Worked solutions: Chapter 10 Oxidation and reduction Cr2O7 2 –(aq) + 14H+(aq ... b No oxidation...

Worked solutions: Chapter 10 Oxidation and reduction

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Section 10.1 Exercises

1 a Redox: copper ions have been reduced to copper atoms.

b Not redox: change of state only.

c Redox: reaction with oxygen is oxidation. Fe atoms have become Fe3+ ions.

d Redox: magnesium atoms oxidized to magnesium ions. Oxygen atoms reduced to oxide ions.

2 a Mg: +2; O: –2

b H: +1; O: –2

c H: +1; P: +6; O: –2

d C: –2; H: +1

e N: +4; O: –2

f K: +1; Mn: +7; O: –2

3 a Cu: +2; copper(II) oxide

b Cu: +1; copper(I) oxide

c Mn: +4; manganese(IV) oxide

d V: +5; vanadium(V) oxide

4 Answers in bold.

Oxide of nitrogen Systematic name N2O Nitrogen(I) oxide NO Nitrogen(II) oxide NO2 Nitrogen(IV) oxide N2O3 Nitrogen(III) oxide N2O5 Nitrogen(V) oxide

5 CH4: ON = –4

C2H6: ON = –3

C2H4: ON = –2

C2H2: ON = –1

C(graphite): ON = 0

CO: ON = +2

CO2: ON = +4




6 a R

b O

c N

d B

7 In C6H12O6: ON(C) = 0 (6x + 12 × +1 + 6 × –2) = 0, so 6x = 0, and x = 0

In CO2: ON(C) = +4 (ON has increased from 0 to +4), so carbon is oxidized.

ON(O) = 0 in O2 and = –2 in CO2 (ON has decreased from 0 to –2), so oxygen is reduced.

8 a CuO(s) + H2(g) → Cu(s) + H2O(l) ON: +2 –2 0 0 +1 –2

H2 has been oxidized (ON increased from 0 to +1).

b Cr2O72–(aq) + 14H+(aq) + 6I–(aq) → 2Cr3

+(aq) + 3I2(l) + 7H2O(l) ON: +6 –2 +1 –1 +3 0 +1 –2

I– has been oxidized (ON increased from –1 to 0).

c MnO2(s) + SO2(aq) → Mn2+(aq) + SO42–(aq)

ON: +4 2 +4–2 +2 +6 –2

S in SO2 has been oxidized (ON increased from +4 to +6).


1 a Oxidation is the loss of electrons. Reduction is the gain of electrons.

b An oxidizing agent accepts electrons. A reducing agent donates electrons.

2 a Oxidizing agent: Fe2+ (ON decreases from +2 to 0 ∴ is reduced)

Reducing agent: Cu (ON increases from 0 to +2 ∴ is oxidized)

b Oxidizing agent: Ag+ (ON decreases from +1 to 0 ∴ is reduced)

Reducing agent: Sn2+ (ON increases from +2 to +4 ∴ is oxidized)

c Oxidizing agent: H in H2O (ON decreases from +1 to 0 ∴ is reduced)

Reducing agent: Ca (ON increases from 0 to +2 ∴ is oxidized)

d Oxidizing agent: O2 (ON decreases from 0 to –2 ∴ is reduced)

Reducing agent: Mg (ON increases from 0 to +2 ∴ is oxidized)

e Oxidizing agent: Cu2+ (ON decreases from +2 to 0 ∴ is reduced)

Reducing agent: Fe (ON increases from 0 to +2 ∴ is oxidized)

f Oxidizing agent: F2 (ON decreases from 0 to –1 ∴ is reduced)

Reducing agent: Ni (ON increases from 0 to +2 ∴ is oxidized)




g Oxidizing agent: Fe3+ (ON decreases from +3 to +2 ∴ is reduced)

Reducing agent: I– (ON increases from –1 to 0 ∴ is oxidized)

