RI 2022 Redox Reactions v2.0

Content Raffles Institution Year 5 H2 Chemistry 2022 Lecture Notes 1 b - Redox Reactions • Redox processes: electron transfer and changes in oxidation number (oxidation state) Learning Outcomes Candidates should be able to: (a) describe and explain redox processes in terms of electron transfer and/or of changes in oxidation number (oxidation state) (b) construct redox equations using the relevant half-equations Lecture Outline References 1 Redox Reactions Similar to those in Lecture Notes 1 a 2 Balancing Redox Equations 3 Redox ntrations -1-

_1 _ __._I_R_e_d_o_x_R_e_a_c_ti_o_n_s ___________________________ , • The term "redox" is used as an abbreviation for the processes of reduction and oxidation which occur simultaneously. A redox reaction is an oxidation-reduction reaction. 1.1. Definitions of oxidation and reduction • There are different ways to define oxidation and reduction. The focus will be on (c) and (d). (a) loss/gain • Oxidation involves gain of oxygen . of oxygen • Reduction involves loss of oxygen . oxidation I i • Example: Fe203 + 3CO 2Fe + 3C02 I t reduction (b) loss/gain • Oxidation involves loss of hydrogen . of hydrogen • Reduction involves gain of hydrogen . • Oxidation of ethanol to ethanal: CH3CH20H CH3CHO • Reduction of ethanal to ethanol: CH3CHO CH3CH20H (c) loss/gain • Oxidation involves loss of electron(s) . of electrons • Reduction involves gain of electron(s} . oxidation I i • Example: Zn + Cu2+ Zn2+ + Cu I t reduction OIL RIG oxidation involves loss of e- reduction involves gain of e- (d) increase/decrease • Oxidation involves an increase in oxidation number. in oxidation number • Reduction involves a decrease in oxidation number . oxidation I + 0 0 +4 - 2 • Example: C + 02 CO2 I t reduction • An oxidising agent (electron acceptor) oxidises the other reagent while it undergoes reduction by receiving the electrons lost by the oxidised species, i.e. an Oxidising Agent is itself Reduced. • A reducing agent (electron donor) reduces the other reagent while it undergoes oxidation by giving electrons to cause the reduction, i.e. a Reducing Agent is itself Oxidised. -2-

1.2. Redox processes in tenns of electron transfer • Consider the reaction between zinc metal and copper(II) sulfate solution. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) Or Zn(s) + Cu2•(aq) Zn2•(aq) + Cu(s) • The overall reaction can be separated into two simpler processes involving electron transfer. Oxidation half-equation: Zn(s) Zn2•(aq) + 2e- Reduction half-equation: Cu2•(aq) + 2e- Cu(s) Overall equation: Zn(s) + Cu2•(aq) Zn2•(aq) + Cu(s) • The two separate equations can be termed ion-electron equations, but they are more commonly known as half-equations. • One half-equation represents the oxidation process while the other represents the reduction process. Addition of the two half-equations gives the overall redox equation. • In the above redox reaction, • Zn acts as a reducing agent. It loses two electrons, and as a result, is itself oxidised to Zn2•. • Cu2• acts as an oxidising agent. It gains two electrons, and as a result, is itself reduced to Cu. Total amount of electrons lost by 1 mol of Zn = Total amount of electrons gained by 1 mol of Cu2• • In a redox reaction involving electron transfer, the total number of electrons transferred is the same for both half-equations. In other words, Total number of electrons lost by the reducing agent = Total number of electrons gained by the oxidising agent • Another example: 2Ag•(aq) + Fe(s) 2Ag(s) + Fe2•(aq) Oxidation half-equation: Fe(s) Fe2•(aq) + 2e- Reduction half-equation: 2Ag•(aq) + 2e- 2Ag(s) In this case, • 1 mol of Fe reacts with 2 mol of Ag•. • Total amount of electrons transferred in each half-equation= 2 mol Note: Total amount of electrons lost by 1 mol of Fe = -3- Total amount of electrons gained by 2 mol of Ag•

