Topic 9/19 Redox Reactions KEY

9.1 Introduction to Oxidation and Reduction

1. A redox reaction is defined as:A reaction in which one species (chemical) gains/loses electrons

2. Oxidation is loss of electrons and reduction is gain of electrons.

3. Which of the following is not a redox reaction?a. CH4 + 2O2 CO2 + 2H2O

C (-4 +4) oxO (0 -2) red

b. H2SO4 + CuO CuSO4 + H2O

c. 2Al + Fe2O3 Al2O3 + FeFe (+3 0) redAl (0 +3) ox

d. PbO + CO Pb + CO2

Pb (+2 0) redC (+2 +4) ox

4. Give oxidation numbers of each element in each of the following species:a. Fe+2 +2 e. P4 0 i. CH3OH -2,+1,-2,+1 m. KMnO4 +1,+7,-2b. SCl4 +4,-1 f. N2O +1,-2 j. IF6

- +5,-1 n. Na2SO4 +1,+6,-2c. Cr2O7 +7,-2 g. BrO- +1,-2 k. Br2 0 d. VO+2 +4,-2 h. SiO2 +4,-2 l. NH4

+ -3,+1

5. For each of the following transformations, give the initial and final oxidation number of the element in bold and state whether it has been oxidized, reduced, or neither (These are not balanced equations, simply “pieces” of equations showing specific transformations). Remember, it is ok to use ½ in oxidation numbers.

a. Cu+2 CuI +2 +1, red d. NO2 N2O4 +4 +4, neitherb. PH3 HPO2

-2 -3 +1, ox e. CH3OH HCOOH -2 +2, oxc. S2O3

-2 S4O6-2 +2 +2.5, ox

6. Hydrazine (N2H4) and dinitrogen tetroxide (N2O4) react violently to form nitrogen gas and water.a. Write a balanced equation.

2 N2H4 + N2O4 3 N2 + 4 H2O

b. Explain why this is a redox reaction by stating what is being oxidized and what is being reduced.

N2H4 (-2 0), oxN2O4 (+4 0), red

c. This has been suggested as a very “environmentally friendly” source of energy. Explain possible reasons why.

N2 and H2O are very abundant in the atmosphere and are inert gases

7. The cyanide ion, CN–, can form two complex ions with iron ions. The formulas of these ions are

[Fe(CN)6]4- and [Fe(CN)6]3-. What is the oxidation number of iron in the two complex ions?

[Fe(CN)6]4- [Fe(CN)6]3-

A. –4 –3

B. +2 +3

C. +3 +2

D. –3 –4

8. Write equations for:

a. the oxidation of the iron (II) ion to the iron (III) ionFe2+ Fe3+ + 1 e-

b. the reduction of the H+ ion2 H+ + 2 e- H2

c. the oxidation of calcium metalCa Ca2+ + 2 e-

d. the reduction of a sodium ionNa+ + e- Na

e. the reduction of a magnesium ionMg2+ + 2 e- Mg

f. the oxidation of chloride ion2 Cl- Cl2 + 2 e-

9. Consider the reaction between nickel and chlorine and the subsequent solvation of nickel (II) chloride in water:

Ni(s) + Cl2(aq) NiCl2(aq) Ni+2(aq) + 2Cl-

(aq)

therefore:

Ni(s) + Cl2(aq) Ni+2(aq) + 2Cl-

(aq)

The overall “half equation” for the reaction of nickel would be written Ni Ni+2 + 2e-

a. Write the half equation for the reaction of chlorine.Cl2 + 2 e- 2 Cl-

b. What is oxidized? nickelc. What is reduced? chlorined. What is the oxidizing agent? chlorinee. What is the reducing agent? nickel