3 a Oxidation: Zn(s) → Zn2+(aq) + 2e–

Reduction: 2H+(aq) + 2e– → H2(g)

b Oxidation: Mg(s) → Mg2+(s) + 2e–

Reduction: O2(g) + 4e– → 2O2–(s)

c Oxidation: Ni(s) → Ni2+(s) + 2e–

Reduction: F2(g) + 2e– → 2F–(s)

d Oxidation: Fe2+(aq) → Fe3+(aq) + e–

Reduction: Cu2+(aq) + 2e– → Cu(s)

e Oxidation: Sn2+(aq) → Sn4+(aq) + 2e–

Reduction: Ag+(aq) + e– → Ag(s)

f Oxidation: Na(s) → Na+(aq) + e–

Reduction: 2H2O(l) + 2e– → 2OH–(aq) + H2(g)

4 a CuO(s) + H2(g) → Cu(s) + H2O(l) ON: +2 –2 0 0 +1 –2

H2 is oxidized to H2O and Cu2+ is reduced to Cu.

H2 is the reducing agent and Cu2+ is the oxidizing agent.

b Cr2O72–(aq) + 14H+(aq) + 6I–(aq) → 2Cr3+(aq) + 3I2(l) + 7H2O(l)

ON: +6 –2 +1 –1 +3 0 +1 –2

I– is oxidized to I2 and Cr2O72–

is reduced to Cr3+.

I– is the reducing agent and Cr2O72–

is the oxidizing agent.

c MnO2(s) + SO2(aq) → Mn2+(aq) + SO42–(aq)

ON: +4 –2 +4–2 +2 +6 –2

S in SO2 is oxidized to SO42– and Mn in MnO2 is reduced to Mn2+.

SO2 is the reducing agent and MnO2 is the oxidizing agent.

5 a H2 has been oxidized (ON has increased from 0 to +1), so this is a redox reaction.

H2 is the reducing agent and O2 is the oxidizing agent.

b No oxidation numbers have changed, so it is not a redox reaction.

c No oxidation numbers have changed, so it is not a redox reaction.

d S in H2S has been oxidized (ON has increased from –2 to 0), so this is a redox reaction.

H2S is the reducing agent and KMnO4 is the oxidizing agent (HCl has no effect).




6 a O2(g) + 4H+(aq) + 4e– → 2H2O(l)

b NO2(g) +H2O(l) → NO3–(aq) + 2H+(aq) + e–

c 2HOCl(aq) + 2H+(aq) + 2e– → Cl2(g) + 2H2O(l)

d MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

e H2S(g) + 2H2O(l) → SO2(g) + 6H+(aq) + 6e–

f H2O2(aq) → O2(g) + 2H+(aq) + 2e–

g H2O2(aq) + 4H+(aq) + 4e– → H2(g) + 2H2O(l)

h C2O42–(aq) → 2CO2(g) + 2e–

7 a Reduction: I2(aq) + 2e– → 2I–(aq)

Oxidation: 2S2O32–(aq) → S4O6

2–(aq) + 2e–

Balanced overall equation:

2S2O32–(aq) + I2(aq) → S4O6

2–(aq) + 2I–(aq)

b Oxidation: Zn(s) → Zn2+(aq) + 2e–

Reduction: NO3–(aq) + 2H+(aq) + e– → NO2(g) + H2O(l)

To balance the two equations, electrons must be balanced

Zn(s) → Zn2+(aq) + 2e–

2NO3–(aq) + 4H+(aq) + 2e– → 2NO2(g) + 2H2O(l)


Zn(s) + 2NO3–(aq) + 4H+(aq) → Zn2+(aq) + 2NO2(g) +2H2O(l)

c Reduction: ClO–(aq) + 2H+(aq) + 2e– → Cl–(aq) +H2O(l)

Oxidation: ClO–(aq) + 2H2O(l) → ClO3–(aq) + 4H+(aq) +4e–

To balance the two equations, electrons must be balanced.

2ClO–(aq) + 4H+(aq) + 4e– → 2Cl–(aq) + 2H2O(l)

ClO–(aq) + 2H2O(l) → ClO3–(aq) + 4H+(aq) +4e–


3ClO–(aq) → 2Cl–(aq) + ClO3–(aq)

d Reduction: Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l)

Oxidation: CH3CH2OH(aq) + H2O(l) → CH3COOH(aq) + 4H+(aq) +4e–

To balance the two equations, electrons must be balanced.

