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

Question

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it? (A) It will not prefer to undergo redox reactions. (B) It will prefer to gain electron and act as an oxidizing agent (C) It will prefer to give away an electron and behave as reducing agent (D) It acts as both, oxidizing and reducing agent.

Tardigrade Admin · Accepted Answer

Cerium exists in two different oxidation state +3, +4 Ce +4+ e - arrow Ce 3+ E 0=+1.61 V Ce +3+3 e - arrow Ce E 0=-2.336 V It shows Ce +4 acts as a strong oxidising agent accepts electron.

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

It will not prefer to undergo redox reactions.

It will prefer to gain electron and act as an oxidizing agent

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

It acts as both, oxidizing and reducing agent.

Cerium exists in two different oxidation state $+ 3$ , $+ 4$
$C e^{+ 4} + e^{-} \to C e^{3 +} E^{0} = + 1.61 V$
$C e^{+ 3} + 3 e^{-} \to C e E^{0} = - 2.336 V$
It shows $C e^{+ 4}$ acts as a strong oxidising agent & accepts electron.

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

It will not prefer to undergo redox reactions.

It will prefer to gain electron and act as an oxidizing agent

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

It acts as both, oxidizing and reducing agent.

Cerium exists in two different oxidation state +3, +4 Ce+4+e−→Ce3+E0=+1.61V Ce+3+3e−→CeE0=−2.336V It shows Ce+4 acts as a strong oxidising agent & accepts electron.

Cerium exists in two different oxidation state $+ 3$ , $+ 4$
$C e^{+ 4} + e^{-} \to C e^{3 +} E^{0} = + 1.61 V$
$C e^{+ 3} + 3 e^{-} \to C e E^{0} = - 2.336 V$
It shows $C e^{+ 4}$ acts as a strong oxidising agent & accepts electron.