
The correct order of \[E_{{M^{2 + }}/M}^0\] values with a negative sign for the four successive elements \[Cr\], \[Mn\], \[Fe\] and \[Co\] is-
A.\[Fe > Mn{\text{ > }}Cr > Co\]
B.\[Cr > Mn > Fe > Co\]
C.\[Mn > Cr > Fe > Co\]
D.\[Cr > Fe > Mn > Co\]
Answer
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Hint: Standard electrode potential is a measurement of the potential for equilibrium. There is a potential difference between the electrode and the electrolyte called the potential of the electrode. When unity is the concentration of all the species involved in a semi-cell, the electrode potential is known as the standard electrode potential. Under standard conditions, the standard electrode potential occurs in an electrochemical cell, say the temperature = 298K, pressure = 1atm, concentration = 1M. The symbol ‘\[E_{cell}^0\]’ represents the standard electrode potential of a cell.
The standard reduction potential is the likelihood that a species will be reduced. It is written in the form of a reduction half reaction. An example can be seen below where ‘M’ is a generic element and ‘n’ is the charge.
\[{M^{n + }} + n{e^ - } \to M\]
Complete answer:
Electronic configuration of given elements are as follows-
Across the first transition series, the negative values for standard electrode potential decreases across the period except for \[Mn\] because it attains the stable half filled \[{d^5}\;\]configuration after losing 2 electrons.
Hence, the correct order is \[Mn > Cr > Fe > Co\]
Option (C) is correct.
Additional information:
All those elements with negative reduction potentials act as strong reducing agents and liberate hydrogen from dilute acids.
Note:
The irregularity in the variation of electrode potentials is due to the irregular variation of the ionisation enthalpies and also the hydration energies of the divalent ions of these elements. The magnitude of electrode potential depends on the following factors, (i) Nature of the electrode, (ii) Concentration of the ions in solution, (iii) Temperature.
The standard reduction potential is the likelihood that a species will be reduced. It is written in the form of a reduction half reaction. An example can be seen below where ‘M’ is a generic element and ‘n’ is the charge.
\[{M^{n + }} + n{e^ - } \to M\]
Complete answer:
Electronic configuration of given elements are as follows-
| \[Cr - \left[ {Ar} \right]{\text{ }}3{d^5}4{s^1}\] | \[C{r^{2 + }} - \left[ {Ar} \right]{\text{ }}3{d^4}\] |
| \[Mn - \left[ {Ar} \right]{\text{ }}3{d^5}\;4{s^2}\] | \[M{n^{2 + }} - \left[ {Ar} \right]{\text{ }}3{d^5}\] |
| \[Fe - \left[ {Ar} \right]{\text{ }}3{d^6}\;4{s^2}\] | \[F{e^{2 + }} - \left[ {Ar} \right]{\text{ }}3{d^6}\] |
| \[Co - \left[ {Ar} \right]{\text{ }}3{d^7}\;4{s^2}\] | \[C{o^{2 + }} - \left[ {Ar} \right]{\text{ }}3{d^7}\] |
Across the first transition series, the negative values for standard electrode potential decreases across the period except for \[Mn\] because it attains the stable half filled \[{d^5}\;\]configuration after losing 2 electrons.
Hence, the correct order is \[Mn > Cr > Fe > Co\]
Option (C) is correct.
Additional information:
All those elements with negative reduction potentials act as strong reducing agents and liberate hydrogen from dilute acids.
Note:
The irregularity in the variation of electrode potentials is due to the irregular variation of the ionisation enthalpies and also the hydration energies of the divalent ions of these elements. The magnitude of electrode potential depends on the following factors, (i) Nature of the electrode, (ii) Concentration of the ions in solution, (iii) Temperature.
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