Answer for Chemistry Module Form 5_Redox Reaction

Answer for Chemistry Module Form 5Redox Reactions

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1(a)Chemicals

Displacement occurs

KI + Cl2/

KCl + Br2X

KBr + KCl

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2(a)(i)The presence of water and oxygen gas (air)1

(a)(ii)3

(b)(i)Iron (II) ions, Fe2+ combine with hydroxide ions, OH- forms iron (II) hydroxide, Fe(OH)2.The iron (II) hydroxide, Fe(OH)2, undergoes further oxidation oxygen gas that continuously dissolves in water to form rust, hydrated iron (III) oxide, Fe2O3.xH2O2Fe(OH)2 ( Fe2O3.xH2O1

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(ii)+2 to+31

(c)(i)Zinc is more electropositive than iron// zinc is located higher than iron in ECS.Zinc metal releases 2 electrons forms zinc ions, Zn2+.

It protects iron from rusting.1

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(ii)Zn ( Zn2+ + 2e1

3(a)(i)Test tube X: green solution turns brown1

(ii)Colourless solution turns brown1

(b)Br2 + 2Fe2+ ( 2Br- + 2Fe3+1

(c)Oxidising agent is a chemical substance that receives electrons.*bcoz oxidising agent undergoes reduction1

(d)(i)0 to -1

* Please refer to Rule 1 and Rule 2 on page 51

(ii)Iron (II) sulphate solution// iron (II) ions1

(e)(i)Sodium bromide// bromide ions1

(ii)2Br- ( Br2 + 2e- 1

(iii)Chlorine water acts as the oxidising agent1

(iv)Acidified potassium manganate (VII)// acidified potassium dichromate solution *acidified condition must be stated1

4(a)Potassium dichromate (VI) powder // manganese (IV) oxide*oxidising agent that produces oxygen gas when heated.1

(b)(i)Oxidation1

(ii)The metal powders (X, Y & Z) gain oxygen form metal oxide.1

(c)Zinc

*yellow when hot, white when cold1

(d) X, Z,Y

( reactivity of metals toward oxygen increase1

(e)*Please refer to the diagram on page 28 Chemis3 Module Redox.

Functional + label correctly1+1

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4(f)Element

Y

O

Mass(g)

4.8

3.2Mole

=4.8/24 = 0.23.2/16 = 0.2Simplest mole

=0.2/0.2 = 10.2/0.2 = 1

Thus, empirical formula

YO

*draw table & show the calculation properly1

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5(a)(i)The arrangement of metals in increasing order of electropositivity is: ( Silver, M, L.

Electropositivity increases

In Experiment 1, L able to displace silver from silver nitrate solution

L is higher than silver in the ECS.

In Experiment 2, M displaces silver from silver nitrate solution

Thus, M is higher than silver in the ECS. In Experiment 3, no change occurs.

M cannot displace L from L nitrate solution.

M is located below L in the ECS.

Max = 6m1

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(ii)Grey deposit is silver,

light blue solution is copper (II) nitrate.11

6(a)Metal X is silver/ copper/ lead/ stanum.Iron is more electropositive than silver/ copper/ lead/stanum.

The iron nail undergoes oxidation by releasing 2 electrons forms iron (II) ions, Fe2+.

The presence of iron (II) ions, Fe2+ is detected by potassium hexacyanoferrate (III) with the formation of dark blue colouration.

Metal Y is magnesium/ aluminium/ zinc.

Magnesium/ aluminium/ zinc is more electropositive than iron, so magnesium undergoes oxidation by releasing 2 electrons forms magnesium ions, Mg2+.11

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(b)*Describe briefly how u could carry out these two conversions.,( EXPERIMENTUse the format of MAPRO.

Describe a test ... ( POC

Answer: [Fe2+ ( Fe3+]Materials & Apparatus: 5 cm3 of iron (II) sulphate solution 0.5 mol dm-3, 5 cm3 of chlorine water/ bromine water [*oxidising agent], 10 cm3 of dilute sulphuric acid [*salt bridge], U-tube, connecting wires and crocodile clip, 2 carbon electrodes.

Procedures:

1. About 10 cm3 of sulphuric acid is poured in a U-tube.

2. 5cm3 of iron (II) sulphate solution is poured at one end of the U-tube.3. 5cm3 of chlorine water/ bromine water at another end of the U-tube.

4. 2 carbon electrodes are dipped in each solution respectively. Both electrodes are connected to a voltmeter with connecting wires.

5. Changes are observed after 15 minutes.

Results & Observation:

1. Green iron (II) sulphate solution turns brown.

Chemical test

P: 2 cm3 of the brown solution produced at the negative electrode is poured in a test tube. Little sodium hydroxide solution is added to the test tube until excess.

O: brown precipitate forms insoluble in excess of sodium hydroxide solution.

C: it confirms the presence of iron (III) ions, Fe3+ 1

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6(b)Answer: [Fe3+ ( Fe2+ ]Materials & Apparatus: 5 cm3 of iron (III) sulphate solution 0.5 mol dm-3, magnesium/ zinc powder, spatula, test tube and test tube rack.

