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The equivalent weight of FeS{{O}_{4}} is half of its molecular weight, when it converts to:

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)



In the case of redox reactions, the n-factor is the net change in the oxidation number of the central metal atom per formula unit of an oxidising or reducing agent. Here, according to the question, the equivalent weight reduces to half, which means the n-factor should be two.

Therefore, FeS{{O}_{4}} converts to Fe{{O}_{2}} 

NOTE: During oxidation, electrons are released by the element undergoing the oxidation process, thus oxidation states increase. Whereas during reduction, electrons are gained by the element, thus oxidation state decreases.

Posted by

Ritika Jonwal

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