Consider the following processes:
$Z{{n}^{2+}}+2{{e}^{-}}\to Zn\left( s \right);~{{E}^{o}}=-0.76V$
$C{{a}^{2+}}+2{{e}^{-}}\to Ca\left( s \right);~{{E}^{o}}=-2.87V$
$M{{g}^{2+}}+2{{e}^{-}}\to Mg\left( s \right);~{{E}^{o}}=-2.36V$
$N{{i}^{2+}}+2{{e}^{-}}\to Ni\left( s \right);~{{E}^{o}}=-0.25V$
The reducing power of the metals increases in the order:
(A) $Ca < Zn < Mg < Ni$
(B) $Ni < Zn < Mg < Ca$
(C) $Zn < Mg < Ni < Ca$
(D) $Ca < Mg < Zn < Ni$
Answer
598.8k+ views
Hint: Attempt this question with the understanding of reducing power of metals (as all the options include metals). Reducing power is defined as the potential of a substance to reduce another substance which can be done either by addition of hydrogen or by gain of electrons. So we will use this concept in the following solution.
Complete answer:
Let us first discuss about the reducing power as follows:-
-Reducing power is defined as the potential of a substance to reduce another substance which can be done either by addition of hydrogen or by gain of electrons. As we go down the group in the periodic table the reducing power increases because the atomic size increases due to which the pull of the nucleus on the valence electron decreases and thus the element loses electrons easily.
-As we know that the reducing power is directly proportional to the standard oxidation potential and inversely proportional to the standard reduction potential and therefore we will use this relation in this solution as follows:-
We are provided with the values of standard reduction potential as in all the reactions, there is the gain of electrons. So let us compare these values: $E_{Ni}^{o}>E_{Zn}^{o}>E_{Mg}^{o}>E_{Ca}^{o}$
So the reducing power of metals will increase in the following order: $Ni < Zn < Mg < Ca$
Hence, the correct option is: (B) $Ni < Zn < Mg < Ca$
Note:
-Remember that oxidizing power is directly proportional to the standard reduction potential and inversely proportional to the standard oxidation potential.
-The standard reduction potential is measured under standard conditions of temperature and pressure and also the metals are in the form of their pure states.
Complete answer:
Let us first discuss about the reducing power as follows:-
-Reducing power is defined as the potential of a substance to reduce another substance which can be done either by addition of hydrogen or by gain of electrons. As we go down the group in the periodic table the reducing power increases because the atomic size increases due to which the pull of the nucleus on the valence electron decreases and thus the element loses electrons easily.
-As we know that the reducing power is directly proportional to the standard oxidation potential and inversely proportional to the standard reduction potential and therefore we will use this relation in this solution as follows:-
We are provided with the values of standard reduction potential as in all the reactions, there is the gain of electrons. So let us compare these values: $E_{Ni}^{o}>E_{Zn}^{o}>E_{Mg}^{o}>E_{Ca}^{o}$
So the reducing power of metals will increase in the following order: $Ni < Zn < Mg < Ca$
Hence, the correct option is: (B) $Ni < Zn < Mg < Ca$
Note:
-Remember that oxidizing power is directly proportional to the standard reduction potential and inversely proportional to the standard oxidation potential.
-The standard reduction potential is measured under standard conditions of temperature and pressure and also the metals are in the form of their pure states.
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