10. For each equation: a) identify what is oxidized b) identify what is reduced c) identify the oxidizing agent (oxidant) d) identify the reducing agent (reductant)

a. Ca + ZnSO4 CaSO4 + Zn0 +2 +6 -2 +2 +6 -2 0Ca (0 +2) oxidized, reductantZn ( +2 0) reduced, oxidant

b. Mg + Cu(NO3)2 Mg(NO3)2 + Cu0 +2 +5 -2 +2 +5 -2 0Mg (0 +2) ox, reductantCu (+2 0) red, oxidant

c. Al + 3AgNO3 Al(NO3)3 + 3Ag0 +1 +5 -2 +3 +5 -2 0Al (0 +3) ox, reductantAg (+1 0) red, oxidant

11. The following reaction takes place in an acidic, aqueous solution.Zn + NO3

- Zn2+ + N2O 0 +5 -2 +2 +1 -2

a. Balance the equation using the half reaction method.Zn (0 +2), oxN (+5 +1), red

Balance atoms Zn Zn2+

2 NO3- N2O

Balance O Zn Zn2+

2 NO3- N2O + 5 H2O

Balance H Zn Zn2+

2 NO3- + 10 H+ N2O + 5 H2O

Add e- Zn Zn2+ + 2 e-

2 NO3- + 10 H+ + 8 e- N2O + 5 H2O

Equalize e- 4 Zn 4 Zn2+ + 8 e-

2 NO3- + 10 H+ + 8 e- N2O + 5 H2O

Overall eq 4 Zn + 2 NO3- + 10 H+ 4 Zn2+ + N2O + 5 H2O

b. What is the oxidizing agent?NO3

-

c. What is the reducing agent?Zn

12. The following reaction takes place in an acidic, aqueous solution.Cr2O7

2- + C2O42- CO2 + Cr3+

+6 -2 +3 -2 +4 -2 +3a. Balance the equation using the half reaction method.

Cr (+6 +3), redC (+3 +4), ox

Balance atoms C2O42- 2 CO2

Cr2O72- 2 Cr3+

Balance O C2O42- 2 CO2 (already balanced)

Cr2O72- 2 Cr3+ + 7 H2O

Balance H C2O42- 2 CO2

Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O

Add e- C2O42- 2 CO2 + 2 e-

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

Equalize e- 3 C2O42- 6 CO2 + 6 e-

Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O

Overall eq 3 C2O42- + Cr2O7

2- + 14 H+ 6 CO2 + 2 Cr3+ + 7 H2O

b. Describe change in appearance as reaction proceeds.Dichromate ion orange chromium ion greenBubbles due to CO2 production

c. Which species is reduced? Which species is oxidized?Cr2O7

2- (red) 3 C2O42- (ox)

d. What is the oxidation number of chromium in the dichromate ion?+6

13. Use the half-equation method to write balanced equations for:a. Nitric acid being reduced to nitrogen dioxide

HNO3 NO2

HNO3 + H+ + e- NO2 + H2O

b. Hydrogen sulfide being oxidized to sulfurH2S SH2S S + 2 H+ + 2 e-

c. Sulfur dioxide being oxidized to sulfate, while reducing iodine to iodide ionsSO2 + I2 SO4

2- + I-

+4 -2 0 +6 -2 -1

SO2 + I2 + 2 H2O SO42- + 2 I- + 4 H+

14. Using oxidation numbers, deduce the complete balanced equation for the reaction showing all the reactants and products.

a. Ag(s) + NO3–(aq) → Ag+(aq) + NO(g)

0 +5 -2 +1 +2 -2Ag ( 0 +1) oxN (+5 +2) red

3 Ag 3 Ag+ + 3 e-NO3

- + 4 H+ + 3 e- NO + 2 H2O

3 Ag + NO3- + 4 H+ 3 Ag+ + NO + 2 H2O

b. Zn(s) + SO42-(aq) Zn2+(aq) + SO2(g)

0 +6 -2 +2 +4 -2

Zn (0 +2) oxS ( +6 +4) red

Zn Zn2+ + 2 e-SO4

2- + 4 H+ + 2 e- SO2 + 2 H2O

Zn + SO42- + 4 H+ Zn2+ + SO2 + 2 H2O

c. I-(aq) + HSO4-(aq) I2(aq) + SO2(g)

