158. How to Answer Question on REDDOX Reactions

Chem Factsheet

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Number 158

How to Answer Questions on Redox Equations This Chemistry Factsheet gives guidance on answering questions on redox reactions. The answers that are provided are “best” answers as indicted by Principal Examiners. However, in some cases, other answers will gain equal / partial credit. In the answers, further explanations are provided – don’t provide them if the question does not ask for them.

Examples

Before starting this Factsheet make sure: (a) you have a Periodic Table

(a) KClO4 : K = + 1 and each of the 4 O atoms = -2 using rule C. Æ (+1) + 4(-2) + Oxn State(Cl) = 0 using rule A i) Æ Oxn State(Cl) = +7 ∴ Potassium chlorate(VII)

(b) you understand the relationship between Group number and the number of outer electrons in the atoms of a Group.

(b) NH 4+NO 3- : The N atoms in the different ions must be considered separately.

(c) you understand how ions (if any) are formed from atoms of elements in each Group

For NH4+ : Each of the 4 H atoms = +1 using rule C

(d) you understand the difference between an “ide” and an “ate”. e.g. sodium sulphide and sodium sulphate

Æ Oxn State(N) + 4(+1) = +1 using rule A ii) Æ Oxn State(N) = -3

(e) you know the formulas of the common acids and their anions.

For NO3- : Each of the 3 O atoms = -2 using rule C.

Æ Oxn State(N) + 3(-2) = -1 using rule A ii) Æ Oxn State(N) = +5 ∴ Ammonium nitrate(V)

Oxidation number rules These rules are used to assign an “oxidation number” to each and every atom in an ion or a compound.

(c) Cr2O72- : Each of the 6 O atoms = -2 using rule C. Æ 6(-2) + Oxn State(Cr) = -2 using rule A ii) Æ Oxn State(Cr) = +6 ∴ Dichromate(VI) ion

The number is then used to desribe the atom’s “oxidation state”. The more positive the number the more oxidised the atom is considered to be. A. The sum of the oxidation states of the atoms in: (i) a compound is 0 (ii) an ion is the charge on the ion.

(d) [CuCl4]2- : Each of the 4 Cl atoms = -1 using rule C. Æ 4(-1) + Oxn State(Cu) = -2 using rule A ii) Æ Oxn State(Cu) = +2 ∴ Tetrachlorocuprate(II) ion (e) CaH2 : Ca (from group 2) = +2 using rule C. Æ +2 + [2 x Oxn State(H)] = 0 using rule A i) Æ Oxn State(H) = -1 (Note : consistent with CaH2 containing hydride, H-, ions.)

B. Atoms of the more electronegative element are given the negative oxidation state. C. Certain oxidation states are assigned as follows (Table 1)

(f) F2O : Each of the 2 F atoms = -1 using rule C. Æ 2(-1) + Oxn State(O) = 0 using rule A i) Æ Oxn State(O) = +2

Table 1. Oxidation states Elements

Gp.1

Gp. 2

Al

F

H

O

0

Always +1

Always +2

Always +3

Always -1

+1 but not with a metal - see B

-2 but not with F -1 but not with F or O - see B. – see B. (in O-O bonds O = -1)

D. Oxidation states for all other atoms are worked out having applied A-C.

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Cl

Chem Factsheet

158. How to Answer Questions on Redox Equations Oxidation, Reduction, Oxidising agents

3. Barium and water

1. Oxidation = increase in oxidation number and / or loss of electrons. 2. Reduction: decrease in oxidation number and / or gain of electrons. 3. Atoms responsible for causing an oxidation (called “oxidants” or “oxidising agents”) have their oxidation number lowered. 4. Oxidising agents / oxidants are electron acceptors. 5. Atoms responsible for causing a reduction (called “reductants” or “reducing agents”) have their oxidation number increased. 6. Reducing agents / reductants are electron donors. 7. Half-equations (H.E.) are balanced equations (atoms and charge) containing electrons and showing either an oxidation or a reduction. 8. In a reduction half-equation, the total decrease in the oxidation number of the atoms that are reduced is equal to the number of electons gained. 9. In an oxidation half-equation, the total increase in the oxidation number of the atoms that are oxidised is equal to the number of electons lost. 10. A redox equation (which may be obtained from the two halfequations) is a balanced equation in terms of atoms and charges and shows an equal amount of oxidation and reduction. NB a redox equation does not contain electrons!

