The {E^0}_{M^{3+}/M^{2+}} values for Cr, Mn, | vedclass

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(C) The ease of oxidation $(M^{2+} 	o M^{3+} + e^-)$ is determined by the oxidation potential,which is the negative of the reduction potential $({E^0

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$Cr$

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(C) The ease of oxidation $(M^{2+} 	o M^{3+} + e^-)$ is determined by the oxidation potential,which is the negative of the reduction potential $({E^0}_{M^{2+}/M^{3+}} = -{E^0}_{M^{3+}/M^{2+}})$. $A$ higher (more positive) oxidation potential indicates that the oxidation process is easier.MetalOxidation Potential $({E^0}_{M^{2+}/M^{3+}})$$Cr$$+0.41 \ V$$Mn$$-1.57 \ V$$Fe$$-0.77 \ V$$Co$$-1.97 \ V$Since $Cr$ has the highest oxidation potential $(+0.41 \ V)$,the change in oxidation state from $+2$ to $+3$ is easiest for $Cr$.

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

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