Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?

Option 1 - It will not prefer to undergo redox reactions.

Option 2 - It will to gain electron and act as an oxidizing agent

Option 3 - It will prefer to give away an electron and behave as reducing agent

Option 4 - It acts as both, oxidizing and reducing agent.

2 Views|Posted 5 months ago

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Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 2

Detailed Solution:

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l} C e^{+ 4} + e^{-} \to C e^{+ 3} E^{0} = 1.61 V \\ C e^{+ 3} + 3 e^{-} \to C e E^{0} = - 2.336 V \end{array}$

It exist as Ce⁺⁴and acts like a strong oxidizing agent by gaining electrons

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