For $M^{2+} / M$ and $M^{3+} / M^{2+}$ systems,the ${E^{\Theta}}$ values for some metals are as follows:
$Cr^{2+} / Cr : -0.9 \ V$
$Cr^{3+} / Cr^{2+} : -0.4 \ V$
$Mn^{2+} / Mn : -1.2 \ V$
$Mn^{3+} / Mn^{2+} : +1.5 \ V$
$Fe^{2+} / Fe : -0.4 \ V$
$Fe^{3+} / Fe^{2+} : +0.8 \ V$
Use this data to comment upon:
$(i)$ The stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}$ and
$(ii)$ The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(i)$ The ${E^{\Theta}}$ value for $Fe^{3+} / Fe^{2+}$ is $+0.8 \ V$,which is higher than that for $Cr^{3+} / Cr^{2+}$ $(-0.4 \ V)$ and lower than that for $Mn^{3+} / Mn^{2+}$ $(+1.5 \ V)$.
Since a higher reduction potential indicates easier reduction,$Mn^{3+}$ is most easily reduced to $Mn^{2+}$,followed by $Fe^{3+}$,and $Cr^{3+}$ is the least easily reduced.
Therefore,the stability of these ions in acid solution follows the order: $Mn^{3+} < Fe^{3+} < Cr^{3+}$.
$(ii)$ The reduction potentials for the $M^{2+} / M$ pairs are: $Mn^{2+} / Mn$ $(-1.2 \ V)$,$Cr^{2+} / Cr$ $(-0.9 \ V)$,and $Fe^{2+} / Fe$ $(-0.4 \ V)$.
Lower reduction potential indicates easier oxidation of the metal to its $M^{2+}$ ion.
Since the reduction potential values are in the order $Mn^{2+} / Mn < Cr^{2+} / Cr < Fe^{2+} / Fe$,the ease of oxidation follows the order: $Fe < Cr < Mn$.

Explore More

Similar Questions

Which ion has the highest ionisation enthalpy?

Which of the following compounds is colourless?

Difficult
View Solution

Which of the metal ion will have the highest number of unpaired electrons?

Transition metals are related to which block?

Which pair of ions is colourless?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo