Balancing redox reactions carol brown saint mary s hall
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Balancing Redox Reactions Carol Brown Saint Mary’s Hall. Assigning oxidation states. An oxidation number is the charge an atom would have if all of its bonds were ionic. Oxidation Number Rules. 1. The oxidation state of a free element is zero.

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Balancing Redox Reactions Carol Brown Saint Mary’s Hall

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Balancing redox reactions carol brown saint mary s hall

Balancing Redox ReactionsCarol BrownSaint Mary’s Hall

  • Assigning oxidation states.

    • An oxidation number is the charge an atom would have if all of its bonds were ionic.


Oxidation number rules

Oxidation Number Rules

  • 1. The oxidation state of a free element is zero.

  • 2. The oxidation state of a monatomic ion is equal to its charge.

  • 3. The algebraic sum of the oxidation states of all of the atoms in a compound is zero.

  • 4. The oxidation state of alkali metals in compounds is 1+; the oxidation state of alkaline earth metals in compounds is 2+.

  • 5. In compounds the more electronegative element is always negative.

  • 6. In compounds, hydrogen is generally 1+ unless it is with an element which is less electronegative than it is.

  • 7. In compounds, oxygen is usually 2-, unless it is a peroxide, in which case it is 1-, or a superoxide when it would have an oxidation state of -1/2.


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o1

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

  • Determine the oxidation state for every atom in the reaction.

  • Determine which atoms are oxidized and which atoms are reduced.


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o2

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

  • Write a half reaction for the reduction reaction.

    • Balance all atoms.

      • Use water to balance oxygens

      • Use H+ to balance hydrogens.

        8 H+ + MnO4- ----> Mn2+ + 4 H2O


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o3

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

  • Balance the charge by adding electrons.

    8 H+ + MnO4- + 5e- ----> Mn2+ + 4 H2O


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o4

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

  • Repeat the process with the oxidation half -reaction.

    2Fe2+ ---> 2Fe3+ + 2e-


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o5

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

  • Multiply the half-reactions so that the same number of electrons are transferred in each one.

    8 H+ + MnO4- + 5e- ---> Mn2+ + 4 H2O

    2Fe2+ ---> 2Fe3+ + 2e-


Feso 4 kmno 4 h 2 so 4 fe 2 so 4 3 k 2 so 4 mnso 4 h 2 o6

FeSO4 + KMnO4 + H2SO4 --> Fe2(SO4)3 + K2SO4 + MnSO4 + H2O

  • Add the two half-reactions together. This will give you the net ionic equation.

    16 H+ + 2MnO4- + 10e- ---> 2 Mn2+ + 8 H2O

    10Fe2+ ---> 10Fe3+ + 10e-

    16 H+ + 2MnO4- + 10Fe2+ ---> 10Fe3+ 2 Mn2+ + 8 H2O


10feso 4 2kmno 4 8h 2 so 4 5fe 2 so 4 3 k 2 so 4 2mnso 4 8h 2 o

10FeSO4 + 2KMnO4 + 8H2SO4 --> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

  • Place coefficients into skeleton equation.

  • Check spectator ions.

  • Count all atoms for recheck counting hydrogens next to last and oxygens last.


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