-1 +1 +6 -2 0 +4 -2

I (-1 0) oxS (+6 +4) red

2 I- I2 + 2 e-HSO4

- + 3 H+ + 2 e- SO2 + 2 H2O

2 I- + HSO4- + 3 H+ I2 + SO2 + 2 H2O

d. NO3-(aq) + Zn(s) NH4

+(aq) + Zn2+(aq)+5 -2 0 -3 +1 +2

N (+5 -3) redZn (0 +2) ox

NO3- + 10 H+ + 8 e- NH4

+ + 3 H2O 4 Zn 4 Zn2+ + 8 e-

NO3- + 10 H+ + 4 Zn NH4

+ + 3 H2O + 4 Zn2+

e. I2(aq) + OCl-(aq) IO3-(aq) + Cl-(aq)

0 -2 +1 +5 -2 -1

I (0 +5) oxCl (+1 -1) red

I2 + 6 H2O 2 IO3- + 12 2 H+ + 10 e-

5 OCl- + 10 H+ + 10 e- 5 Cl- + 5 H2O

5 OCl- + I2 + H2O 5 Cl- + 2 IO3- + 2 H+

f. MnO4-(aq) + H2SO3(aq) Mn2+(aq) + SO4

2-(aq)+7 -2 +1 +4 -2 +2 +6 -2

Mn (+7 +2) redS (+4 +6) ox

2 MnO4- + 16 H+ + 10 e- 2 Mn2+ + 8 3 H2O

5 H2SO3 + 5 H2O 5 SO42- + 20 4 H+ + 10 e-

2 MnO4- + 5 H2SO3 2 Mn2+ + 3 H2O + 5 SO4

2- + 4 H+

15. Consider relative reactivity’s of the halogens as well as their colors in molecular form.a. What would you see when chlorine gas is bubbled through aqueous sodium bromide?

Cl is more powerful oxidizing agent (gets reduced easier) than Br

Cl2 + NaBr NaCl + Br2

Yellow/greem red/brown color change

b. What would you see when bromine gas is bubbled through aqueous sodium chloride?Br is less powerful oxidizing agent, cannot take e- away from Cl, no reaction, color stays red/brown

16. Use the two reactivity series given to predict whether reactions will occur between the following reactants and write equations where relevant.

a. CuCl2(aq) + Ag(s) NR (Cu is stronger reducing agent)

b. Fe(NO3)2(aq) + Al(s) Al(NO3)3 + Fe (Al is stronger reducing agent, causing Fe to become oxidized)

c. NaI(aq) + Br2(aq) 2 NaBr + I2 (Br is stronger oxidizing agent, able to oxidize I)

d. KCl(aq) + I2(aq) NR (Cl is more reactive, stays in ion form)

17. The following information is given about reactions involving the metals X, Y and Z and solutions of their sulfates.

X(s) + YSO4(aq) no reactionZ(s) + YSO4(aq) Y(s) + ZSO4(aq)

When the metals are listed in decreasing order of reactivity (most reactive first), what is the correct order?

a. Z>Y>X Z is able to reduce Y but X cannotb. X>Y>Zc. Y>X>Zd. Y>Z>X

18. Which equations represent reactions that occur at room temperature?1. 2Br-(aq) + Cl2(aq) 2Cl-(aq) + Br2(aq)2. 2Br-(aq) + I2(aq) 2I-(aq) + Br2(aq)3. 2I-(aq) + Cl2(aq) 2Cl-(aq) + I2(aq)

a. I and II onlyb. I and III only I) Cl2 more reactive than Br III) Cl2 more reactive than Ic. II and III onlyd. I, II, and III

19. A bag of ‘road salt’, used to melt ice and snow from roads, contains a mixture of calcium chloride CaCl2, and sodium chloride, NaCl. A 2.765 g sample of the mixture was analyzed by first converting all the calcium into calcium oxalate, CaC2O4. This was then dissolved in H2SO4, and titrated with 0.100 mol dm -3 KMnO 4 solution. The titration required 24.65 cm 3 of KMnO 4(aq) and produced Mn2+

(aq), CO2(g), and H2O(l).a. What would be observed at the equivalence point of the titration?

