In acidic medium, one mole each of MnO_4^- an | vedclass

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(D) In acidic medium, the reduction half-reactions are as follows:$1$. For $MnO_4^-$: $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$. Here, $1

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$11$

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(D) In acidic medium, the reduction half-reactions are as follows:$1$. For $MnO_4^-$: $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$. Here, $1 	ext{ mole of } MnO_4^- 	ext{ requires } 5 	ext{ moles of } e^-$.$2$. For $Cr_2O_7^{2-}$: $Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$. Here, $1 	ext{ mole of } Cr_2O_7^{2-} 	ext{ requires } 6 	ext{ moles of } e^-$.Ferrous ions $(Fe^{2+})$ are oxidized to ferric ions $(Fe^{3+})$: $Fe^{2+} ightarrow Fe^{3+} + e^-$.Thus, $1 	ext{ mole of } Fe^{2+} 	ext{ provides } 1 	ext{ mole of } e^-$.Therefore, $x = 5 	ext{ moles}$ and $y = 6 	ext{ moles}$.The sum $x + y = 5 + 6 = 11$.

In acidic medium, one mole each of $MnO_4^-$ and $Cr_2O_7^{2-}$ is reduced by $x$ and $y$ moles of ferrous ions $(Fe^{2+})$ respectively. The sum of $x$ and $y$ is:

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