
An iron wire is immersed in a solution containing \[ZnS{O_4}\] and \[NiS{O_4}\]. Predict giving reasons which of the following reactions is likely to proceed?
(i) Iron reduces \[Z{n^{2 + }}\] ions (ii) Iron reduces \[N{i^{2 + }}\] ions.
Given: \[{E_{Z{n^{2 + }}/Zn}}\, = - 0.76\;\,volt\] \[{E_{F{e^{2 + }}/Fe}}\, = - 0.44\;\,volt\] and \[{E_{N{i^{2 + }}/Ni}}\, = - 0.25\;\,volt\]
Answer
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Hint:To predict the feasibility of a reaction you need to check the reactivity of elements from the reduction potential. In electrochemical series elements with less negative value cannot reduce the element with a more negative value.
Complete step by step answer:
To displace an element from its solution, the other element has to be more reactive. It means that it should come above that element in the electrochemical series.
In case of \[ZnS{O_4}\] the metal is zinc and when you dip an iron wire in it so in order to reduce zinc, iron needs to be more reactive. Since the reduction potential of iron is $ - 0.44\,v$ and of zinc is $ - 0.76$. From this, we can deduce that the reduction potential of iron is less negative than zinc so it won’t be able to displace zinc from the \[ZnS{O_4}\] solution.
As a result of this, the reaction will not proceed further.
In case \[NiS{O_4}\] of the metal is Nickel and when you dip an iron wire in it so in order to reduce Nickel. Iron needs to be more reactive. Since the reduction potential of iron is $ - 0.44\,v$ and of nickel is $ - 0.25\,v$. From this, we can deduce that the reduction potential of iron is more negative than nickel so it will be able to displace nickel from the solution \[NiS{O_4}\].
As a result of this, the reaction will proceed further.
Note:
When $Fe$ is placed in a \[ZnS{O_4}\] solution no color change is found because no reaction takes place while we place $Fe$in \[NiS{O_4}\] solution the color of the solution changes because iron displaces nickel in the solution.
Complete step by step answer:
To displace an element from its solution, the other element has to be more reactive. It means that it should come above that element in the electrochemical series.
In case of \[ZnS{O_4}\] the metal is zinc and when you dip an iron wire in it so in order to reduce zinc, iron needs to be more reactive. Since the reduction potential of iron is $ - 0.44\,v$ and of zinc is $ - 0.76$. From this, we can deduce that the reduction potential of iron is less negative than zinc so it won’t be able to displace zinc from the \[ZnS{O_4}\] solution.
As a result of this, the reaction will not proceed further.
In case \[NiS{O_4}\] of the metal is Nickel and when you dip an iron wire in it so in order to reduce Nickel. Iron needs to be more reactive. Since the reduction potential of iron is $ - 0.44\,v$ and of nickel is $ - 0.25\,v$. From this, we can deduce that the reduction potential of iron is more negative than nickel so it will be able to displace nickel from the solution \[NiS{O_4}\].
As a result of this, the reaction will proceed further.
Note:
When $Fe$ is placed in a \[ZnS{O_4}\] solution no color change is found because no reaction takes place while we place $Fe$in \[NiS{O_4}\] solution the color of the solution changes because iron displaces nickel in the solution.
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