The solution changes from purple to colorless (since MnO4- is purple and Mn2+ ions

are colorless)b. Write the half-equation for the oxidation reaction, starting with C2O4

2-.C2O4

2- 2 CO2 + 2 e-

c. Write the half-reaction for the reduction reaction, starting MnO4-.

MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O

d. Write the overall equation for the redox reaction.5 C2O4

2- + 2 MnO4- + 16 H+ 10 CO2 + 2 Mn2+ + 8 H2O

e. Determine the number of moles of C2O42-.

c=mV 0.100∗0.02465=0.002465 mols MnO4

-

0.002465 mols x 5/2 = 0.0061625 mols C2O42-

f. Deduce the number of moles of Ca2+ in the original sample.Ca2+: C2O4

2- 1:1 0.0061625 moles Ca2+

g. What was the percentage by mass of CaCl2 in the road salt?m(CaCl2) = 0.0061625 x 110.98 g mol- = 0.684 g/2.765 g x 100 = 24.7%

20. Alcohol levels in blood can be determined by a redox titration with potassium dichromate, K2Cr2O7, according to the following equation:

C2H5OH(aq) + 2 Cr2O72- (aq) + 16 H+(aq) 2 CO2(g) + 4 Cr3+ (aq) + 11 H2O(l)

a. Determine the alcohol percentage in the blood by mass if a 10.000 g sample of blood requires 9.25 cm 3 of 0.0550 mol dm -3 K 2Cr2O7 solution to reach equivalence.

m(K2Cr2O7) = 0.0550 mol dm-3 x 0.00925 dm3 = 5.0875 x 10-4 mols K2Cr2O7

5.0875 x 10-4 mols / 2 = 2.54 x 10-4 mols C2H5OH

2.54 x 10-4 mols x 46.08 g mol-1 = 0.0117 g / 10.000 g x 100 = 0.117%

b. Describe the change in color that would be observed during the titration.K2Cr2O7 is orange, Cr3+ is dark green

9.2/19.1 Electrochemical Cells

Voltaic Cells (SL)

21. Use the metal reactivity series given earlier to predict which electrode will be the anode and which will be the cathode when the following half-cells are connected. Write half-equations for the reactions occurring at each electrode.

a. Zn/Zn2+ and Fe/Fe2+

Zn/Zn2+ - oxidized at anode Zn(s) Zn2+ (aq) + 2 e-

Fe/Fe2+ - reduced at cathode Fe2+ (aq) + 2 e- Fe(s)

b. Fe/Fe2+ and Mg/Mg2+

Mg/Mg2+ - oxidized at anodeMg(s) Mg2+ (aq) + 2 e-

Fe/Fe2+ - reduced at cathode Fe2+ (aq) + 2 e- Fe(s)

c. Mg/Mg2+ and Cu/Cu2+

Mg/Mg2+ - oxidized at anodeMg(s) Mg2+ (aq) + 2 e-

Cu/Cu2+ - reduce at cathode Cu2+ (aq) + 2 e- Cu(s)

22. Draw a voltaic cell with one half-cell consisting of Mg and a solution of Mg2+ ions and the other consisting of Zn and a solution of Zn2+ ions. Label the electrodes with name and charge, the direction of electron and ion movement and write equations for the reactions occurring at each electrode.

23. Predict what would happen if an iron spatula was left in a solution of copper sulfate overnight. Fe is a stronger reducing agent than Cu, therefore will be oxidized.

