Lecture 7: Types of Chemical Reactions

  Lecture 7: Types of Chemical Reactions ƒ

  Oxidation-Reduction Reactions

  1 Oxidation-Reduction (Redox) Reactions Reactions involving the transfer of electrons. eg: in the formation of an ionic compound

2 Na + Cl →

  2NaCl (s) 2(g) (s)

  Half reactions: _{- + -}

  

2Na(s) + 2e electrons lost: oxidation

→ 2Na ^- Cl (g) + 2e → 2Cl electrons gained: reduction ₂ # of electrons lost = # of electrons gained

  In this case 2 electrons are transferred

• O
• H

  2 O (g) How can we tell if electrons have been transferred?

  , Br ^- …)

  group IA: +1 (Na ⁺ , K ⁺ , …) group IIA: +2 (Ca ²⁺ , Mg ²⁺ , …) group VII -1 (F

• ^- , Cl ^-

  

2. The oxidation state of a monatomic ion is the same as its

charge:

  eg: Ag, H ₂ , P ₄ , Cl ₂ , He

  zero :

  Keeping Track of Electrons: Oxidation States Assigning Formal Oxidation States (oxidation numbers)

  2(g)

  2(g) → CO

  2 H 6(g)

  Another example of an oxidation reduction : C

  

In reactions between metals and non metals, generally,

metals are oxidized and nonmetals are reduced .

  Electrons lost by one species must be gained by another.

  3 Redox Oxidation and reduction always occur together.

• system for keeping track of electrons
• assign “formal” charges to the atoms in a compound/molecule

1. The oxidation number of an atom in its elemental state is

  Assigning Oxidation States

  3. Oxygen is (usually) assigned an oxidation state -2 eg: in its covalent compounds: H O, CO

  2

  2

exceptions: -can be positive (bonded to fluorine): OF

  2 2-

• 1 in peroxides (contain O ) :

  2 H O , Na O

  2

  2

  2

  2

• -0.5 in superoxides (contain O )

  2 NaO

  2

4. Hydrogen is usually assigned an oxidation state of +1 eg: in its covalent compounds: H O, NH , CH ..

  2

  3

  4 exceptions: in metal hydrides (H is -1 ):

  NaH, CaH

  2 ₅ Assigning Oxidation States

5. For atoms in covalent compounds or polyatomic ions:

• assign formal charge as if the most

  electronegative element controls both electrons in a shared pair.

  Generally: just treat “ as if” these compounds were ionic and assign “charges” and oxidation states based on the previous rules

eg: CO - oxygen is more electronegative so treat it as

  2 a negative ion and assign its oxidation state as –2

C must balance charge of 2 O atoms so assign C +4

6. The sum of the oxidation states must equal:

• zero, for an electrically neutral compound or -the overall charge, for an ionic species.

  7 Assigning Oxidation States

  Examples: assign oxidation states to each atom CO 3 2-

  KMnO

  4 CH

  3 OH S

  2 O

  8 S

  4 O

  6 2- HCN

  Identifying a Reaction as a Redox Reaction ƒ

  Transfer of electrons must take place ƒ

  

Look for changes in oxidation states of atoms

involved in the reaction ƒ

  Something must gain electrons: this is reduction and will show a decrease in oxidation state

  ƒ Something must lose electrons: this is oxidation and will show an increase in oxidation state Oxidizing and Reducing Agents

  → Na + Cl NaCl _{(s) 2(g) (s)}

  Na is oxidized

  ƒ Reducing Agent

• supplies electrons to Cl -is or contains the element being oxidized
• increase in oxidation number from 0 t
• contains the atom that shows an increase in oxidation number

  reducing agent

• Na is a

  ƒ Oxidizing Agent

  Cl is reduced

• is or contains the element
• gains the electrons supplied being reduced by Na - decrease in oxidation nu
• contains the atom that shows a from 0 to –1

  decrease in oxidation number oxidizing agent

• Cl is an
₂ Balancing Oxidation Reduction Reactions

  ƒ often too complex to balance by inspection

  ƒ take advantage of fact that

  # of electrons lost = # of electrons gained Two methods 1) Oxidation state method (for all redox equations, best for non aqueous reactions)

2) Ion-electron or half reaction method (best for

aqueous redox reactions)

  11 Oxidation State Method For Aqueous and Non Aqueous reactions 1. assign oxidation states to all atoms

  

2. determine which element is oxidized and its

increase in oxidation state 3. determine which atom is reduced and its decrease in oxidation state

4. choose coefficients for the species containing

the atom oxidized and the atom reduced such that the total increase in oxidation state equals the total decrease in oxidation state

  

