Consider these two entries from a fictional table of standard reduction potentials:A: X^{2+} + 2e^- → X(s)E^ext{°} = -0.50 VB: Y^{+} + e^- → Y(s)E^ext{°} = +0.30 VWhich species is the stronger oxidizing agent under standard conditions?

• A. Y^{+}
• B. X(s)
• C. Y(s)
• D. X^{2+}

Answer: (A)

Verified step by step guidance

1. Recall that the strength of an oxidizing agent is related to its tendency to gain electrons, which corresponds to the standard reduction potential (E^{\circ}). The species with the higher (more positive) standard reduction potential is the stronger oxidizing agent.
2. Look at the given half-reactions and their standard reduction potentials: \(X^{2+} + 2e^- \rightarrow X(s) \quad E^{\circ} = -0.50\ \text{V}\) \(Y^{+} + e^- \rightarrow Y(s) \quad E^{\circ} = +0.30\ \text{V}\)
3. Compare the two E^{\circ} values: -0.50 V for X^{2+} and +0.30 V for Y^{+}. Since +0.30 V is more positive, Y^{+} has a greater tendency to be reduced (gain electrons).
4. Because the oxidizing agent is the species that gets reduced, the stronger oxidizing agent is the one with the higher standard reduction potential, which is Y^{+}.
5. Therefore, under standard conditions, Y^{+} is the stronger oxidizing agent compared to X^{2+}.