# Topic 9 Oxidation and Reduction Introduction Oxidation numbers Redox equations Reactivity Voltaic cells Electrolytic cells.

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Topic 9 Oxidation and Reduction Introduction Oxidation numbers Redox equations Reactivity Voltaic cells Electrolytic cells

9.1 Introduction to oxidation and reduction Definition: Oxidation: Particle that loses electron(s). Na  Na + + e - Reduction: Particle that gains electron(s). Cl 2 + 2e -  2 Cl - or If something is oxidised, something else must be reduced in a chemical reaction.

OILRIG Oxidation Is Loss, Reduction Is Gain … of electrons

Half reactions Consider the redox reaction: 2 Na + Cl 2  2 NaCl This reaction can be split into two “half reactions”: – Oxidation reaction: 2 Na  2 Na + + 2 e - – Reduction reaction: Cl 2 + 2e -  2 Cl -

Oxidation number It’s sometimes difficult to see if it is a redox reaction or if a particle has been oxidised/reduced. Then you have to find out the oxidation numbers or oxidation state of the atoms in the compound. The oxidation number is placed above the atom symbol often in roman figures but not in IB => Use normal figures e.g +2 (Charge: 2+)

Rules for determining oxidation number: Elements in their element state: 0 The total oxidation state = the charge of the compound H 2 O total oxidation state = 0 H 3 O + total oxidation state = 1 “Atomic ions”: the same oxidation number as its charge Cl - oxidation number = -1 element Na + oxidation number = +1 Al 3+ oxidation number = +3

Some elements often have the same oxidation number Fluorine: -1 Hydrogen: +1 (except in hydrides = -1) Oxygen: -2 (except in peroxides = -1, bonded to F= pos.) The halogens: -1 (except when bonded to oxygen or a more electronegative halogen.)

Find the oxidation number in following compounds Cl 2 0 => element Cu 2+ +2 => atomic ions: the same oxidation number as its charge CH 4 H: +1 C: -4 => total oxidation state = the charge of the compound H 2 SO 3 H: +1 => total +2 O: -2 => total –6 S: +4 ( 2 + 4 - 6 = 0) NO 3 - O: -2=> total -6 N: 5 (5 – 6 = -1)

9.2 In a chemical reaction Oxidation: the atom that has an increase in oxidation number. Reduction: the atom that has a decrease in oxidation number. No change in oxidation number for any atom = no redox reaction.

Is following compounds being oxidised or reduced? NO  NO 3 - N: +2  N: +5 => oxidation N 2 O  NH 3 N: +1  N: -3 => reduction HNO 2  NO 2 - N: +3  N: +3 => no redox

Naming compounds: Oxidation number is also often used in naming compounds: FeCl 2 : Iron(II)chloride (2 = Ox. no. of Fe) CuO Copper(II)oxide Cu 2 O Copper(I)oxide

Redox reactions Balance and put together half-reactions: by calculating electrons Silver ions react with magnesium metal: – Oxidation reaction: Mg  Mg 2+ + 2 e - – Reduction reaction: 2 Ag + + 2 e -  2 Ag Total rection: Mg + 2 Ag +  Mg 2+ + 2 Ag

Reducing agent An element like sodium, Na, is eager to become an ion through oxidation: Na  Na + + e - Then some other particle, X, must be reduced (X + e -  X - ) Sodium is then said to be a reducing agent A reducing agent reduces a compound by self being oxidised

Oxidising agent An element like chlorine, Cl, is eager to become an ion through reduction: Cl + e -  Cl - Then some other particle must become oxidised Chlorine is then said to be a oxidizing agent An oxidising agent oxidises a compound by self being reduced

Balance redox reactions An easy example to start with: Cu + Ag +  Cu 2+ +Ag Divide into Half equations 1. Balance one of the reactants; Cu  Cu 2+ + 2e - 2. Balance the other; Ag + + e-  Ag Have to be multiplied by 2 to get the same number of electrons. 3. Add the reactions; Cu + 2 Ag +  Cu 2+ + 2 Ag 4. Check atoms and charge.