Reacts with water at room temperature. Fizzes - colourless gas (H2) and colourless solution (Ba(OH)2) formed.

Again, the metal is oxidised since it loses electrons. Hence: Oxidation H.E. : Ba - 2e- → Ba2+ - - - (1). The water molecules accept the electrons from the metal. Hence: Reduction H.E. : H2O + e- → OH- + ½H2 - - - (2). (1) + 2x(2) causes the electrons to cancel giving the overall redox equation. i.e. Ba + 2H2O → Ba2+ + 2OH- + H2.

4. Halogen (e.g. Br2) with a halide ion (e.g. I- from NaI) Orange solution (Br2(aq)) + colourless solution (NaI(aq)) of produces brown solution ( I2(aq)).

The halide ion is oxidised since it loses electrons. Hence: Oxidation H.E. : I- - e- → ½I2 - - - (1). The halogen molecules accept the electrons from the halide ions.

Examples of Half-Equations and Redox Equations

Hence: Reduction H.E. : Br2 + 2e- → 2Br- - - - (2). 2x(1) + (2) causes the electrons to cancel giving the overall redox equation. i.e. Br2 + 2I- → I2 + 2Br-

(State symbols have been left out of the half-equations and redox equations.)

5. Concentrated sulfuric acid reduced by iodide ions (e.g. I- from NaI). The NaI initially reacts with the acid to produce white fumes (HI). Then follows a series of reductions of the acid by the HI (or oxidation of the HI by the acid!) to produce a pungent smell (SO2), a yellow solid (S) and a bad-egg smell (H2S) along with a brown colouration (I2).

Note The writing of half-equations is sometimes a matter of personal choice. For example, the oxidation of bromide to bromine may be represented by any of the following: e.g. 2Br- - 2e- → Br2, Br- - e- → ½Br2, 2Br- → Br2 + 2eor Br- → ½Br2 + e-

The initial is NOT a redox reaction. NaI + H2SO4 Æ HI + NaHSO4 The hydrogen iodide is then oxidised since it loses electrons. Hence: Oxidation H.E. : HI - e- → ½I2 + H+- - - (1).

1. Metal (e.g. magnesium) and a non-metal (e.g. oxygen) React at high temperature. Bright white flame and white solid (MgO) formed.

The acid molecules accept various numbers of electrons from the halide ions producing the various reduction products.

Metals are oxidised since they lose of electrons. Hence: Oxidation H.E. : Mg - 2e- → Mg2+ - - - (1). The non-metal accepts the electrons from the metal. Hence: Reduction H.E. : ½O2 + 2e- → O2- - - - (2). (1) + (2) causes the electrons to cancel giving the overall redox equation. i.e. Mg + ½O2 → MgO.

(a) Consider the reduction of the acid to sulfur dioxide. H2SO4 (S = + 6) reduced to SO2 (S = +4). ∴2e- will be in the half-equation and, since all O’s (and H’s) form water, this gives: Reduction H.E. : H2SO4 + 2e- + 2H+ → SO2 + 2H2O - - - (2a).

2. Metal (e.g. zinc) and dilute acid (e.g.HCl)

(b) Consider the reduction of the acid to sulfur. H2SO4 (S = + 6) reduced to S (S = 0).

React at room temperature. Slow fizzing - colourless gas (H2) and colourless solution (e.g. ZnSO4) formed.

∴6e- will be in the half-equation and, since all O’s (and H’s) form water, this gives: Reduction H.E. : H2SO4 + 6e- + 6H+ → S + 4H2O - - - (2b).

“Full equations” are often used to describe such reactions. e.g. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) and Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g).

(c) Consider the reduction of the acid to hydrogen sulfide. H2SO4 (S = + 6) reduced to H2S (S = -2).

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However, in these reactions the chloride ions (Cl ) and the sulphate ions (SO42-) have not reacted as they stay in aqueous solutions – they are spectator ions. Hence, in the ionic redox equations they are omitted and so the ionic redox equation is the same for any acid with zinc.

∴8e- will be in the half-equation and, since all O’s (and H’s) form water, this gives:

Reduction H.E. : H2SO4 + 8e- + 8H+ → H2S + 4H2O - - - (2c). Hence, combining 2x(1) with (2a) gives : 2HI- + H2SO4 → I2 + SO2 + 2H2O

Again, the metal is oxidised since it loses electrons. Hence: Oxidation H.E. : Zn - 2e- → Zn2+ - - - (1).

and combining 6x(1) with (2b) gives : 6HI- + H2SO4 → 3I2 + S + 4H2O

The hydrogen ions of the acid accept the electrons from the metal. Hence: Reduction H.E. : 2H+ + 2e- → H2 - - - (2).

and combining 8x(1) with (2c) gives : 8HI- + H2SO4 → 3I2 + H2S + 4H2O.

