In an acidic medium, the reduction half-reaction of permanganate ion is given by: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ This shows that 1 mole of $MnO_4^-$ requires 5 moles of electrons for its complete reduction to $Mn^{2+}$. The number of moles of electrons provided is calculated by dividing the given number of electrons by Avogadro's number ($N_A = 6.022 \times 10^{23}$): Number of moles of electrons = $\frac{4.517 \times 10^{24}}{6.022 \times 10^{23}} = 7.5$ moles Since 5 moles of electrons reduce 1 mole of $MnO_4^-$, the number of moles of $MnO_4^-$ reduced by 7.5 moles of electrons will be: $\frac{1}{5} \times 7.5 = 1.5$ moles.