Balance redox reactions in acidic solutions Cu + HNO 3 +….  Cu(NO 3 ) 2 + NO + ….. 1. Half reactions: (1) Cu  Cu 2+ + 2e - (2) HNO 3  NO 2. Check and balance half reactions for O and H O not balanced in (2) => add H 2 O so oxygen balances (2) HNO 3  NO + 2 H 2 O ; H not balanced in (2) => add H as H + (acidic solution, remember…) (2) HNO 3 + 3 H +  NO + 2 H 2 O

Balance redox reactions in acidic solutions (II) (1) Cu  Cu 2+ + 2e - (2) HNO 3 + 3 H +  NO + 2 H 2 O 3. Check and balance half reactions for charges If charges not balanced =>add e - (2) HNO 3 + 3 H + + 3 e -  NO + 2 H 2 O 4. Check and balance half reactions so they use the same no of e - Multiply (1) with 3 and (2) with 2 => 6 e - produced and consumed in each half-reaction: (1) 3 Cu  3 Cu 2+ + 6 e - (2) 2 HNO 3 + 6 H + + 6 e -  2 NO + 4 H 2 O 5. Add the reactions: 3 Cu +2 HNO 3 + 6 H + + 6 e -  3 Cu 2+ + 6 e - + 2 NO + 4 H 2 O Check atoms and e - : Equation balanced

Balance redox reactions in acidic solutions (III) 6. Remove electrons. 3 Cu +2 HNO 3 + 6 H +  3 Cu 2+ + 2 NO + 4 H 2 O 7. Fix the reaction (if needed). Add 6 NO 3 - on both side: 3 Cu +2 HNO 3 + 6 H + + 6 NO 3 -  3 Cu 2+ + 2 NO + 4 H 2 O + 6 NO 3 - 8. Simplify: 3 Cu +8 HNO 3  3 Cu(NO 3 ) 2 + 2 NO + 4 H 2 O Check atoms and charges.

9.3 Reactivity Redox couple: A species that gain or lose electron(s), e.g. Na + + e - → Na If a compound is a good reducing agent (easily oxidized/lose e - ), the “other form” will be a bad oxidising agent and vice versa Redox couples can be arranged in a reactivity series

Redox reactivity series Redox couple Oxidised form Reduced form Na + + e -  Na Good red. agent Mg 2+ + 2e -  Mg Fe 2+ + 2e -  Fe 2 H + + 2e -  H 2 Cu 2+ + 2e -  Cu I 2 + 2e -  2 I - Br 2 + 2e -  2 Br - Good ox. agent F 2 + 2e -  2 F -

In the upper part of the redox reactivity series the redox couple prefer to be on the left side (oxidised form) In the lower part of the table the redox couple prefer to be on the right side (reduced form) Eg. sodium is rather Na + than Na, flourine is rather F 2 than F -

You can use the redox reactivity series to predict if a reaction is possible or not by comparing redox couples If the redox couple standing above in the table is going to the left and the redox couple standing below is going to the right, then a reaction will occur. – Mg 2+ + 2e -  Mg – Br 2 + 2e -  2 Br - => Mg + Br 2  MgBr 2 is possible If the above couple going to the right and the pair below to the left no reaction will occur. E.g. F 2 + 2I -  2F - + I 2 OK but I 2 + F -  no reaction

9.4 Voltaic cell The power of oxidising and reduction can be given in volts (V) You can measure the potential between two metals and their ions in a galvanic cell In a galvanic/voltaic cell chemical energy is converted into electrical energy It’s a spontaneous reaction

The less noble metal (Zn) will oxidise: Zn  Zn 2+ +2e -. This will be the negative pole. The electrons will pass the voltmeter and reach the copper metal and copper ions will be reduced: Cu 2+ +2e -  Cu. This will be the positive pole. In the salt bridge ions will go from Copper half cell to Zinc half cell. It balances the charges and will give you a closed circuit.

Cell diagram - Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) + Negative electrode, anode Positive electrode, cathode

Standard electrode potentials In the table “standard electrode potential” you find potentials of different redox couples compared to a “standard hydrogen electrode”. If you want to calculate the potential, E, you take the difference between the positive half cell and the negative half cell:E = E + -E - Use of Voltaic cells: Batteries

9.5 Electrolytic cell In electrolysis electrical energy is converted to chemical energy. It’s a non-spontaneous reaction. Electrolysis can be done in ionic aqueous solution (= electrolyte) or in molten salt (= electrolyte).

Electrolysis of CuBr 2 + pole/electrode = Anode: 2 Br -  Br 2 + 2e - Oxidation Reduction -pole/electrod = Cathode: Cu 2+ + 2e -  Cu

Electrolysis of molten salt: e.g. NaCl (l) Use of electrolysis: Electroplating, analysis, chargeable batteries, purification of metals

Differences between Voltaic cells and Electrolysis Voltaic cell Spontaneous Chemical energy  Electrical energy + electrode: Reduction (Cathode) - electrode: Oxidation (Anode) Electrolysis Non-spontaneous Electrical energy  Chemical energy + electrode: Oxidation (Anode) - electrode: Reduction (Cathode)

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