In an MCQ, the correct statement regarding the abovementioned compound is that its solution is used as an oxidizing agent. I do not understand why that is so.

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    $\begingroup$ calculate the oxidation state of Ce in the compound. $\endgroup$ – M. Farooq Nov 23 '19 at 5:49
  • $\begingroup$ I think it's +4. Which means it has achieved Xe configuration. $\endgroup$ – Novice chemist Nov 23 '19 at 9:47

Most lanthanides display stability in their +3 oxidation states. Cerium does the same, which is why it has a very high reduction potential. The reduction potential for the $\ce{Ce^{4+}|Ce^{3+}}$ couple is 1.61 V (for comparison, $\ce{Cl2}$ has a potential of 1.32 V)

Cerium in the form of ceric ammonium nitrate is also quite stable compared to other oxidizing agents and can be stored on the shelf for a longish period of time. It's also quite usable: the bright orange colour is decolourized to a pale yellow on reduction of $\ce{Ce^{4+}}$. This is why it's used as a common oxidizing agent.

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  • $\begingroup$ Thanks. This clears it up for me. $\endgroup$ – Novice chemist Nov 23 '19 at 17:06
  • $\begingroup$ No problem! Please mark it as the correct answer if it answers your question. $\endgroup$ – Aniruddha Deb Nov 23 '19 at 17:31

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