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Introduction to Chemistry

Redox Reactions

SCIE1000

Redox Reactions: Introduction to Chemistry SCIE1000 Lecture Objectives Explain oxidation and reduction as an electron transfer process. Oxidation is the process where an atom or ion loses an electron. Reduction is the process where an atom or ion gains an electrons. Oxidation and Reduction reactions occur simultaneously. Examples: Mg + Pb2+ → Mg2+ + Pb

in this reaction Mg is being oxidized and Pb 2+ is reduced

2Mg + O2 → 2MgO in this reaction Mg is being oxidised because it is losing electrons to ions in the products and O2 is being reduced because it gains electrons

become Mg

Name the species oxidised or reduced and identify the oxidising agent (oxidant) and reducing agent (reductant) in redox reactions. Oxidising agent (oxidant) is the substance in a redox reaction that causes the oxidation of another substance by accepting electrons. The reducing agent is the substance in a redox reaction that causes the reduction of another substance by losing electrons Examples: Mg + Pb2+ → Mg2+ + Pb In this reaction Pb2+ accepts e- from Mg therefore Pb2+ is the oxidising agent. Mg loses e- which causes the reduction of Pb2+ therefore Mg is the reducing agent Determine oxidation numbers. The oxidation number of a substance is a number assigned to an atom or ion following a set of rules. They are used to determine if a substance is being oxidised or reduced when it is not obvious what is occurring. Rules: 1. Elements have an oxidation number of zero a. E.g. Mg, O2, Cl2, P4, Al, 2. For simple monatomic ions the oxidation number is equal to the charge a. E.g. Cl- = -1 Mg2+ = +2 O2- = -2 3. Combined oxygen is -2 unless it is part of a peroxide compound like NaO or H 2O2 where it is -1 or F2O where it is +2 4. Combined hydrogen is +1 except in metal hydride where it is -1 a. H2O = +1 MgH2 = -1 5. For polyatomic ions the sum of the oxidation number is equal to the charge on the ion a. For MnO4- the charge is = -1. We know that oxygen is -2 and so to work out Mn we must -1 = Mn + 4(O) i. -1 = Mn + 4(-2) ii. -1 = Mn - 8 iii. -1 + 8 = Mn = +7 6. The sum of the oxidation number in a neutral molecule must = 0 a. e.g. In H2S the ON(S) + (2 x ON (H)) = 0 7. Some elements, especially the transition metals, can exhibit a variety of oxidation numbers a. e.g. Mn2+ Mn4+ Fe2+ Fe3+ Cu+ Cu2+

Introduction to Chemistry

Redox Reactions

SCIE1000

Assign the oxidation numbers to the bolded element: O2

MgH2

Na2O2

SO42-

HCl

CrO42-

CO32-

NH4+

HNO3

ZnO

ClO4-

C2O42-

Identify oxidation-reduction reactions using oxidation numbers. Oxidation numbers help to identify redox reactions from non-redox reactions and identify the species being oxidised or reduced. A substance that is oxidised experiences an increase in oxidation number in a reaction. A substance that is reduce experiences a decrease in oxidation number in a reaction. Reactions in which there is no change in oxidation numbers are not redox reactions eg

Mg + Pb2+ → Mg2+ + Pb 0

+2

+2

0

Mg goes from 0 to +2 and therefore it is oxidised, Pb2+ goes from +2 to 0 and therefore it is reduced Identify common oxidising and reducing agents.

Represent redox reactions with half-equations and balanced redox equations. Half-equations show the oxidation part and the reduction part of a redox reaction individually. In half equations an e is either loss of gained to form part of the equation. The method used will be dependent on the complexity of the equation and the conditions under which it takes place. 1. The electron method Redox Equation: Zn2+ + Mg → Zn + Mg 2+ Oxidation Half-equation: Mg → Mg2+ + 2 eReduction half-equation: Zn2+ + 2e-  Zn

Introduction to Chemistry

Redox Reactions

SCIE1000

Writing half equations from unbalanced redox reactions: Cu2+ + Ag → Cu + Ag+ +2

0

0

+1

Oxidation: Ag  Ag+ + eReduction: Cu2+ + 2e-  Cu Balancing Redox Equations The overall equation for a redox reaction is called a redox equation. These consist of the combination of the oxidation and reduction half equations without expressing the eExample 1

Na + Cl2 → NaCl

Oxidation

Na → Na+ + e-

Reduction

Cl2 + 2 e- → 2 Cl-

Write the half-equations

The number of e- in the 2 half-equations are not balanced so: 2 Na → 2 Na+ + 2 e-

x2

Cl2 + 2 e- → 2 ClRedox

2 Na + Cl2 → 2 NaCl

Balance the e- (number lost must equal number gained) Cancel out the e- and combine the reactants and products from both equations

Write balanced redox equations for redox reactions in acidic conditions. Ion-electron method (for more complicated half-equations) Steps in acidic solution: 1. 2. 3. 4. 5. 6.

Step 1:

Write down reactants and products in an equation and assign the oxidation numbers Balance the atom being oxidised or reduced using coefficients as normal Balance O atoms by adding H2O Balance H atoms by adding H+ Balance charge by adding ePut in (g), (aq), (s) etc

2+ MnO4 → Mn +7 -2

+2

Mn is being reduced

Step 2:

already balanced

Step 3:

2+ MnO4 → Mn + 4 H2O

Step 4:

8H

Step 5:

5e

Step 6:

5e

+

-

+ MnO + 8H

+

2+ + 4 H2O 4 → Mn 2+ + MnO4 → Mn + 4 H2O

+ 2+ + 4 H2O(l) + 8 H (aq) + MnO4 (aq) → Mn (aq)

Introduction to Chemistry

Redox Reactions

SCIE1000

**The above notes and diagrams were collated from the lecture slides and course notes from Kerinne Hindle, in the SCIE1000 Introduction to Chemistry Course in semester 1 of 2020**...