2Cr2O72–(aq) + 28H+(aq) + 12e– → 4Cr3+(aq) + 14H2O(l)

3CH3CH2OH(aq) + 3H2O(l) → 3CH3COOH(aq) + 12H+(aq) + 12e–





2Cr2O72–(aq) + 3CH3CH2OH(aq) +16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)


1 a An electronic conductor; the site of oxidation and reduction reactions in a voltaic cell.

b An electrolytic conductor; the solution through which ions flow in a voltaic cell.

c The electrode at which oxidation occurs.

d The electrode at which reduction occurs.

e A combination of an oxidizing agent and its conjugate reducing agent arranged to include an electrode and an electrolyte.

2 It must have high solubility and be non-reactive with any component or product of the cell.

3 Ni2+/Ni half-equation: Ni(s) → Ni2+(aq) + 2e–

Pb2+/Pb half-equation: Pb2+(aq) + 2e– → Pb(s)

Electron flow is from Ni to Pb.

Cation flow is from Ni half-cell to the Pb half-cell.

Ni is the negatively charged anode, Pb is the positively charged cathode.

4 Salt bridge may dry out. Pb2+ ions may all be reacted.

5 A2+/A half-equation: A2+(aq) + 2e–→ A(s)

B2+/B half-equation: B(s) → B2+(aq) + 2e–

Electron flow is from B to A.

Positive ions flow is from B half-cell to A half-cell.

A is the positively charged cathode, B is the negatively charged anode.

Equation: A2+(aq) + B(s) → A(s) + B2+(aq)

6 Answers in shaded cells. Reaction at anode (–)

Reaction at cathode (+)

Overall reaction Oxidizing agent

Electrode material at cathode

Cu(s) → Cu2+(aq) + 2e– Ag+(aq) + e– → Ag(s) Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

Ag+(aq) Ag

Mg(s) → Mg2+(aq) + 2e– Pb2+(aq) + 2e– → Pb(s) Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(aq)

Pb2+(aq) Pb

2I–(aq) → I2(aq) + 2e– Fe3+(aq) + e– → Fe2+(aq) 2Fe3+(aq) + 2I–(aq) → I2(aq) + 2Fe2+(aq)

Fe3+(aq) Pt (inert)

Al(s) → Al3+(aq) + 3e– Ni2+(aq) + 2e– → Ni(s) 3Ni2+(aq) + 2Al(s) → 3Ni(s) + 2Al3+(aq)

Ni2+(aq) Ni




7 A salt bridge is used to complete the circuit in a voltaic cell by allowing the flow of positive ions into the half-cell containing the positive electrode (the cathode) and negative ions into the half-cell containing the negative electrode (the anode).

8 a Electrons travel from the negative electrode to the positive electrode through the conducting wires.

b Potassium ions travel through the KNO3 salt bridge towards the half-cell containing the positive electrode. Nitrate ions travel through the KNO3 salt bridge towards the half-cell containing the negative electrode.

9 a b c

10 a cathode

b from right to left as shown

c from right to left as shown

d negative

e Pt or graphite

f Fe3+(aq) + e– → Fe2+(aq)

g 2I–(aq) → I2(aq) + 2e–

h 2Fe3+(aq) + 2I–(aq) → I2(aq) + 2Fe2+(aq)


1 a They are highly reactive metals that easily donate electrons.

b They are highly reactive non-metals that readily gain electrons.

c Most metals are sufficiently reactive to react with oxygen and other non-metals present in the atmosphere or crust.

2 Reactivity: Z > X > W > Y

3 a Fe(s) + Ni2+(aq) → Fe2+(aq) + Ni(s)

b No reaction

c Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s) (There may also be a slight reaction between the Mg and water.)




4 a No reaction occurs.

b A reaction will occur according to the following equation:

Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)

c No reaction occurs.

d A reaction will occur according to the following equation:

2Na(s) + 2H2O(l) → 2Na+(aq) + H2(g) + 2OH–(aq)

5 a Ag+

b Na+

c Na

d Ag

6 A metal coating will form when the metal cation is a sufficiently strong oxidizing agent to oxidize the iron nail. In these cases reaction will occur with all four species. (Sn4+ will react to produce Sn2+, which in turn may react to deposit Sn.)

