Question

In the reaction, $F{e_2}{O_3} + 2Al \to 2Fe + A{l_2}{O_3} + heat$ , the substance that is oxidised is _____________ and the substance that is reduced is ______________.
A. \[Al,{\text{ }}Fe\]
B. \[\;Fe,\;Al\]
C. \[Al,{O_2}\]
D. \[{O_2},{\text{ }}Fe\]

Answer 1

Hint: We know that redox reaction is a reaction where oxidation and reduction takes place simultaneously. In the process of oxidation electrons are lost which results in the increase of the oxidation number of that element. In the same way if the oxidation number reduces then the element is reduced or a reduction process takes place.

Complete step by step answer:
In the given reaction ferric oxide is reacting with aluminium to give aluminium oxide and iron ,that is $F{e_2}{O_3} + 2Al \to 2Fe + A{l_2}{O_3} + heat$ ; here the substance which is oxidised is aluminium ($Al$) and the substance which is reduced is iron ($Fe$). It is because of the change in the oxidation number of aluminium and iron that we are able to easily identify the substance which is oxidising and which is reducing. In free state any element possesses zero oxidation number and we can calculate the oxidation number of iron and aluminium in the compound which is equal to $ + 3$ . The oxidation number of aluminium increases from $0$ to $ + 3$ and the oxidation number of iron decreases from $ + 3$ to $0$. Hence the correct answer of this problem is option A which is \[Al,{\text{ }}Fe\]. Thus , the substance that is oxidised is aluminium and the substance that is reduced is iron.
So, the correct answer is “Option A”.

Note: Thus we can conclude that oxidation is the loss of electrons from an atom molecule or an ion whereas the reduction is the gain of electrons from an atom, molecule or an ion. In oxidation , oxidation state or number increases. In reduction Oxidation state or number decreases.