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Sm^(2+), Eu^(2+) and Yb^(2+) act as good...

`Sm^(2+), Eu^(2+)` and `Yb^(2+)` act as good reducing agents in aqueous solution but `Ce^(4+)` is a good oxidising agent. Why ?

Text Solution

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The most stable oxidation state for for the lanthanoid is + 3. Evidently `Sm^(2+), Eu^(2+)` and `Yb^(2+)` easily oxidises to their + 3 oxidation state and act as good reducing agent. For similar reason, `Ce^(4+)` reduces to the more stable + 3 state.
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