Question

Rusting of iron is a chemical change because:
A. It is temporary
B. It can be cleaned and the original substance recovered
C. It is a slow reaction
D. A new substance is formed

Answer 1

Hint: The most common example of corrosion is rusting of iron. Rust is hydrated ferric oxide with the chemical formula $F{e_2}{O_3}.x{H_2}O$. Where x is the number of water molecules.

Complete step by step answer: As we know, rusting of iron is example of corrosion and corrosion is the process of slowly eating away of the metal like iron ,silver etc. due to attack of atmospheric gases on the surface of metal resulting into formation of compounds such as oxides , carbonates, sulphides, sulphates etc. As there is formation of new compounds so we can say rusting of iron is a chemical change.

The mechanism of the rusting of iron is based on electrochemical theory as the formation of rust is explained on the basis of the formation of electrochemical cells on the surface of iron metal.
The steps involved in formation of rust are:
Step 1: The water vapours on the surface of the metal dissolve $C{O_2}$ and $O_2$ from the air as a result the surface of metal is covered by carbonic acid.
${H_2}O + C{O_2} \to {H_2}C{O_3}$
This act as electrolyte of the cell and dissociates to small extent as:
${H_2}CO\underset {} \leftrightarrows 2{H^ + } + CO_3^{2 - }$
H+ ion is also produced by dissolution of other acidic oxide such as so2.
Step2: Next, Iron in contact with dissolved O2 and in presence of H+ ions undergoes oxidation as:
$Fe \to F{e^{2 + }} + 2{e^ - }$ (1)
The site where the above reaction takes place is known as anode.
Step 3: The electrons lost by iron migrate through the metal to another site and reduce dissolved oxygen in the presence of H+ ions and this site is known as cathode.
${O_2} + 4{H^ + } + 4{e^ - } \to 2{H_2}O$ (2)
On adding equation (1) and (2) we get:
$2Fe(s) + {O_2}(g) + 4{H^ + }(aq) \to 2F{e^{2 + }}(aq) + 2{H_2}O(l)$
Step 4: The ferrous ion formed in above step thus react with oxygen to form ferric ion as:
$4F{e^{2 + }} + {O_2} + 4{H_2}O \to 2F{e_2}{O_3} + 8{H^ + }$
Then ferric oxide undergoes hydration to form rust.
$F{e_2}{O_3} + x{H_2}0 \to F{e_2}{O_3}.x{H_2}O$

Hence the correct answer is option D.

Note: Generally, cathode is the area having the largest amount of dissolved oxygen and rust is often formed here. Talking about properties of rust, it is a non-sticking element and it peels off exposing fresh iron surface.