Question

Number of moles of MnO4 required to oxidize one mole of ferrous oxalate completely in acidic medium will be

Answer 1

• The reduction half-reaction is given as: $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{+2}+4H_{2}O$
• The oxidation half-reaction is given as:

  $C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$

  $F{e}^{2+}\to F{e}^{3+}+1e^{-}$

  $Hence,thenethalfoxidationreactionis:$

  $C_2{O}_4^{2-}+F{e}^{2+}\to 2C{O}_2+F{e}^{3+}+3e^{-}$

• Hence the net Redox reaction can be written as: $3Mn{O}_4^{-}+24H^{+}+5C_2{O}_4^{2-}+5F{e}^{2+}⟶10C{O}_2+5F{e}^{3+}+3M{n}^{2+}+12H_{2}O$

• From the equation, it is clear that 3 moles of $Mn{O}_4^{-}$ require 5 moles of $Fe{C}_2{O}_4$

• Hence, 1 mole of $Fe{C}_2{O}_4$ will require $\frac{3}{5}$ moles of $Mn{O}_4^{-}$