If stoichiometric quantities of a KMnO_4 and | vedclass

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(B) In an acidic medium,both $KMnO_4$ and $K_2Cr_2O_7$ act as oxidizing agents to convert $Fe^{2+}$ to $Fe^{3+}$.$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn

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(B) In an acidic medium,both $KMnO_4$ and $K_2Cr_2O_7$ act as oxidizing agents to convert $Fe^{2+}$ to $Fe^{3+}$.$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ (n-factor = $5$)$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ (n-factor = $6$)Since the question specifies that stoichiometric quantities are added,the amount of $Fe^{2+}$ oxidized depends on the number of moles of electrons provided by each oxidant.However,$KMnO_4$ is a stronger oxidizing agent than $K_2Cr_2O_7$ due to its higher reduction potential in acidic medium ($E^\circ_{MnO_4^-/Mn^{2+}} = 1.51 \ V$ vs $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$).Therefore,$KMnO_4$ is more effective at oxidizing $Fe^{2+}$.

If stoichiometric quantities of a $KMnO_4$ and $K_2Cr_2O_7$ mixture are added for the oxidation of $Fe^{2+}$ to $Fe^{3+}$ in an acidic medium,then $Fe^{2+}$ will be oxidized:

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