Two iron rod $A$ and $B$ are taken. $A$ is dipped in dilute $H_{2} S O_{4}$ . $B$ is first dipped in conc. $H N O_{3}$ and later in dilute $H_{2} S O_{4}$ . What will be the status of iron rods after some time? Justify.

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Solution

Generally, Iron rod when dipped in concentrated $H_{2} S O_{4}$ , then ferric sulphate forms.
$3 F e + 8 H_{2} S O_{4} \to F e S O_{4} + F e_{2} (S O_{4})_{3} + 4 S O_{2} + 8 H_{2} O$
The rod A is directly dipped. So A forms $F e_{2} (S O_{4})_{3}$
But rod B does not form anything with concentrated $H_{2} S O_{4}$ . Because a passive film of oxide is formed on the surface by reacting rod B with concentrated $H N O_{3}$

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