For Cr, Mn, Fe, and Co,the E^{0}_{M^{3+}/M^{2 | vedclass

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(C) The change from $+2$ to $+3$ oxidation state involves the oxidation of the metal ion: $M^{2+} \rightarrow M^{3+} + e^-$. The standard electrode po

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$Cr$

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(C) The change from $+2$ to $+3$ oxidation state involves the oxidation of the metal ion: $M^{2+} \rightarrow M^{3+} + e^-$. The standard electrode potential given is for reduction: $M^{3+} + e^- \rightarrow M^{2+}$. The oxidation potential is the negative of the reduction potential: $E^{0}_{M^{2+}/M^{3+}} = -E^{0}_{M^{3+}/M^{2+}}$. For $Cr$: $E^{0}_{Cr^{2+}/Cr^{3+}} = -(-0.41 \ V) = +0.41 \ V$. For $Mn$: $E^{0}_{Mn^{2+}/Mn^{3+}} = -(+1.57 \ V) = -1.57 \ V$. For $Fe$: $E^{0}_{Fe^{2+}/Fe^{3+}} = -(+0.77 \ V) = -0.77 \ V$. For $Co$: $E^{0}_{Co^{2+}/Co^{3+}} = -(+1.97 \ V) = -1.97 \ V$. $A$ higher (more positive) oxidation potential indicates that the oxidation is easier. Since $Cr$ has the highest positive oxidation potential $(+0.41 \ V)$,the change from $+2$ to $+3$ is easiest for $Cr$.

For $Cr, Mn, Fe,$ and $Co$,the $E^{0}_{M^{3+}/M^{2+}}$ values are $-0.41, +1.57, +0.77,$ and $+1.97 \ V$ respectively. For which metal is the change in oxidation state from $+2$ to $+3$ the easiest?

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