For Cr, Mn, Fe, and Co,the E_{{M^{3+}}/{M^{2+ | vedclass

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(C) The oxidation of a metal from the $+2$ to the $+3$ state is represented by the half-reaction: $M^{2+} \rightarrow M^{3+} + e^-$. This process is f

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$Cr$

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(C) The oxidation of a metal from the $+2$ to the $+3$ state is represented by the half-reaction: $M^{2+} \rightarrow M^{3+} + e^-$. This process is favored when the reduction potential $E_{{M^{3+}}/{M^{2+}}}^{\circ}$ is low. $A$ lower (or more negative) value of $E_{{M^{3+}}/{M^{2+}}}^{\circ}$ indicates that the $M^{2+}$ ion is more easily oxidized to the $M^{3+}$ ion. Comparing the given values: $Cr: -0.41 \ V$ (Note: The value for $Cr^{3+}/Cr^{2+}$ is actually negative,$-0.41 \ V$) $Mn: +1.57 \ V$ $Fe: +0.77 \ V$ $Co: +1.97 \ V$ Since $Cr$ has the lowest (most negative) reduction potential,it is the most easily oxidized from the $+2$ to the $+3$ state.

Column-$I$	Column-$II$
$(a) \ Co^{3+}$	$(i) \ \sqrt{8} \ B.M.$
$(b) \ Cr^{3+}$	$(ii) \ \sqrt{35} \ B.M.$
$(c) \ Fe^{3+}$	$(iii) \ \sqrt{3} \ B.M.$
$(d) \ Ni^{2+}$	$(iv) \ \sqrt{24} \ B.M.$
	$(v) \ \sqrt{15} \ B.M.$

For $Cr, Mn, Fe,$ and $Co$,the $E_{{M^{3+}}/{M^{2+}}}^{\circ}$ values are $0.41, +1.57, +0.77,$ and $+1.97 \ V$ respectively. Which of these metals is most easily oxidized from the $+2$ to the $+3$ oxidation state?

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