The correct order of E^o M^{2+}/M values with | vedclass

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(B) The standard electrode potential ($E^o$ $M^{2+}/M$) values for the first transition series generally become less negative as we move from left to

Vedclass · Accepted Answer

$Mn > Cr > Fe > Co$

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(B) The standard electrode potential ($E^o$ $M^{2+}/M$) values for the first transition series generally become less negative as we move from left to right,with some exceptions.$Mn^{2+}$ has a stable $d^5$ configuration,which makes its reduction potential more negative than expected.The electronic configurations of the $M^{2+}$ ions are:$Cr^{2+}$: $[Ar] 3d^4$$Mn^{2+}$: $[Ar] 3d^5$ (half-filled,extra stable)$Fe^{2+}$: $[Ar] 3d^6$$Co^{2+}$: $[Ar] 3d^7$Due to the extra stability of the $Mn^{2+}$ ion,the $E^o$ value for $Mn^{2+}/Mn$ is more negative than that of $Cr^{2+}/Cr$. The overall order of the magnitude of negative values is $Mn > Cr > Fe > Co$.

The correct order of $E^o$ $M^{2+}/M$ values with negative sign for the four successive elements $Cr$,$Mn$,$Fe$ and $Co$ is

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