7 The metal is oxidized by HCl and so is a stronger reducing agent than H2. It cannot be oxidized by Zn2+ and so is a weaker reducing agent than zinc. As it reacts with Ni2+ it must be a stronger reducing agent than nickel. By inspection of the electrochemical series the metal could be Co, Cd, Fe or Cr.

8 a Oxidation: Sn(s) → Sn2+(aq) + 2e–

Reduction: Ag+(aq) + e– → Ag(s)

b Anode is Sn electrode; cathode is Ag electrode.

c Sn electrode is the negative electrode; Ag electrode is the positive electrode.

d Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s)

e Oxidizing agent is Ag+; reducing agent is Sn.

f Electrons flow from Sn to Ag through the connecting wires.

g Positive ions in the salt bridge flow into the Ag half-cell, while negative ions flow into the Sn half-cell.

9 The electrochemical series predicts whether or not a reaction should occur under standard conditions of temperature, pressure and concentration. It provides no information about the rate of reaction. In this case the rate is very slow as aluminium has a uniform, impervious and unreactive oxide layer that protects it from reaction.

10 a An inert electrode is used when one of the conjugate redox pair is a gas, if both are ions in aqueous form, or if the solid does not conduct electricity. In such cases there is no solid form of the oxidizing agent or reducing agent that can form an electrode and conduct electricity.




b For example:

Cl2(g) + 2e– → 2Cl–(aq)

2H+(aq) + 2e– → H2(g)

Fe3+(aq) + e– → Fe2+(aq)

c Graphite is much cheaper than platinum.

11 Chlorine and fluorine will displace bromine from a solution of sodium bromide.

Cl2(g) + 2NaBr(aq) → Br2(aq) + 2NaCl(aq)

F2(g) + 2NaBr(aq) → Br2(aq) + 2NaF(aq)


1 The power source supplies the energy required to drive the non-spontaneous endothermic redox reactions in the forward direction. The electrical energy from the power source is converted into chemical potential energy stored in the products of reaction.

2 Reduction is the gain of electrons. The negative electrode in an electrolytic cell is the source of electrons, so positive species are attracted to the negative electrode and gain electrons there (are reduced).

3 a At the anode (positive): 2Cl–(l) → Cl2(g) + 2e–

b The products of the electrolysis are chlorine gas and lithium.

c Overall equation: Li+(l) + 2Cl– (l) → Li(l) + Cl2(g)

4 i Electrons flow from right to left.

ii The electrode on the right is the cathode.

iii S2– is the reductant.

iv The electrode on the left is the anode.

v S2–(l) → S(l) + 2e–

vi Mg2+(l) + 2e– → Mg(l)

5 a

b 2F–(l) → F2(g) + 2e–; Al3+(l) + 3e– → Al(l)




c Anode is positive electrode; cathode is negative electrode.

d Oxidizing agent is Al3+; reducing agent is F–.

e 2Al3+(l) + 6F–(l) → 2Al(l) + 3F2(g)

6 Ions move through the electrolyte towards the anode or cathode. Electrons move through the wire that connects the anode and cathode. Positive ions move towards the cathode and negative ions towards the anode.

Chapter 10 Review questions

1 Oxidizing agent Undergoes reduction during a redox reaction

Reducing agent An electron donor

Oxidation Happens to a reducing agent during a reaction

Reduction Involves the gain of electrons

2 Gain of oxygen, loss of electrons, loss of hydrogen

3 a N +4, O –2

b K +1, Cr +6, O –2

c H +1, S +6, O –2

d H +1, N +3, O –2

e H +1, P +5, O –2

f S +2, O –2

g Ga +3, C +4, O –2

h Zn +2, Br +5, O –2

4 a MnO4– +7, MnO2 +4, Mn2+ +2

b Reduction: MnO4– + 4H+ + 3e– → MnO2 + 2H2O

Oxidation: Mn2+ + 2H2O → MnO2 + 4H+ + 2e–

5 ClO– + 2H+ + 2e– → Cl– + H2O

ClO– + 3H2O → ClO3– + 6H+ + 6e–

6 a C6H12O6 + 2H+(aq) + 2e– → C6H8(OH)6

b C6H8(OH)6 → C6H8O6 + 6H+(aq) + 6e–

c C2O42– → 2CO2 + 2e–

d CH3CH2OH(l) + H2O(l) → CH3COOH(l) + 4H+(aq) + 4e–

e SO32–(aq) + 8H+(aq) + 6e– → H2S(g) + 3H2O(l)

f HNO3(aq) + H+(aq) + e– → NO2 + H2O(l)




