Science, asked by smrutipadhi, 1 year ago

1) What is reactivity series? How does the reactivity series of metals help in predicting the relative activities of various metals?

2) Suggest different chemical processes for obtaining a metal from its oxides for metals in the middle of the reactivity series and metals towards the top of the reactivity series. Support your answer with one example each.

Answers

Answered by piyushkumar22
1


(a) The arrangement of metals in a series according to the decreasing order of their reactivity is called reactivity series. The metals that are placed above hydrogen are more reactive and those below hydrogen are less reactive. A metal placed above another metal can displace lower metal from its salt solution. Thus a metal placed above in series is more reactive than metals placed below it.

(b) Metals placed in the middle of the reactivity series are obtained from their oxides by reduction method. Suitable reducing agents are used. For example, zinc metal can be obtained from its oxide by using carbon as reducing agent. 

ZnO+C → Zn + CO 

Metals placed at top of series can be extracted by using electrolysis method. Example - Aluminium is obtained by electrolytic reduction of aluminium oxide.


piyushkumar22: please mark it as brainliest
Answered by BendingReality
0

Answer:

1 ) .Reactivity series :

It is series of decreasing order of metals' reactivity. Hydrogen is place in middle of series. Above hydrogen metals are more reactive in nature below are less. Metal which is more reactive it displaces a less reactive metal from its salt solution.

2 ) .

In order to obtaining a metal from its  oxides for metals in the middle of the reactivity series then metals towards oxides of such metals can be reduced with coke (carbon) which acts as a reducing agent.

For instance :

2 Fe₂O₃ + 3 C  ⇒ 4 Fe + 3 CO₂.

In order to obtaining a metal from its  oxides for metals in the high of the reactivity series their oxides are reduced to metals by the process of electrolysis

For instance :

Electrolysis of sodium chloride :

Na⁺ + e⁻  ⇒ Na   [ At cathode ]

2 Cl⁻ ⇒ Cl₂ + 2 e⁻   [ At anode ]

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