If the number of moles of Fe^{2+} ions oxidiz | vedclass

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(B) The oxidation of $Fe^{2+}$ to $Fe^{3+}$ involves the loss of $1$ electron: $Fe^{2+} \rightarrow Fe^{3+} + e^-$.For $MnO_4^-$ in acidic medium: $Mn

Vedclass · Accepted Answer

$\frac{6x}{5}$

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(B) The oxidation of $Fe^{2+}$ to $Fe^{3+}$ involves the loss of $1$ electron: $Fe^{2+} \rightarrow Fe^{3+} + e^-$.For $MnO_4^-$ in acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Thus, $1$ mole of $MnO_4^-$ oxidizes $5$ moles of $Fe^{2+}$. Given $x = 5$.For $Cr_2O_7^{2-}$ in acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$. Thus, $1$ mole of $Cr_2O_7^{2-}$ oxidizes $6$ moles of $Fe^{2+}$.Since $x = 5$, the number of moles of $Fe^{2+}$ oxidized by $Cr_2O_7^{2-}$ is $6$. Expressing $6$ in terms of $x$: $6 = \frac{6x}{5}$.

If the number of moles of $Fe^{2+}$ ions oxidized by one mole of acidified $MnO_4^{-}$ is $x$, the number of moles of $Fe^{2+}$ ions oxidized by one mole of acidified $Cr_2O_7^{2-}$ is:

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