7 a C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Oxidation number of C increases from +2 in C2H5OH to +4 in CO2, so this is a redox reaction.

b 2H2O2(aq) → 2H2O(l) + O2(g)

Oxidation number of O decreases from –1 in H2O2 to –2 in H2O and increases from –2 to 0 in O2, so this is a redox reaction.

8 a Al → Al3+ + 3e–

2H+ + 2e– → H2

2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)

b Cu → Cu2+ + 2e–

NO3– + 4H+ + 3e– → NO + 2H2O

3Cu(s) + 2NO3–(aq) + 8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

c SO2(g) + 2H2O(l) → SO42–(aq) + 4H+(aq) + 2e–

MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

5SO2(g) + 2H2O(l) + 2MnO4–(aq) → 5SO4

2–(aq) + 4H+(aq) + 2Mn2+(aq)

d ClO–(aq)+ 2H+ + 2e– → Cl–(aq) + H2O(l)

2I–(aq) → I2(aq) + 2e–

ClO–(aq) + 2I–(aq) + 2H+(aq) → Cl–(aq) + I2(aq) + H2O(l)

9 a Ni(s) → Ni2+(aq) + 2e–

b 2H+(aq) + 2e– → H2(g)

c Ni(s) + 2H+(aq) → Ni2+(aq) + H2(g)

d H+(aq)

10 a Oxidizing agent: Pb2+; reducing agent: Zn

b Oxidizing agent: O2; reducing agent: Mg

c Oxidizing agent: Cu2+; reducing agent: H2

d Oxidizing agent: Cl2; reducing agent: I–

11 a Br2(aq) is the oxidizing agent and Mg(s) is the reducing agent.

b Ag+(aq) is the oxidizing agent and Sn2+(aq) is the reducing agent.

c Cu2+(aq) is the oxidizing agent and Pb(s) is the reducing agent.

d MnO4–(aq) is the oxidizing agent and Fe2+(aq) is the reducing agent.

e H2O(l) is the oxidizing agent and Na(s) is the reducing agent.




12 Zn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s)

13 a reduced

b reduced

c neither oxidized nor reduced

14 a Cu2+(aq) + Cd(s) → Cu(s) + Cd2+(aq)

b There is no reaction.

15 Answers are shaded. Reaction at anode (–)

Reaction at cathode (+) Overall reaction equation Ox. Red.

Ni(s) → Ni2+(aq) + 2e–

Cu2+(aq) + 2e– → Cu(s) Cu2+(aq) + Ni(s) → Cu(s) + Ni2+(aq) Cu2+ Ni

K(s) → K+(aq) + e–

2H2O(l) + 2e– → H2(g) + 2OH–(aq)

2H2O(l) + K(s) → H2(g) + 2OH–(aq) + K+(aq)

H2O K

Fe2+(aq) → Fe3+(aq) + e–

MnO4–(aq) + 8H+(aq) + 5e–→

Mn2+(aq) + 4H2O(l) MnO4

–(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

MnO4– Fe2+

Sn(s) → Sn2+(aq) + 2e–

Sn4+(aq) + 2e– → Sn2+(aq) Sn4+(aq) + Sn(s) → 2Sn2+(aq) Sn4+ Sn

16 Fe: anode, negative electrode, Fe(s) → Fe2+(aq) + 2e–

Sn: cathode, positive electrode, Sn2+(aq) + 2e– → Sn(s)

Electron flow is from Fe to Sn.

Cation flow is from Fe half-cell to Sn half-cell. Sn2+/Sn electrode is made of Sn.

17 a i Zn

ii Zn

iii Zn

b An example is KNO3(aq).

18 The following reaction would occur, dissolving the Cu container:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

19 a At the anode (–): H2O2(aq) → O2(g) + 2H+(aq) + 2e–

At the cathode (+): H2O2(aq) + 2H+(aq) + 2e– → 2H2O(l)

Overall equation: 2H2O2(aq) → 2H2O(l) + O2(g)

b It is used to slow the rate at which the H2O2 reacts with itself and so decomposes.

c The MnO2 acts a catalyst and significantly increases the rate of reaction.

