To determine the strongest oxidizing agent, we need to understand the concept of standard reduction potential (E°). The standard reduction potential measures the tendency of a chemical species to gain electrons and be reduced. A higher (more positive) E° value indicates a greater tendency to be reduced, making it a stronger oxidizing agent.

Given the standard reduction potentials:

• For O₂/H₂O: E° = +1.23 V
• For S₂O₈²⁻/SO₄²⁻: E° = +2.05 V
• For Br₂/Br⁻: E° = +1.09 V
• For Au³⁺/Au: E° = +1.4 V

Comparing these values:

${\mathrm{E}}_{{\mathrm{S}}_2{\mathrm{O}}_8^{2-}/{\mathrm{SO}}_4^{2-}}^{\mathrm{r}}=2.05\mathrm{V}$ is the highest, meaning S₂O₈²⁻ has the strongest tendency to gain electrons and be reduced. Therefore, it is the strongest oxidizing agent among the given options.

Final Answer: S₂O₈²⁻ (peroxodisulfate ion) is the strongest oxidizing agent.

Related Topics

Electrochemical Series: The series that arranges elements and ions in order of their standard reduction potentials. Oxidizing agents with higher E° values are stronger.

Oxidizing and Reducing Strength: A species with high E° is a strong oxidizing agent, while one with low (or highly negative) E° is a strong reducing agent.

Key Formulae

Standard reduction potential (E°) directly indicates oxidizing strength: Higher E° = Stronger oxidizing agent.

For a reduction half-reaction: Oxidized form + ne⁻ → Reduced form, the E° value defines the tendency to undergo reduction.