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Reaction of Acids with Metals

Four metals ...

Question

Four metals $Z n, F e, C u$ and $A l$ are taken and added to the following solutions one by one. The results obtained are tabulated below:
Metal $Z n S O_{4} (a q)$ $F e S O_{4} (a q)$ $C u S O_{4} (a q)$ $A l_{2} (S O_{4})_{3} (a q)$
$Z n$ - Displaced Displaced No reaction
$F e$ No reaction - Displaced No reaction
$C u$ No reaction No reaction - No reaction
$A l$ Displaced Displaced Displaced -
From the above data, the decreasing order of reactivity of metals is-

A l > C u > F e > Z n

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A l > Z n > F e > C u

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A l > Z n > C u > F e

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A l > F e > C u > Z n

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Solution

The correct option is B $A l > Z n > F e > C u$
From the given results, the decreasing order of reactivity can be written as:
$A l > Z n > F e > C u$
Al is the most reactive followed by zinc metal. Fe is less reactive than zinc but more reactive than copper.

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L i

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C a

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A l

Z n

F e

P b

C u

H g

A g

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Z n (s) + C u {({N O}_{3})}_{2} (a q) \to C u (s) + Z n {({N O}_{3})}_{2} (a q)

I I

A l, C u, F e, M g

Z n

E_{A l^{3} + / A l}^{0} + - 1.66 V, E_{C u^{2} + / C u}^{0} = + 0.34 V

E_{F e^{2} + / F e}^{0} = - 0.44 V, E_{M g^{2} + / M g}^{0} = - 1.40 V

E_{Z n^{2} + / Z n}^{0} = - 0.76 V

What will happen when metals such as Zn, Fe, Cu and Al react with nitric acid?

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