V

Quizplus · Accepted Answer

Fluorine has a higher reduction potential (2.87 V) than bromine (1.09 V), meaning it is more likely to be reduced. Therefore, fluorine gas will oxidize bromide ions to bromine, resulting in the spontaneous reaction: F2(g) + 2 Br-(aq) -> 2 F-(aq) + Br2(l).

Utilizing the Half-Cell Potential Information Provided in Table 10 →\rightarrow→ 2 Br -(aq) Potential = 1

Utilizing the Half-Cell Potential Information Provided in Table 10 $\rightarrow$ 2 Br ^-_(aq)Potential = 1