1. Copyright Sautter 2003

2. REVIEW OF ELECTROCHEMISTRY All electrochemical reactions involve oxidation and reduction. Oxidation means the loss of electrons (it does not always involve oxygen). Reduction means the gain of electrons (gaining of negatives, that is electrons, reduces the oxidation number of an atom. We will discuss oxidation number latter in this program). Whenever electrons are lost by one substance they must be gained by another. The substance that loses electrons is referred to as a reducing agent (it lets another substance be reduced, that is, gain the electrons). The substance that gains electrons is referred to as an oxidizing agent (it lets another substance be oxidized, that is, lose the electrons). These terms are important in electrochemistry!!

3. REVIEW OF OXIDATION & REDUCTION OXIDATION: Zn (s)  Zn +2 (aq) + 2 e - Metallic zinc is oxidized to zinc ion. Metallic zinc is serving as a reducing agent.(electron loser) REDUCTION: Cu +2 (aq) + 2 e -  Cu (s) Copper ion is reduced to copper metal. Copper ion is serving as an oxidizing agent (electron gainer) In the overall reaction two electrons are transferred from the zinc metal to the copper ion. Zn (s) + Cu +2 (aq)  Zn +2 (aq) + Cu (s)

4. REVIEW OF OXIDATION NUMBERS In the previous reactions the loss and gain of electrons was obvious. Zn (s) must lose two negative electrons in order to form a Zn +2 (aq) ion. And Cu +2 (aq) must gain two negative electrons to form a Cu (s) atom. What about more complicated reactions? How do we tell which substance loses and which gains electrons and how many? For example: H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O Here we need oxidation numbers!

5. OXIDATION NUMBERS Before discussing the mechanics of oxidation numbers it is important to realize that: (1) oxidation numbers have no physical significance. They are merely a way to tell which substances gain or lose electrons and how many. (2) oxidation numbers are assigned to atoms never molecules. Molecules contain atoms with oxidation numbers but they themselves cannot be assigned oxidation numbers. (3) there a several rules used to assign oxidation numbers. They must be observed carefully.

6. OXIDATION NUMBERS RULES (1) Elements have oxidation numbers of zero. For example Cu has an oxidation number of zero. In Cl 2 each atom of chlorine has an oxidation number of zero. (2) The oxidation number of a monatomic ion (consisting of one atom) is the charge on the ion. In Cu +2 the oxidation number is +2. In Cl - the oxidation number is –1. (3) The oxidation number of combined oxygen is always –2 except in peroxides such as hydrogen peroxide, H 2 O 2. In CO 2 each oxygen atom has an oxidation number of –2. In SO 3 each oxygen is –2.

7. OXIDATION NUMBERS RULES (4) Combined hydrogen is always +1 except in hydrides (metal atoms and hydrogen like NaH). In water H 2 O each hydrogen has an oxidation number of +1. In methane, CH 4, each hydrogen is +1. (5) The sum of all of the oxidation numbers of the atoms in a molecule equal its charge. If no charge is shown the sum must equal zero. This allows us to determine the oxidation numbers of elements not discussed in the first four oxidation number rules! Oxidation numbers of other elemental groups: Elements of Column I in a compound = +1 Elements of Column II in a compound = +2 It is important to realize that the oxidation number of an atom can change when it combines differently during reaction. This is the basis for deciding how electrons are transferred during a chemical reaction!

9. FINDING UNKNOWN OXIDATION NUMBERS Problem: What is the oxidation number of Mn in MnO 2, the permanganate ion? Solution: Mn is unknown Each O is –2 (rule 3) All of the oxidation numbers must add up to zero (rule 5) Mn + 2(-2) = 0, Mn = +4 in permanganate Problem: What is the oxidation number of C in CN -, cyanide ion? Solution: The oxidation number rules will not allow us to do this since there is no rule for N ! However we can find the number of electrons transferred in a half reaction involving cyanide using the ion electron method.

10. USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS Using oxidation numbers we will find the oxidizing agent, the reducing agent, the number of electrons lost and gained, the oxidation half-reaction, the reduction half-reaction and the final balanced equation for: H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O Step I – Find the oxidation number of each atom. H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O +1 (+7, -2) -1 +2 0 (+1, -2) Step II – Determine which oxidation numbers change MnO 4 -  Mn +2 2 Cl -  Cl 2 +7  +2 2(-1)  0 5 electrons gained 2 electrons lost (MnO 4 - is the oxidizing agent. It gained electrons from Cl - ) (Cl - is the reducing agent. It lost electrons to MnO 4 - ) If oxidization # s decrease, Reduction occurs If oxidization # s increase, Oxidation occurs

11. USING OXIDATION NUMBERS TO DETERMINE ELECTRON TRANSFERS Step III - Electrons lost by a reducing agent must always equal electrons gained by the oxidizing agent! Therefore: 2 (MnO 4 - + 5e -  Mn +2 ) 2MnO 4 - + 10e -  2Mn +2 (reduction half-reaction*) And 5(2Cl -  Cl 2 + 2e - ) 10 Cl -  5Cl 2 + 10e - (oxidation half-reaction) Step IV -Completing the first half-reaction with H + ions and H 2 O molecules: 16 H + + 2MnO 4 - + 10e -  2Mn +2 + 8 H 2 O Step V - Adding the oxidation half-reaction and reduction half- reaction the 10 electrons gained and lost cancel to give the overall reaction: 16 H + + 2MnO 4 - + 10 Cl -  2 Mn +2 + 5 Cl 2 + 8 H 2 O * Half-reaction refers to the reaction showing either the electron gain (reduction) or the electron loss (oxidation) step of the reaction.

