Rank these species by their ability to act as an oxidizing agent.
Na^+
I2
Cr^3+
F2
Put best oxidizing agent first and continue to have best reducing agent last.
Cr+3, Na+, I2, F2
Oxidized- lose electrons
Wrong answer, its backwards
jvkmgvkuyku
To rank these species by their ability to act as an oxidizing agent, we need to consider their reduction potentials or standard electrode potentials (E°). The higher the reduction potential, the stronger the oxidizing agent.
The reduction potential values for these species can be obtained from tables of standard electrode potentials. For the purpose of this explanation, let's assume the following values:
Na⁺: -2.71 V
I₂: -0.54 V
Cr³⁺: -0.74 V
F₂: +2.87 V
Note that reduction potential values are negative for oxidizing agents, as they are willing to accept electrons.
Based on these values, we can rank the species as follows:
1. F₂ (+2.87 V): F₂ has the highest reduction potential among the given species, making it the strongest oxidizing agent.
2. I₂ (-0.54 V): I₂ has a less positive reduction potential compared to F₂, but it still has a relatively higher value, making it a stronger oxidizing agent than the remaining species.
3. Cr³⁺ (-0.74 V): Cr³⁺ has a lower reduction potential compared to F₂ and I₂, making it a weaker oxidizing agent.
4. Na⁺ (-2.71 V): Na⁺ has the lowest reduction potential among the given species, making it the weakest oxidizing agent.
Therefore, the rank from best oxidizing agent to best reducing agent is: F₂ > I₂ > Cr³⁺ > Na⁺.