Question

How many moles of MnO4' are required to oxldise 2
moles of Fe^+2 Fe^+3 in acidic medium?​

Answer 1

Answer: 0.4 mol $MnO_{4} ^{-}$

Explanation: If the reaction takes place in an acidic medium, the acid is donating H+ ions, so H+ is one of the reactants in the chemical equation. Therefore, the equation to balance is:

__$MnO_{4} ^{-}$ + __$Fe^{+2}$ + __$H^{+}$ --> __$Mn^{+2}$ + __$Fe^{+3}$ + __$H_{2} O$

Find the oxidation number of each element:

Reactants:

• $Mn$: +7
• $O$: -2
• $Fe^{+2}$: +2
• $H$: +1

Products:

• $Mn^{+2}$: +2
• $Fe^{+3}$: +3
• $H$: +1
• $O$: -2

Since manganese (Mn) becomes less positive, it gains electrons and is reduced. We can write the following reduction half reaction:

$Mn^{+7}$ + 5e- --> $Mn^{+2}$

Because iron (Fe) gains a greater positive charge, it loses electrons and is oxidized. We can write the following oxidation half reaction:

$Fe^{+2}$ --> $Fe^{+3}$ + e-

To balance the half reactions, there must be 5e- on the left side of the oxidation half reaction. Therefore, the oxidation half reaction must be:

$5Fe^{+2}$ --> $5Fe^{+3}$ + 5e-

Put these new coefficients on the initial chemical equation to check our work:

$MnO_{4} ^{-}$ + $5Fe^{+2}$ + $8H^{+}$ --> $Mn^{+2}$ + $5Fe^{+3}$ + $4H_{2} O$

For every 5 moles of $Fe^{+2}$ 1 mole of $MnO_{4}^{-}$is required to oxidize it. Use the ratio to solve for the moles of $MnO_{4} ^{-}$ required.

$(2molFe^{+2} )(\frac{1molMnO_{4}^{-} }{5molFe^{+2} } )$=$0.4 mol MnO_{4} ^{-}$

Excellent question! It helped me review old concepts and make connections. I hope my response is helpful. :)