20 a Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)

At the anode (–): Cu(s) → Cu2+(aq) + 2e–

At the cathode (+): Ag+(aq) + e– → Ag(s)




b Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

At the anode (–): Mg(s) → Mg2+(aq) + 2e–

At the cathode (+): 2H+(aq) + 2e– → H2(g)

c No reaction occurs with these reactants.

d Cl2(g) + SnCl2(aq) → SnCl4(aq)

Relevant half equations are:

At the anode (–): Sn2+(aq) → Sn4+(aq) + 2e–

At the cathode (+): Cl2(g) + 2e– → 2Cl–(aq)

e H2S(g) + CuSO4(aq) → H2SO4(aq) + S(s) + Cu(s)

At the anode (–): H2S(g) → S(s) + 2H+(aq) + 2e–

At the cathode (+): Cu2+(aq) + 2e– → Cu(s)

21 Zinc is the only metal in this list that is a sufficiently strong reducing agent to displace metal Q from the solution.

22 The first statement indicates that C2+(aq) reacts with B, so C2+(aq) is a stronger oxidizing agent than B2+(aq).

The second statement indicates that A2+(aq) will not react with C, so C2+(aq) is a stronger oxidizing agent than A2+(aq).

The third statement indicates that B2+(aq) does not react with A(s), so A2+(aq) is a stronger oxidizing agent than B2+(aq).

A partial electrochemical series for these three redox conjugate pairs would therefore be: B2+(aq) + 2e– → B(s)

A2+(aq) + 2e– → A(s)

C2+(aq) + 2e– → C(s)

The order from strongest reducing agent to weakest of these metals is B >A > C.

23 The first experiment indicates that the metal cation is a weaker oxidizing agent than Cu2+.

The second experiment indicates that the metal cation is a sufficiently strong oxidizing agent to oxidize iron metal (i.e. it is between Cu2+ and Fe2+ on the electrochemical series).

The third experiment states that the pH decreases, which means that H+(aq) ions must be formed. The metal cation is a stronger oxidizing agent than H+(aq) and so, by inspecting the electrochemical series, it must be Sn4+(aq). Note that no reaction is observed as Sn4+(aq) is reduced to Sn2+(aq) by the hydrogen gas, and Sn2+(aq) has no colour.

Answer is Sn4+(aq) and so Sn(NO3)4.




24

Voltaic cells Electrolytic cells Spontaneous reactions produce energy. Non-spontaneous reactions require energy. Comprises two half-cells, connecting wires and a salt bridge of KNO3.

Comprises one cell only, connecting wires and a power source.

Oxidation occurs at the anode, reduction at the cathode.

Oxidation occurs at the anode, reduction at the cathode.

The anode has a negative charge, the cathode has a positive charge.

The anode has a positive charge, the cathode has a negative charge.

The polarity of electrodes is determined by the reactions occurring there.

The polarity of electrodes is determined by the power source.

25 a

b Calcium is formed at the negative electrode (cathode) and chlorine is formed at the positive

electrode (anode).

c At anode: 2Cl–(l) → Cl2(l) + 2e–

At cathode: Ca2+(l) +2e– → Ca(l)

d Overall equation: Ca2+(l) + 2Cl–(l) → Ca(s) + Cl2(l)

Chapter 10 Test

Part A: Multiple-choice questions

Question Answer Explanation 1 C Oxidation involves an increase in oxidation number.

+4 –2 +1 –1 +2 –1 0 +1 –2 MnO2 + 4HCl → Mn2+ + 2Cl– + Cl2 + 2H2O Thus, Cl– in HCl was oxidized to Cl2.

2 D In electrolysis, positive ions are attracted to the negative electrode (cathode), where they undergo reduction.

3 C +1 + 2S + (5× –2) = –1 ∴ 2S = +8, S = +4 4 D A more reactive metal will displace a less reactive metal ion from solution. Thus,

Zn is more reactive than Ag+, Co is more reactive than Ag+, but Co is less reactive than Zn2+. This leads to Zn being the most reactive and Ag being the least reactive.