(1) + (2) causes the electrons to cancel giving the overall redox equation. i.e. Zn + 2H+ → Zn2+ + H2

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Chem Factsheet

158. How to Answer Questions on Redox Equations 6. Potassium manganate(VII) solution (acidified with H2SO4) with a solution of an iron(II) salt (e.g. iron(II) sulphate)

2. The following redox equation occurs when conc. nitric acid is added to copper: Cu + 4H+ + 2NO3- → Cu2+ + 2NO2 + 2H2O

The purple solution (MnO4-) is decolourised when added to pale green solution (Fe2+).

(a) Give the oxidation states of nitrogen in NO3- and NO2. (2 marks) (b) Write the half-equation for the formation of : (i) NO2 from NO3(ii) Cu2+ from Cu. (2 marks)

The iron(II) is oxidized to iron(III) by the manganate(VII) in the presence of hydrogen ions. Oxidation H.E. : Fe2+ - e- → Fe3+ - - - (1)

3. Consider the redox reaction: H2S(aq) + 4H2O(l) + 8Au+(aq) → 8Au(s) + SO42-(aq) + ?

The oxidation state of the manganese changes from (+7) to (+2). Hence, there are 5 electrons in the reduction half equation with O’s and H’s forming water.

(a) What is represented by ? in the equation? (2 marks) (b) Identify the oxidizing agent and reducing agent in this reaction. (2 marks)

Reduction H.E. : MnO4- + 8H+ + 5e- → Mn2+ + 4H2O - - - (2) Hence, 5x(1) + (2) gives : 5Fe2+ + MnO4- + 8H+ → Fe3+ + Mn2+ + 4H2O

(c) Write a half-equation for the conversion of hydrogen sulfide into sulfate ions. (1 mark)

7. Reduction of potassium dichromate(VI) in dilute sulphuric acid to chromium(II) by zinc.

4. When potassium manganate(VII) reacts with iron(II) ions in the presence of excess acid the following reaction occurs: 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O. However, if there is insufficient acid, a brown precipitate of manganese(IV) oxide forms.

Orange solution (Cr2O7 ) turns to a green solution (Cr ) then to a blue solution (Cr2+). 3+

2-

The Cr(VI) is reduced, firstly to Cr(III) and then to Cr(II) by the zinc in the presence of hydrogen ions.

(a) What is the formula of manganese(IV) oxide? (1 mark)

Oxidation H.E. : Zn - 2e- → Zn2+ - - - (1)

(b) Write a half-equation for the oxidation of iron(II) ions into iron(III) ions. (1 mark)

First of all, the oxidation state of the chromium changes from (+6) to (+3). Hence, there are 6 electrons in the first reduction half equation since dichromate contains two Cr atoms, with O’s and H’s forming water.

(c) Write a half-equation for the reduction of manganate(VII) ions to manganese(IV) oxide and explain why this is reduction. (2 marks)

Reduction H.E. : Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O - - - (2a)

(d) Write a redox equation for the conversion of manganate(VII) ions to manganese(IV) oxide by iron(II) ions. (1 mark)

Hence, 3x(1) + (2a) gives : 3Zn + Cr2O72- + 14H+ → 3Zn2+ + 2Cr3+ + 7H2O ——(3)

5. Write full equations from the following half-equations:

Since zinc is such a strong reductant, the Cr(III) can then be reduced further to Cr(II).

(a) As2O3 + 6H+ + 6e- → 2As + 3H2O and Cu + Cl- - e- → CuCl. (2 marks)

Reduction H.E. : Cr + e → Cr - - - (2b) Hence, (1) + 2x(2b) gives : Zn + 2Cr3+ → Zn2+ + 2Cr2+ ———(4) 3+

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2+

(b) Cr2O72- + 8H+ + 6e- → Cr2O3 + 4H2O and NH4 + → ½N2 + 4H+ + 3e- (2 marks)

Alternatively, equations (3) and (4) can be combined so that the “2Cr3+“ cancel, to show the overall reduction of Cr(VI) to Cr(II).

