1 mole of Fe^{2+} ions are oxidised to Fe^{3+ | vedclass

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Question

(A) The oxidation half-reaction is: $Fe^{2+} \rightarrow Fe^{3+} + e^-$. The reduction half-reaction for $KMnO_4$ in acidic medium is: $MnO_4^- + 8H^+

Vedclass · Accepted Answer

$1/5$ moles of $KMnO_4$

Vedclass · Answer

(A) The oxidation half-reaction is: $Fe^{2+} \rightarrow Fe^{3+} + e^-$. The reduction half-reaction for $KMnO_4$ in acidic medium is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. To balance the electrons,multiply the oxidation half-reaction by $5$: $5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$. Adding the two half-reactions gives the balanced redox equation: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$. From the stoichiometry,$5$ moles of $Fe^{2+}$ require $1$ mole of $KMnO_4$. Therefore,$1$ mole of $Fe^{2+}$ requires $1/5$ moles of $KMnO_4$.

$1$ mole of $Fe^{2+}$ ions are oxidised to $Fe^{3+}$ ions with the help of $KMnO_4$ in acidic medium. How many moles of $KMnO_4$ are required?

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