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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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?

Similar Questions

What is the equivalent weight of $SnCl_{2}$ in the following reaction: $SnCl_{2} + Cl_{2} \longrightarrow SnCl_{4}$?

When copper is heated with concentrated $HNO_3$,it produces:

$H^{+} + Cr_2O_7^{2-} + SO_3^{2-} \to Cr^{3+} + SO_4^{2-} + H_2O$. In the balanced reaction,the coefficients of $H^{+}$ and $SO_4^{2-}$ will be respectively:

For the redox reaction $MnO_4^- + C_2O_4^{2-} + H^+ \longrightarrow Mn^{2+} + CO_2 + H_2O$,the correct coefficients of the reactants for the balanced equation are $MnO_4^-$ - $C_2O_4^{2-}$ - $H^+$

Equivalent mass of the oxidizing agent in the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module