
Given,${E^o}$ for $M{g^{2 + }}/Mg = - 2 \cdot 37V,Z{n^{2 + }}/Zn = - 0 \cdot 76V$ and $F{e^{2 + }}/Fe = - 0 \cdot 44V$. Which of the following statements is correct?
A) ${\text{Zn}}$ reduces $F{e^{2 + }}$
B) ${\text{Zn}}$ reduces $M{g^{2 + }}$
C) ${\text{Mg}}$ oxidizes ${\text{Fe}}$
D) ${\text{Zn}}$ oxidizes ${\text{Fe}}$
Answer
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Hint: Analyze the given reactions carefully and determine the oxidation potential and reducing the potential of each element. One can compare their reducing potentials with each other and relate them with the given options to see which statement is correct.
Complete answer:1) First of all we will analyze the given conversions and ${E^o}$ values and decide what exactly happens in those reactions.
$M{g^{2 + }}/Mg = - 2 \cdot 37V$
$Z{n^{2 + }}/Zn = - 0 \cdot 76V$
$F{e^{2 + }}/Fe = - 0 \cdot 44V$
2) The order of reduction potential will be as below,
${\text{Fe > Zn > Mg}}$
The order of oxidation potential will be as below,
${\text{Mg > Zn > Fe}}$
3) In the conversion of $F{e^{2 + }}/Fe$ there is reduction potential value is higher than all other conversions which are $ - 0 \cdot 44$ . The higher the reduction potential, the strong oxidizing agent it will be. Hence this means the ${\text{Fe}}$ can oxidize ${\text{Zn}}$ and ${\text{Mg}}$.
4) Now the oxidation potential ${\text{Mg}}$ is higher than the other elements. This means that the ${\text{Mg}}$ is a good reducing agent than other elements. Hence, ${\text{Mg}}$ can reduce ${\text{Zn}}$ and ${\text{Fe}}$.
5) Now that we have understood the concepts let's analyze the given options. Option A states that ${\text{Zn}}$ reduce $F{e^{2 + }}$ which is true as we have seen the oxidation potential of ${\text{Zn}}$ is greater than the oxidation potential of ${\text{Fe}}$ which means it can reduce $F{e^{2 + }}$.
6) Now in option B states as ${\text{Zn}}$ reduces $M{g^{2 + }}$. As the oxidation potential of ${\text{Mg}}$is greater than oxidation potential of ${\text{Zn}}$ which means ${\text{Mg}}$ can not reduce ${\text{Zn}}$. Hence, this statement is incorrect.
7) option C states as ${\text{Mg}}$ oxidizes ${\text{Fe}}$. As the reduction potential of ${\text{Fe}}$ is greater than ${\text{Mg}}$ it can not oxidize ${\text{Fe}}$. Hence, this statement is an incorrect choice. Option D which states as ${\text{Zn}}$ oxidizes ${\text{Fe}}$. The reduction potential of ${\text{Fe}}$ is greater than ${\text{Zn}}$ it can not oxidize ${\text{Fe}}$. Hence, this statement is incorrect.
Therefore, the statement ${\text{Zn}}$ reduces $F{e^{2 + }}$ is correct which shows option A as the correct choice.
Note:It is very important to remember that the higher the reduction potential the stronger oxidizing agent it will be and vice-versa. In an oxidation there is the loss of electrons and in case of reduction there is the gain of electrons.
Complete answer:1) First of all we will analyze the given conversions and ${E^o}$ values and decide what exactly happens in those reactions.
$M{g^{2 + }}/Mg = - 2 \cdot 37V$
$Z{n^{2 + }}/Zn = - 0 \cdot 76V$
$F{e^{2 + }}/Fe = - 0 \cdot 44V$
2) The order of reduction potential will be as below,
${\text{Fe > Zn > Mg}}$
The order of oxidation potential will be as below,
${\text{Mg > Zn > Fe}}$
3) In the conversion of $F{e^{2 + }}/Fe$ there is reduction potential value is higher than all other conversions which are $ - 0 \cdot 44$ . The higher the reduction potential, the strong oxidizing agent it will be. Hence this means the ${\text{Fe}}$ can oxidize ${\text{Zn}}$ and ${\text{Mg}}$.
4) Now the oxidation potential ${\text{Mg}}$ is higher than the other elements. This means that the ${\text{Mg}}$ is a good reducing agent than other elements. Hence, ${\text{Mg}}$ can reduce ${\text{Zn}}$ and ${\text{Fe}}$.
5) Now that we have understood the concepts let's analyze the given options. Option A states that ${\text{Zn}}$ reduce $F{e^{2 + }}$ which is true as we have seen the oxidation potential of ${\text{Zn}}$ is greater than the oxidation potential of ${\text{Fe}}$ which means it can reduce $F{e^{2 + }}$.
6) Now in option B states as ${\text{Zn}}$ reduces $M{g^{2 + }}$. As the oxidation potential of ${\text{Mg}}$is greater than oxidation potential of ${\text{Zn}}$ which means ${\text{Mg}}$ can not reduce ${\text{Zn}}$. Hence, this statement is incorrect.
7) option C states as ${\text{Mg}}$ oxidizes ${\text{Fe}}$. As the reduction potential of ${\text{Fe}}$ is greater than ${\text{Mg}}$ it can not oxidize ${\text{Fe}}$. Hence, this statement is an incorrect choice. Option D which states as ${\text{Zn}}$ oxidizes ${\text{Fe}}$. The reduction potential of ${\text{Fe}}$ is greater than ${\text{Zn}}$ it can not oxidize ${\text{Fe}}$. Hence, this statement is incorrect.
Therefore, the statement ${\text{Zn}}$ reduces $F{e^{2 + }}$ is correct which shows option A as the correct choice.
Note:It is very important to remember that the higher the reduction potential the stronger oxidizing agent it will be and vice-versa. In an oxidation there is the loss of electrons and in case of reduction there is the gain of electrons.
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