Question

On the basis of Eo

values identify which amongst the following is the strongest

oxidising agent

Cl2(g) + 2 e– 2Cl-

Eo= +1.36 V,

MnO4–

+ 8H+ + 5e– Mn2+ + 4H2O Eo= +1.51 V

Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O Eo= +1.33 V​

Answer 1

Answer:

MnO4–+ 8H+ + 5e– Mn2+ + 4H2O Eo= +1.51 V

Explanation:

MnO⁴- is the strongest oxidising agent, since Anhydrous MnO4- or more commonly known as permanganate ion is a strong oxidizing agent.

Anhydrous MnO4- or more commonly known as permanganate ion is a strong oxidizing agent.If the oxidation state of an element increases, its electronegativity also increases. And this is why, permanganate is a strong oxidizing agent.

Anhydrous MnO4- or more commonly known as permanganate ion is a strong oxidizing agent.If the oxidation state of an element increases, its electronegativity also increases. And this is why, permanganate is a strong oxidizing agent.For example, Mn ion has an oxidation state of +7 in permanganate anion which is why it is highly electronegative in nature.

Hope u get it...

Answer 2

MnO4- is the strongest oxidizing agent among Cl2, MnO4-, and Cr2O72-.

• The oxidizing power of a species is related to its ability to gain electrons, i.e., it has a high tendency to get reduced.
• The more positive the reduction potential (Eo) of a species, the more likely it is to act as a strong oxidizing agent.

Here, we can compare the Eo values of the given species to determine which is the strongest oxidizing agent:

Cl2(g) + 2 e– → 2Cl- Eo = +1.36 V

MnO4– + 8H+ + 5e– → Mn2+ + 4H2O Eo = +1.51 V

Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O Eo = +1.33 V

• As we can see, the species with the highest Eo value is MnO4– (+1.51 V), which means it has the highest tendency to gain electrons and is the strongest oxidizing agent among the given options.
• Therefore, MnO4- is the strongest oxidizing agent among Cl2, MnO4-, and Cr2O72-.

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