Li- K- Ca- Na- Mg- Al- Zn- Fe- Pb- Cu- Hg- Ag I- The reaction 3Fes-2AlNO33aq- 2Als-3FeNO32aq will occur- II- Iron is higher on the activity series than aluminum-

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Reactivity Series

Li, K, Ca...

Question

$L i$ , $K$ , $C a$ , $N a$ , $M g$ , $A l$ , $Z n$ , $F e$ , $P b$ , $C u$ , $H g$ , $A g$
$I$ . The reaction $3 F e (s) + 2 A l {({N O}_{3})}_{3} (a q) \to 2 A l (s) + 3 F e {({N O}_{3})}_{2} (a q)$ will occur.
$I I$ . Iron is higher on the activity series than aluminum.

Statement

I

is true, Statement

I I

is true and is a correct explanation of the phenomena described in

I

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Statement

I

is true, Statement

I I

is false

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Statement

I

is false, Statement

I I

is true

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Statement

I

is false, Statement

I I

is false

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Solution

The correct option is D Statement $I$ is true, Statement $I I$ is true and is a correct explanation of the phenomena described in $I$
By using the reactivity series of metals, one can predict the products of displacement reactions . Each element in the reactivity series can be replaced from a compound by any of the elements above it. For ex: $M g$ can replace $Z n$ from the solution.
$A l$ is higher on the activity series than $F e$ .
Therefore $F e$ can not displace $A l$ from $A l {(N O_{3})}_{3}$
$∴ 3 F e + 2 A l {(N O_{3})}_{3} ⟶ 2 A l + 3 F e {(N O_{3})}_{2}$ will not occur.

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Li, K, Ca, Na, Mg, Al, Zn, Fe, Pb, Cu, Hg, AgI. The reaction 3Fe(s)+2Al(NO3)3(aq)→2Al(s)+3Fe(NO3)2(aq) will occur.II. Iron is higher on the activity series than aluminum.

$L i$ , $K$ , $C a$ , $N a$ , $M g$ , $A l$ , $Z n$ , $F e$ , $P b$ , $C u$ , $H g$ , $A g$
$I$ . The reaction $3 F e (s) + 2 A l {({N O}_{3})}_{3} (a q) \to 2 A l (s) + 3 F e {({N O}_{3})}_{2} (a q)$ will occur.
$I I$ . Iron is higher on the activity series than aluminum.