Electrode potentials (E°) are given below :
$$Cu^+/Cu = +0.52$$ V,
$$Fe^{3+}/Fe^{2+} = +0.77$$ V,
$$\frac{1}{2}I_2(s)/I^- = +0.54$$ V,
$$Ag^+/Ag = +0.88$$ V.
Based on the above potentials, strongest oxidizing agent will be :

We are given the standard electrode potentials (E°) for several half-cells:

• For the reduction reaction: Cu⁺ + e⁻ → Cu, E° = +0.52 V
• For the reduction reaction: Fe³⁺ + e⁻ → Fe²⁺, E° = +0.77 V
• For the reduction reaction: ½I₂(s) + e⁻ → I⁻, E° = +0.54 V
• For the reduction reaction: Ag⁺ + e⁻ → Ag, E° = +0.88 V

An oxidizing agent is a species that accepts electrons and gets reduced. The strength of an oxidizing agent is determined by its tendency to gain electrons and be reduced. A higher (more positive) standard electrode potential (E°) indicates a greater tendency for reduction, meaning that species is a stronger oxidizing agent.

Comparing the given E° values:

• Cu⁺: E° = +0.52 V
• Fe³⁺: E° = +0.77 V
• I₂ (from ½I₂(s)): E° = +0.54 V
• Ag⁺: E° = +0.88 V

We see that Ag⁺ has the highest E° value of +0.88 V. This is followed by Fe³⁺ at +0.77 V, then I₂ at +0.54 V, and Cu⁺ at +0.52 V.

Therefore, Ag⁺ has the strongest tendency to be reduced, making it the strongest oxidizing agent among the given options.

Now, looking at the choices:

• A. Cu⁺
• B. Fe³⁺
• C. Ag⁺
• D. I₂

The strongest oxidizing agent is Ag⁺, which corresponds to option C.

Hence, the correct answer is Option C.