Balanced Equation : MnO4- + 8 H+ + 5 Fe2+ - - > Mn2+ + 4 H2O + 5 Fe3+ ( 5 e- Transferred in Redox Reaction )
Mole of Fe2+ = Mole of MnO4- x 5
First Titration = = = >>> Original Fe2+ are oxidized to Fe3+
Zinc Reduction = = = >>> Fe3+ are reduced to Fe2+
Second Titration = = = >>> Total Fe ions ( Reduced to Fe2+ ) are oxidized to Fe3+
Mole of Fe2+ in 25 mL Solution = Mole of MnO4- x 5 = 0.0200 Mole / L x 23 mL x ( 1 L / 1000 mL ) x 5 = 0.0023 Mole
Mole of Total Fe in 25 mL Solution = Mole of MnO4- x 5 = 0.0200 Mole / L x 40 mL x ( 1 L / 1000 mL ) x 5 = 0.0040 Mole
Mole of Total Fe in 25 mL Solution = Mole of MnO4- x 5 = 0.0200 Mole / L x 40 mL x ( 1 L / 1000 mL ) x 5 = 0.0040 Mole
Mole of Fe3+ in 25 mL Solution = Mole of Total Fe in 25 mL Solution - Mole of Fe2+ in 25 mL Solution = 0.0040 Mole - 0.0023 Mole = 0.0017 Mole
Concentration of Fe2+ = 0.0023 Mole / 25 mL x ( 1000 mL / L ) = 0.092 Mole / L = 0.092 M
Concentration of Fe3+ = 0.0017 Mole / 25 mL x ( 1000 mL / L ) = 0.068 Mole / L = 0.068 M
Balance a Net Ionic Equation ; Then modify to Full Equation
Steps :
Identify " Reduced Element " ( or Oxidizing Agent ) _ and _ " Oxidized Element " ( or Reducing Agent ) >> Write the 2 Half Reactions ( Reduction Side _ and _ Oxidation Side ) >> Balancing the " Transferred Electron " >> Balanced Redox Part >> Balancing the Number of Atom of All Elements _ and _ Net Charge on Both Sides of equation ( without changing the ratio of balanced Redox Part >> Balanced Net Ionic Equation >> Add " Counter Cation(s) _ and _ Counter Anion(s) " >> Final Balancing ( If Needed ) >> Balanced Full Redox Equation
" Manganese " in MnO4- is reduced from Oxidation No. = ( +7 ) to Oxidation No. = ( +2 ) in Mn++
[ Oxidation Number of Atoms in MnO4- : Mn = ( +7 ) ; O = ( -2 ) ; Sum = ( +7 -8 ) = ( -1 ) = Charge of MnO4- ]
[ Oxidation Number of " Manganese " in Mn++ = ( +2 ) = Charge of Mn++ ]
or
" Manganese " in KMnO4 is reduced from Oxidation No. = ( +7 ) to Oxidation No. = ( +2 ) in MnSO4
[ Oxidation Number of Atoms in KMnO4 : K = ( +1 ) ; Mn = ( +7 ) ; O = ( -2 ) ; Sum = ( +1 ) + ( +7 ) + ( -2 x 4 ) = ( 0 ) ]
[ Oxidation Number of Atoms in MnSO4 : Mn = ( +2 ) ; S = ( +6 ) ; O = ( -2 ) ; Sum = ( +2 ) + ( +6 ) + ( -2 x 4 ) = 0 ]
" Iron " in FeSO4 is oxidized from Oxidation No. = ( +2 ) to Oxidation No. = ( +3 ) in Fe2(SO4)3
[ Oxidation Number of Atoms in FeSO4 : Fe = ( +2 ) ; S = ( +6 ) ; O = ( -2 ) ; Sum = ( +2 ) + ( +6 ) + ( -2 x 4 ) = 0 ]
[ Oxidation Number of Atoms in Fe2(SO4)3 : Fe = ( +3 ) ; S = ( +6 ) ; O = ( -2 ) ; Sum = ( +3 x 2 ) + ( +6 x 3 ) + ( -2 x 4 x 3 ) = 0 ]
Balancing the equation :
Reduction Side :
MnO4- + 5 e- ----> [ Mn++ ] __ ( R Side )
( Focus on MnO4- and Mn++ only ; Ignore the fact that atom and charge may not be balanced )
Oxidation Side :
[ Fe++ ] ----> [ Fe+++ ] + e- __ ( O Side )
( Focus on Fe++ and Fe+++ only ; ignore the fact that atom and charge may not be balanced )
1 x ( R Side ) + 5 x ( O Side ) [[ to compensate electron ]]
[ MnO4- ] + 5 [ Fe++ ] ----> [ Mn++ ] + 5 [ Fe+++ ]
The Redox Part ( Electron Transfer ) already balanced
General Equation Balancing : ( Atoms and Positive / Negative Charges )
Add [ H+ ] on left side - - ( To compensate the positive charges on right side )
and
Add [ H2O ] on right side - - ( To enable balancing of H and O on both sides )
Balance [ H+ ] on Left Side -- vs -- H2O on Right Side
{ but do not change the ratio of [ MnO4- , Fe++ , Mn++ , Fe+++ ] ; since they are already balanced at Redox Part }
( at this step ; we may have to multiply the " Balanced Redox Part " with 2 or 3 or 4 or 5 . . . )