1.3. Redox processes in tenns of change in oxidation number (a) Oxidation number • An oxidation number is a number which is assigned to an element in a substance to show its state of oxidation. Note: The oxidation number of an element in a substance is related to the number of electrons lost, gained, or shared as a result of chemical bonding. • An atom is oxidised if its oxidation number increases. An atom is reduced if its oxidation number decreases. • Oxidation occurs when the oxidation number of an atom increases. Reduction occurs when the oxidation number of an atom decreases. (b) Rules for assigning oxidation numbers • The oxidation number of an atom in the elemental state is 0. Examples: Oxidation number of oxygen in 02 is 0. Oxidation number of phosphorus in p4 is 0. • The oxidation number of hydrogen in all compounds, except metal hydrides, is +1 . In metal hydrides (e.g. NaH, MgH2), the oxidation number of hydrogen is -1 . • The oxidation number of fluorine in all compounds is -1 . • The oxidation number of oxygen is -2 in all compounds, except in peroxides, superoxides and OF2. In peroxides (e.g. Na2O2}, the oxidation number of oxygen is -1 . In superoxides (e.g. KO2), the oxidation number of oxygen is- ½. In OF2, the oxidation number of oxygen is +2. • In any compound, the more electronegative atom has the negative oxidation number while the less electronegative atom has the positive oxidation number. Example: For BrC/, the oxidation number of Br is +1 and the oxidation number of C/ is -1 . • The electro negativity of an atom is the ability of the atom in a molecule to attract shared electrons in a bond. • Electronegativity generally increases across a period and decreases down a group. • To search for electronegativity values, visit: http://www.rsc. orolperiodic-tableltrends • In monatomic ions, the oxidation number is simply the charge on the ion. Examples: Oxidation number of Na in Na• is +1 Oxidation number of C/ in C, is -1 • In polyatomic ions, the algebraic sum of the oxidation numbers equals the charge on the ion. Example: For Mno4-, (+7) + (4)(-2) = -1 • In a compound (e.g. CO2, NaCl, AJCfa), the algebraic sum of the oxidation numbers of the atoms is O. Examples: For CO2, (+4) + (2)(-2) = 0 For NaCl, (+1) + (-1) = 0 Note: Oxidation number of Mg in Mg2• = +2 (the "+" sign precedes the number "2") Charge on Mg2• ion = 2+ (the "+" sign is written after the number "2") The oxidation number of Mg in MgSQ4 is +2. Mg exists in an oxidation state of +2 or displays an oxidation state of +2 in MgSO4. -4-

Worked Example 1 // What is the oxidation number of uranium (symbol: U) in l<JUFs? Solution Let the oxidation number of U in l<JUFs be n. (3)(+1) + n + (6)(-1) = o n = +3 Hence oxidation number of U is +3. Worked Example // An element can exhibit different oxidation states in different compounds. Determine the oxidation number of (a) nitrogen in NH3 N2H4 N2O NO NF3 NO2 N2Os -3 -'"1. +1 .+). 13 --\ 'i -t) (b) chlorine in ChO1 C/O3 NaC/O3 C/O2 KC/O2 ChO C/2 NaCl -;1 1\. , t"7 -1'\, -\ l, t\ 0 -1 ilbis question 'is"""'kart of RedoxReactioiis]juiz'i.oJ1iW, An element can exhibit different oxidation states in different compounds. Determine the oxidation number of sulfur in: H2S SO3 sol- SC/2 H2SO3 H2SO4 - "l. -t~ -t4 -t"'L- -t-4- --\- \. (c) Oxidation numbers of atoms in complicated structures • In complicated molecules or ions, it is helpful to deduce the oxidation states of the elements based on the structures. We will revisit this method in topic on 'Introduction to Organic Chemistry'. -5-

(d) Examples of redox reactions • In general, to decide whether a reaction is a redox reaction, assign oxidation numbers to the atoms in the reactants and products, and check for any changes in oxidation numbers. The following are examples of redox reactions. +1 0 +2 0 2HC/(aq) + Mg(s) MgC/2(aq) + H2(g) 0 0 +3 -1 2Fe(s) + 3C/i(g) 2FeC/3(S) 0 +1 +2 0 Ca(s) + 2H20(I) Ca(OH)2(aq) + H2(g) 0 -1 0 -1 Ch(aq) + 2Br-(aq) -+ Br2(aq) + 2Cf(aq) Note: In a redox