12. USING THE ION ELECTRON METHOD Find the balanced half reaction for: H + + MnO 4 - + Cl -  Mn +2 + Cl 2 + H 2 O Step I – we see that when MnO 4 - converts to Mn +2 the H and O atoms are not balanced so we add water to the right and H ions to the left 8H + + MnO 4 -  Mn +2 + 4H 2 O Step II - using the concept that total charges on the left must balance the total charges on the right, we have 8 + on the left and 2 + on the right. Adding 5 electrons on the left (negatives) will balance the charges. 8H + + MnO 4 - + 5e -  Mn +2 + 4H 2 O When Cl - converts to Cl 2 2Cl -  Cl 2 there are 2 negatives on the left and no charges on the right Adding 2 electrons on the right will balance the charge 2Cl -  Cl 2 + 2e -

13. REDOX REACTIONS IN BASIC SOLUTION The reaction we have balanced so far have occurred in acidic solutions (those containing H ions and water. Now we will balance basic solution reactions ( those containing OH - and water). MnO 4 - + CN -  MnO 2 + CNO - MnO 4 -  MnO 2 balance O with H ion and water as before 4H + + 3e - + MnO 4 -  MnO 2 + 2 H 2 O add 3 electrons to the left Now since the solution is basic only OH - and water can be in the equation, so we will add 4 OH - ions to each side to neutralize the H + 4OH - + 4H + + 3e - + MnO 4 -  MnO 2 + 2 H 2 O+ 4OH – The 4OH - + 4H + gives 4 H 2 O on the right 4 H 2 O + 3e - + MnO 4 -  MnO 2 + 2 H 2 O+ 4OH – Cancel 2 waters from each side gives 2H 2 O + 3e - + MnO 4 -  MnO 2 + + 4OH –

14. REDOX REACTIONS IN BASIC SOLUTION Now we will balance the other half reaction. We will again use the ion electron method since oxidation numbers cannot be assigned CN -  CNO - balance O with H ion and water as before H 2 O + CN -  CNO - + 2H + + 2e - add 2 electrons to the right Now, again since the solution is basic only OH - and water can be in the equation, so we will add 2OH - ions to each side to neutralize the H + 2OH - + H 2 O + CN -  CNO - + 2H + + 2e - + 2 OH – The 2OH - + 2H + gives 2H 2 O on the right 2OH - + H 2 O + CN -  CNO - + 2H 2 O + 2e - Cancel 1 water from each side gives 2OH - + CN -  CNO - + H 2 O + 2e -

15. REDOX REACTIONS IN BASIC SOLUTION We now have: 2H 2 O + 3e - + MnO 4 -  MnO 2 + 4OH – 2OH - + CN -  CNO - + H 2 O + 2e - Multiplying both half reactions to obtain 6e - and 6e - gained 2 (2H 2 O + 3e - + MnO 4 -  MnO 2 + 4OH – ) 3 (2OH - + CN -  CNO - + H 2 O + 2e - ) 4H 2 O + 6e - + 2MnO 4 -  2MnO 2 + 8OH – 6OH - + 3CN -  3CNO - + 3H 2 O + 6e Adding half reactions we get H 2 O + 2MnO 4 - + 3CN  3CNO - + 2MnO 2 + 2OH –

16. DISPROPORTIONATION REACTIONS In disproportionation reactions a single reactant serves as both the oxidizing and reducing agent. Cl 2 + OH − → Cl − + ClO 3 − + H 2 O In this reaction some chlorine atoms gain electrons (are reduced) to become Cl − and others lose electrons (are oxidized) forming ClO 3 − Cl 2 + 2e - → 2Cl − 6 H 2 O + Cl 2 → 2ClO 3 − + 12H + + 10e - adding 12OH − gives 12OH − + 6H 2 O + Cl 2 → 2ClO 3 − + 12H + + 12OH − + 10e - which reduces to 12OH − + Cl 2 → 2ClO 3 − + 6H 2 O + 10e - Multiplying to get electrons equal 5(Cl 2 + 2e - → 2Cl − ) 12OH − + Cl 2 → 2ClO 3 − + 6H 2 O + 10e -

17. DISPROPORTIONATION REACTIONS 5Cl 2 + 10e - → 10Cl − 12OH − + Cl 2 → 2ClO 3 − + 6H 2 O + 10e - adding the half reactions gives 6Cl 2 + 12OH − → 10Cl − + 2ClO 3 − + 6H 2 O Dividing by 2 gives 3Cl 2 + 6OH − → 5Cl − + ClO 3 − + 3H 2 O