Procedures:

1. 5 cm3 of iron (III) sulphate solution is added into a test tube.

2. A spatula of magnesium/ zinc powder is added to the solution.

3. Changes is observed and recorded.

Results & Observation:

2. Brown iron (III) sulphate solution turns green.

Chemical test

P: 2 cm3 of the green solution produced is poured into another test tube. Little sodium hydroxide solution is added to the test tube until excess.

O: Green precipitate forms insoluble in excess of sodium hydroxide solution.

C: It confirms the presence of iron (II) ions, Fe2+ Max= 10 marks1

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7(a)(i)Oxidation number of aluminium is +3Oxidation number of copper is +111

(a)(ii)Al2O3- aluminium oxide

Cu2O copper (I) oxide1

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(iii)Aluminium is a metal in Group 13. No need to mention its oxidation number in its name.

Copper is a transition element, the oxidation number of copper +1 should be mention in its name by using a roman numeral copper (I).1

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7(b)

(i)*Tips to answer U-tube qstn:

1st - identify oxidizing agent, acts as a +ve terminal.

2nd- draw the flow of electron, from ve to +ve terminal.3rd - determine redox reaction, write equations and state the observations.

Oxidising agent is acidified potassium manganate (VII) solution 1

(ii)Half equation at the negative terminal:

Fe2+ ( Fe3+ + e-

Oxidation occurs.

Half equation at the positive terminal:

MnO4- + 8H+ + 5e- ( Mn2+ + 4H2O *[correct formula + balance equation]

Reduction occurs1

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(iii)Graphite electrode dipped in iron (II) sulphate solution acts as a negative terminal.

Iron (II) sulphate undergoes oxidation,Iron (II) ions, Fe2+ releases 1 electron form iron (III) ions, Fe3+.

Green solution turns brown.

Iron (II) sulphate/ Iron (II) ion is a reducing agent.

Graphite electrode dipped in acidified potassium manganate (VII) solution is the positive terminal.

Manganate (VII) ions undergo reduction.

Manganate (VII) ions receive 5 electrons form mangan (II) ions, Mn2+.

Purple solution turns colourless.

Manganate (VII) ions act as an oxidising agent.11

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8(a)Oxidation occurs when the oxidation number of a substance increases.Reduction occurs when the oxidation number of a substance decreases.11

(b)(i)NaOH + HCl ( NaCl + H2O

+1 +1-1 +1 -1 +1

[*this is just a guidance to answer, no mark will be given]

The neutralisation between sodium hydroxide and hydrochloric acid forms sodium chloride and water(Oxidation number of sodium, Na (in NaCl and in HCl) remains unchanged, +1. (The oxidation number of H in (HCl and H2O) is +1, also does not changed. (Therefore, the neutralisation is not a redox reaction( as the oxidation number of the substances involved does not change1

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(b)(ii)AgNO3 + NaCl -( AgCl + NaNO3+1 +1 +1 +1The reaction between silver nitrate and sodium chloride produces silver chloride and sodium nitrate. (The oxidation number of silver in both silver nitrate and silver chloride does not change, +1. (The oxidation number of sodium is also the same, +1 in both sodium chloride and sodium nitrate. (Thus, precipitation is not a redox reaction. 1

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9(a)Oxidation occurs when a reacting substance releases electrons.Reduction occurs whe a reacting substance receives electrons.11

(b)(i)*Based on the equation, it shows that the displacement occurs. Metal M is more e+ve than copper.

Metal M is magnesium/ zinc/ stanum/ lead.1

b(ii)Based on the information;

Metal M (magnesium/ zinc/ stanum/ lead) displaces copper from copper to sulphate solution.

Metal M is located higher than copper in Electrochemical Series.

Metal M is a reducing agent, while copper (II) sulphate is an oxidising agent.

b(iii)Magnesium undergoes oxidation,

as its oxidation number increases from 0 to +2.

Copper (II) ions undergo reduction,

As its oxidation number decreases from +2 to 0.

(c)i. Diagram of set up apps: (XO + C // Y2O3 + C) ( Y, C, X

ii. Procedures:

1. A spatula of carbon powder is placed in a crucible.

2. A spatula of X oxide, XO powder is added to the crucible, the mixture is mixed well.

3. The mixture is heated strongly.

4. The vigorousness and the colour of the flame are observed.

5. For the second experiment, a mixture of carbon powder and metal Y oxide, Y2O3 is placed in a crucible and mixed well.6. The mixture is heated strongly and the changes are observed.iii. Observations:

Experiment

ObservationsCarbon + XO powder

The mixture burns with a bright flame

Black residue forms.Carbon + Y2O3 powderNo change occurs

Based on the observations:

Carbon is located above X in Reactivity Series of Metals. Carbon able to remove oxygen from X oxide to form carbon dioxide and X2XO+ C ( 2X + CO2Carbon is located below Y in Reactivity Series of Metal. So that carbon cannot remove oxygen from Y oxide.

Thus, it is proved that the position of carbon is above metal X and below metal Y in the Reactivity Series of metal.

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Carbon + metal oxide

heat

2010 Chemistry Panel of SM Sains Kota Tinggi 2