Question Answer Explanation 5 B Sn4+ is reduced to Sn2+ as there is a decrease in oxidation number. Thus, H2SO3

is oxidized to HSO4– (oxidation number increases from +4 to +6). Reducing agents

are oxidized. 6 A In a molten salt, ions are able to move continually. 7 C +1 + (2 × +1) + P + (4 × –2) = 0; P = +5 8 C Positive ions are attracted to the negative electrode and are reduced,

i.e. Mg2+ ions become Mg. 9 C In a reduction reaction, the oxidation number decreases or there is a gain of

electrons. In reaction C the oxidation number of S in SO42– is +6 and the oxidation

number of S in SO32– is +4.

10 B To balance this equation, the two half reactions need to be combined. Oxidation: Ce4+ + e– → Ce3+ Reduction: SO3

2– + H2O → SO42– + 2H+ + 2e–

Overall reaction: 2Ce4+ + SO32– + H2O → SO4

2– + 2H+ + 2Ce3+

Part B: Short-answer questions

1 a Oxidation numbers: Cu+ +1, Cu 0, Cu2+ +2 (1 mark)

b Cu+ → Cu2+ + e– (1 mark)

c Cu+ + e– → Cu (1 mark)

2 Oxidation numbers: Cl– –1, Cl2 0, MnO4– +7, Mn2+ +2

Half-equations: [2Cl– → Cl2 + 2e–] × 5

[MnO4 + 8H+ + 5e– → Mn2+ + 4H2O]

10Cl–(aq) + 2MnO4–(aq) + 16H+(aq) → 2Mn2+(aq) + 5Cl2(g) + 8H2O(l)

Cl– is the reducing agent. (5 marks)




3 a

Bromine formed at +ve electrode.

Potassium formed at –ve electrode. (3 marks)

b Electrons flow through connecting wires.

Ions move through liquid to electrodes and lose or gain electrons. (2 marks)

c At cathode: K+ + e– → K

At anode: 2Br– → Br2 + 2e– (2 marks)

Part C: Data-based question

a Copper would react only with silver nitrate, as Ag+ is a stronger oxidizing agent than Cu2+. (1 mark)

b Copper would undergo oxidation: Cu → Cu2+ + 2e– (2 marks)

c Solution turns blue due to formation of Cu2+ ions and a black solid would form as Ag is deposited.

(2 marks)

Part D: Extended-response question

a i The order of decreasing reactivity is Mg, Fe, Cu, Ag. (1 mark)

ii Oxidation is the loss of electrons:

e.g. Cu(s) → Cu2+ + 2e– (2 marks)




iii Reduction is indicated by a decrease in oxidation number.

Cu2+ + 2e– → Cu(s) ON: +2 0

(2 marks) iv Mg is the strongest reducing agent (the most reactive metal).

We know this because Mg reduces Fe2+, Fe reduces Cu2+ and Cu reduces Ag+. (2 marks)

v Ag+ is the strongest reducing agent because every metal present can reduce it to Ag. (2 marks)

vi A silver coin will not react with MgCl2(aq). It is a very poor reducing agent. (2 marks)

b The voltaic cell will have the Mg electrode as the anode (negative) and Cu electrode as the cathode (positive).

Electrons travel from Mg to Cu.

At the anode: Mg(s) → Mg2+(aq) + 2e–

At the cathode: Cu2+(aq) + 2e– → Cu(s) (4 marks)

c The cathode is negative (is connected to the negative terminal of the power supply). Positive ions are attracted to the cathode and gain electrons (are reduced). In this case copper ions, Cu2+, are attracted to the cathode and copper metal is formed.

The anode is positive (is connected to the positive terminal of the power supply). Negative ions are attracted to the anode and lose electrons (are oxidized). In this case chloride ions Cl– are attracted to the anode and are oxidized to form Cl2.

(5 marks)

Worked solutions: Chapter 10 Oxidation and reduction Cr2O7 2 –(aq) + 14H+(aq ... b No oxidation...

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Transcript of Worked solutions: Chapter 10 Oxidation and reduction Cr2O7 2 –(aq) + 14H+(aq ... b No oxidation...