Chapter 8 Redox Reactions

Redox Reactions Oxidation number Balancing redox equation Voltaic cells Standard voltages

Oxidation number 1.The concept of Oxidation number It is used to refer to the charge an atom would have if the bonding electrons were assigned arbitrarily to the more electronegative element. For example: HCl

H : Cl

Oxidation number of Cl is -1, Oxidation number of H is +1,

Note: The Oxidation number of an atom in a molecule is an artificial concept, unlike the charge of an ion, Oxidation number can not be determined experimentally. For ionic compounds, the oxidation number is the charge of ion. For example: NaCl (Na:+1, Cl:-1)

How to determine the oxidation number of elements?

2. Rules for Assigning oxidation numbers a. The oxidation number of an element in an elementary substance is 0. e.g. O2, H2, N2,P4 b. The oxidation number of an element in a monatomic ion is equal to the charge of that ion. e.g. Na+, Cl-, Fe3+ , O2- , Mg2+

c. Certain elements have the same oxidation number in all or almost all their compounds. e.g. Group 1 metals always exist as +1 ions in their compounds, so the oxidation number is +1. Group 2 : +2. Fluorine, the most electronegative of all elements, has an oxidation number of -1 in all of its compounds.

O: -2, but, in H2O2 O: -1 H have different oxidation number in different compounds in H2O (H:+1), in NaH (H:-1) d. The sum of the oxidation numbers of all the atoms in a neutral species is 0; in an ion, it is equal to the charge of that ion. e.g. NaCl, Fe2O3, Mg(OH)2, KCl, OH-, CN-, NH4+

3. Oxidation and Reduction Oxidation is defined as an increase in oxidation number, Reduction is defined as a decrease in oxidation number. e.g. 2Na + Cl2 = 2NaCl

Na oxidized(0--+1)

Cl reduced(0---1) 4Fe + 3O2 = 2Fe2O3 Fe oxidized(0--+3) O reduced(0---2)

An element which loses electrons must increase in oxidation number, the gain of electrons always results in a decrease in oxidation number. 4. Oxidizing and Reducing Agents In a redox reaction, we usually have at lease two reactants, one of these is referred to as the oxidizing agent, another as the reducing agent.

An oxidizing agent brings about the oxidation of another species. To do this , it must take electrons away from that species and the oxidation number decrease, hence the oxidizing agent is itself reduced in the reaction.

An reducing agent brings about the reduction of another species. To do this , it must give up electrons to that species and the oxidation number increase, hence the reducing agent is itself oxidized in the reaction.

e.g. Na + Cl = NaCl Here Na is oxidized, Na is reducing agent. Cl is reduced, Cl is oxidizing agent. 4Fe + 3O2 = 2Fe2O3 Here Fe is oxidized, Fe is reducing agent. O2 is reduced, O2 is oxidizing agent.

Balancing redox equation The half-equation method The steps: 1) Breaking the overall equation down into two half-equations, one is oxidation, the other is a reduction; 2) Balancing the two half-equations separately; 3) Combining them in such a way so as to obtain an overall equation.

e.g. Reaction between MnO4- and Cl- (acidic solution) MnO4- + Cl- + H+ = Mn2+ + Cl2 +H2O 1)Recognize which species undergo oxidation and reduction; Oxidation: Cl- -- Cl2 Reduction: MnO4- + H+ -- Mn2+ +H2O

(a) (b)

2)Balance the two half-equations with respect to mass;

2Cl- -- Cl2 +2e MnO4- + 8H+ +5e -- Mn2+ + 4H2O 3)the two half-equations are combined so as to eliminate electrons from the final equation. To do this, a least common multiple of 10 electrons is used, thus a is multiplied by 5 and b by 2, producing 10e on each side: 5× a: 10Cl- -- 5Cl2 +10e 2× b:

2MnO4- + 16H+ +10e -- 2Mn2+ + 8H2O 2MnO4- + 16H+ + 10Cl- = 2Mn2+ + 5Cl2 + 8H2O

e.g. Reaction between MnO4- and I- (basic solution) MnO4- + I- -- MnO2 + l2 1) Oxidation: l- -- l2 Reduction: MnO4- -- MnO2

(a) (b)

2) In basic solution: more OH-, H2O 2l- -- l2 +2e MnO4- + 2H2O +3e -- MnO2 + 4OH3) 3a+2b 3a

6l- -- 3l2 +6e

2b

2MnO4- + 4H2O +6e -- 2MnO2 + 8OH2MnO - + 4H O + 6l- = 2MnO + 8OH- + 3l

Voltaic cells Redox reactions: Spontaneous

2Na + Cl2 = 2NaCl

Nonspontaneous

2NaCl = 2Na + Cl2

a Voltaic cells : a kind of electrical cell, a spontaneous redox reaction takes place , and it produces electrical energy. Like dry cells used in flashlights and calculators The lead storage battery in an automobile

the Zn electrode

the Cu electrode

salt

How a voltaic cell works? The Zn-Cu cell A piece of zinc is added to a Cu2+ solution, the following redox reaction takes place: Zn + Cu2+ = Zn2+ + Cu We can see that electrons transfer from Zn atom to Cu2+ ion.

To design an electrical cell 1)At the Zinc anode, electrons are produced by the oxidation half-reaction. This electrode is marked as the negative pole of the cell. 2)Electrons move through the external circuit, this pat of the circuit may be a simple resistance wire, a light bulb or some other divice that consumes electrical energy.

3)Electrons pass from the external circuit to the copper cathode where they are used in the reduction of Cu2+ ions in the surrounding solution. The copper electrode is considered to be the positive pole of the cell.

4)To complete the circuit, ions must move through the aqueous solutions in the cell. Movement of ions occurs through a salt bridge, which consist of an inverted U-tube, the tube is filled with a solution of salt which takes no part in the electrode reactions.

In the notation Left: the anode reaction (oxidation) Zn -- Zn2+ + 2e Middle: “II” the salt bridge Right: the cathode reaction (reduction) Cu2+ + 2e – Cu So Zn-Cu cell can express the following: (-)ZnIZn2+ IICu2+ ICu(+)

the Zn electrode

the Cu electrode salt

e.g.1

Ni + Cu2+ = Ni2+ + Cu (-)NiINi2+ IICu2+ ICu(+) Zn + 2H+ = Zn2+ + H2

e.g.2

Anode: Zn -- Zn2+ + 2e Cathode: 2H++2e -- H2 (-)ZnIZn2+ IIPt(H+)IH2(+)(Pt: an inert platinum cathode)