How would you account for the increasing oxidising power in the series $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The oxidising power of these oxoanions depends on the stability of the lower oxidation states to which they are reduced.
As we move from $VO_2^+$ to $MnO_4^-$,the stability of the reduced species increases,which makes the reduction of the parent oxoanion more favorable.
Specifically,the reduction potentials increase in the order $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$,reflecting the increasing ease with which these species accept electrons.

Explore More

Similar Questions

The highest Mn fluoride is $MnF_4$ whereas the highest oxide is $Mn_2O_7$, because . . . . . . .

$(a)$ Transition metals can act as catalysts because they can change their oxidation state. How does $Fe(III)$ catalyze the reaction between iodide and persulphate ions?
$(b)$ Mention any three processes where transition metals act as catalysts.

Difficult
View Solution

Which is a stronger reducing agent,$Cr^{2+}$ or $Fe^{2+}$,and why?

Ferrous sulphate on heating gives

Which of the following is the correct order of acidic character?

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