5. balance any other atoms by inspection without

changing the coefficients established in step 4 In class Example

  C ₂ H _6(g) + O _2(g) →

  CO _2(g) + H ₂ O _(g) Oxidation states:

  C ₂ H ₆ : H +1 O ₂ : O 0 CO ₂ : O -2 H ₂ O: H +1 C -3 C +4 O -2

  Changes in oxidation # C -3 → +4 change of +7 , carbon is oxidized

  O 0 → -2 change of –2, oxygen is reduced

  13 In class example As

• Cl
• H
>HCl

• +3 -2 0 +1 -2 +1 +5 -2 +1 -1

  4 O

  6

  2

  2 O → H

  3 AsO

  4 Balancing Redox Reactions ƒ

  Half Reaction Method

• – for redox reactions occurring in aqueous solution
• – write separate equations for the oxidation process and the reduction process
• – balance these individually and then add them together for the overall balanced equation

  15 Reactions Occurring in Acidic Solution ƒ

  Write the individual oxidation and reduction half reactions

  ƒ

  For each half reaction:

• – balance all elements except hydrogen and oxygen
• – balance oxygen by adding H O

  2

• – balance hydrogen by adding H
• – balance the charge by adding electrons (e )

  ƒ

  Where applicable, multiply one or both balanced half reactions by an appropriate integer so that the # of electrons lost = # of electrons gained

  ƒ

  Add the two half reactions and cancel out equivalent species on both sides of the equation

  ƒ

  Check that elements and charges balance

• – Example in acidic solution

  2+ - -

Br + MnO → Br + Mn

  4 2(l)

• 1 +7 -2 0 +2

  oxidation half reaction reduction half reaction ^{- + - - 2+}

  10 Br + 2MnO + 16H +10e → 5 Br _{4 2}

• _{- 2+ - +}
• 10 e + 2Mn O + 8H
₂

  10Br + 2MnO + 16H → 5 Br + 2Mn + 8H O _{4 2 2 17} Reactions Occurring in Basic Solution

  ƒ

  Balance each half reaction following the same steps as for acidic solutions

  ƒ

  Add the equations and eliminate common species on both sides

  ƒ

  To of the equation add OH for H

  each side one each

  in the equation

  ƒ

  Combine H and OH ions, on the same side, to form H O

  2 ƒ

  Where possible, eliminate equal numbers of H O when

  2

  they appear on both sides

  ƒ

  Check that the elements and charges balance

• S
• S

• +7 -2 -2 +2 -2 0

  combine OH ^- and H ⁺ to form water: if possible, eliminate equal numbers of waters from both sides

  add OH ^- :

  → 5S _(s) + 2MnS _(s) + 8H ₂ O

  16H ⁺

  7S ^2- + 2MnO ₄ ^- +

  oxidation half reaction reduction half reaction Basic solution

  7S ^2- + 2MnO ₄ ^- + 16H ⁺ → 5S _(s) + 2MnS _(s) + 8H ₂ O

  5S ^2- + 2MnO ₄ ^- + 2S ^2- + 16H ⁺ + 10 e ^- → 5S + 10e- + 2MnS + 8H ₂ O

  (s)

  (s)

  2- → MnS

  4

  19 Example – in basic solution MnO

• 16OH ^-
• 16OH ^-

  7S ^2- + 2MnO ₄ ^- + 8 H ₂ O → 5S _(s)

• 2MnS _(s)
• 16OH ^-
When given an equation to balance: ƒ

  check to see if the equation represents an oxidation reduction reaction (look for a change in oxidation state)

  ƒ

  if it is not a redox equation then balance by inspection

  ƒ

  if it is a redox equation use one of the redox balancing methods which method? - (these are guidelines only not rules)

• – if all or most of the reactants and products are in molecular form use the oxidation state method
• – if all or most of the reactants and products are (aq) or in ionic form use the half reaction method
• – if it is specified that the reaction occurs in either acidic or basic solution, use the half reaction method
• – if oxygen or hydrogen appear on one side but not on the other, use the half reaction method
₂₁ Redox Titrations eg: Addition of an oxidizing agent of known concentration to a solution of a reducing agent of unknown concentration _{- + - 2+}

  → MnO + 5e Mn + 4H O ₄ + 8H ₂ dark purple colourless oxidizing agent _{2+ - 3+}

  5Fe → 5Fe + 5e ₊ reducing agent _{2+ - 3+ 2+}

  MnO + 8H + 5 Fe → 5 Fe + Mn _{4 2+} + 4H O ₂ As KMnO is added to a solution of Fe , the purple colour _{4 2+} disappears with the formation of Mn . When there is no more ₂₊ Fe left in the solution, then one extra drop of KMnO will not ₄ be reduced and the solution will turn slightly pink.

  This colour change is the equivalence point of the redox titration