6. Fe3+ ions are oxidized to FeO42- in alkaline conditions by Cl2. The Cl2 is reduced to Cl- ions. Write half-equations for the oxidation of Fe3+ ions and for the reduction of Cl2. Hence write the overall equation for the reaction. (3 marks)

Hence : 4Zn + Cr2O72- + 14H+ → 4Zn2+ + 2Cr2+ + 7H2O

Practice Questions 1. (a) Consider the following redox reaction: 2NO + 12H+ + 10I- → 2NH4+ + 2H2O + 5I2 (i) Deduce the oxidation state of nitrogen in NO and of nitrogen in NH4+ (ii) Identify the species formed by oxidation in this reaction.(3 marks)

7. Arsenic (As) reacts with nitric(V)acid to produce arsenic(V) acid (H3AsO4), nitrogen(IV) oxide and water. Write both half-equations and the full redox equation. (3 marks)

(b) When chlorine gas is bubbled into sulphur dioxide(aq), hydrogen ions, sulphate ions and chloride ions are formed. (i) Write a half-equation for the formation of chloride ions from chlorine. (ii) Write a half-equation for the formation of hydrogen ions and sulphate ions from sulphur dioxide and water. (iii) Hence, deduce an overall equation for the reaction which occurs when chlorine is bubbled into aqueous sulphur dioxide.(3 marks)

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Chem Factsheet

158. How to Answer Questions on Redox Equations

Answers (Don’t provide explanations if they are not asked for in the question!) 1. (a) (i) + 2, -3 (ii) I2 (b) (i) Cl2 + 2e- → 2Cl-, ii) SO2 + 2H2O - 2e- → SO42- + 4H+ (- 2e- since S = + 4 to S = + 6, realise that H+ come from water and 2 H2O as 2 O’s are needed.) (iii) SO2 + 2H2O + Cl2 → 2Cl- + SO42- + 4H+ - by adding the half-equations. 2. (a) + 5 / + 4. (b) (i) 2H+ + NO3- + e- → NO2 + H2O - - - (1) ( Copy all species in the provided equation apart from Cu and Cu2+and divide by 2.) (ii) Cu - 2e- →Cu2+ - - - (2). 3. (a) 10H+(aq) (since no H’s on right and charges must balance at 8+ overall. ) (b) Au+ / H2S (c) H2S(aq) + 4H2O(l) → SO42-(aq) + 10H+(aq) + 8e-. (Copy all species from the redox equation except 8Au+ and 8Au. In H2S, S = - 2 and in SO42-, S = + 6, an increase of 8, ∴ 8e-.) 4. (a) MnO 2 (b) Fe2+ - e- → Fe3+ (c) MnO4- + 4H+ + 3e- → MnO2 + 2H2O (MnO2 has 2 O from original 4 ∴ 2H2O. MnO4- (Mn + 7) to MnO2 (Mn + 4), a decrease of 3 ∴ + 3e-.) reduction because MnO2 has gained electrons. (d) 3Fe2+ + MnO4- + 4H+ → 3Fe3+ + MnO2 + 2H2O ( by 3 (b) + c)). 5. (a) As2O3 + 6H+ + 6Cl- + 6Cu → 2As + 3H2O + CuCl. (Reduction shows 6e- and the oxidation shows 1 e- ∴ 6 × oxidation needed.) (b) Cr2O72- + 2NH4+ → Cr2O3 + 4H2O + N2 (Reduction shows 6e- and the oxidation shows 3e- ∴ 2 × oxidation needed.) 6. Fe3+ + 8OH- - 3e- → FeO42- + 4H2O. (Fe3+ (+ 3) → FeO42- (+ 6) = up 3 (oxidation) ∴ - 3e-. ∴ 8OH- to balance charges ∴4H2O.) e.g. ½Cl2 + e- → Cl-. Fe3+ + 8OH- + 1½Cl2 → FeO42- + 4H2O + 3Cl- (or double to remove 1½) 7. As + 4H2O - 5e- → H3AsO4 + 5H+. (- 5e- as ox. st. of As increases by 5; 4H2O to balance O’s in H3AsO4; 5H+ to balance H’s & charges.) HNO3 + e- + H+ → NO2 + H2O (+ e- as ox. st. of N decreases by 1; H+ to balance H’s & charges.) As + 4H2O + 5HNO3 → H3AsO4 + 5NO2 + 5H2O. (5 × reduction + oxidation ; note – 5 H+ cancel).

Acknowledgements: This Factsheet was researched and written by Bob Adam. Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. ChemistryFactsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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