Iron spatula will slowly dissolve as it is being oxidizedCopper metal will precipitate out. The solution will become less blue as Cu2+ leaves the solution

Standard Electrode Potentials (HL)

24. Consider a voltaic cell that utilizes the following reaction at standard conditions:Mg + Sn2+ Mg2+ + Sn

a. Draw and label the voltaic cell including anode, cathode, salt bridge, voltmeter, direction of electron flow, cations in each half cell.

b. Write the oxidation half reaction.Mg Mg2+ + 2 e- EΘ = -2.37 V

c. Write the reduction half reaction.Sn2+ + 2 e- Sn EΘ = -0.14 V

d. Calculate the standard cell potentialEΘcell = EΘ

reduction cell - EΘoxidation cell

EΘSn - EΘ

Mg

-0.14 – (-2.37)+2.23 V

e. Is the reaction spontaneous as written? Calculate ∆ G. Mg + Sn2+ Mg2+ + Sn EΘ is postivie number, spontaneouse

ΔGΘ = -nF EΘ = -(2)(96500)(+2.23) = 430390 J mol- = 430.4 kJ mol-

f. How would an increase in the concentration of Sn2+ affect the voltage? ExplainIncrease Sn2+ you increase the concentration of reactant pushing the reaction forward to make more product = increase in voltage.

g. If a 5.0 amp current is supplied with sufficient voltage to reverse the reaction for 8.5 hours, what would be the mass change for the magnesium electrode?

C = Axs = (5.0)(8.5 hrs x 3600 s) = 153,000 C

F = 153000 C / 96500 C mol- = 1.59 mol e-

Mole Mg = 1.59 mol e- x ½ = 0.79 mol Mg

g Mg 0.79 mol x 24.39 g mol- = 19.34 g Mg

25. Which is a feature of the standard hydrogen electrode?A. hydrogen gas at 1.01×105 Pa (1 atm) pressureB. 1.0 mol dm-3 sulfuric acidC. a temperature of 273 KD. a magnesium electrode

26. The standard electrode potentials for two half-cells involving iron are given below.

Fe2+ (aq) + 2e- → Fe(s) Eο = –0.44 V (oxidized)Fe3+ (aq) + e- → Fe2+ (aq) Eο = +0.77 V (reduced)

What is the equation and the cell potential for the spontaneous reaction that occurs when the two half-cells are connected?

A. 3Fe2+ (aq) → Fe(s) + 2Fe3+ (aq) Eο = +1.21 V

B. Fe2+ (aq) + Fe3+ (aq) → 2Fe(s) Eο = +0.33 V

C. Fe(s) + 2Fe3+ (aq) → 3Fe2+ (aq) Eο = +0.33 V

D. Fe(s) + 2Fe3+(aq) → 3Fe2+ (aq) Eο = +1.21 V

27. Consider the following reactions.

Cu2+ (aq) + 2e– Cu(s) Eο = +0.34 V

Mg2+ (aq) + 2e– Mg(s) Eο = –2.36 V

Zn2+ (aq) + 2e– Zn(s) Eο = –0.76 VWhich statement is correct?A. Cu2+ (aq) will oxidize both Mg(s) and Zn(s).B. Zn(s) will reduce both Cu2+ (aq) and Mg2+ (aq).C. Mg2+ (aq) will oxidize both Cu(s) and Zn(s).D. Cu(s) will reduce both Mg2+ (aq) and Zn2+ (aq).

28. Consider the standard electrode potentials of the following reactions.

Cr3+ (aq) + 3e– →Cr(s) –0.75 V (oxidized)Cd2+ (aq) + 2e– → Cd(s) –0.40 V (reduced)

What is the value of the cell potential (in V) for the following reaction?2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd(s)

A. –0.35B. –1.15C. +0.30D. +0.35

29. From the given standard electrode potentials which statement is correct?

Ca2+ (aq) + 2e– Ca(s) EӨ = –2.87 V

Ni2+ (aq) + 2e– Ni(s) EӨ = –0.23 V

Fe3+ (aq) + e– Fe2+ (aq) EӨ = +0.77 VA. Ca2+ (aq) can oxidize Ni(s)B. Ni2+ (aq) can reduce Ca2+ (aq)C. Fe3+ (aq) can oxidize Ni(s)D. Fe3+ (aq) can reduce Ca2+ (aq)

30. The standard electrode potential for the reaction Al(s) + Cr3+(aq) Al3+(aq) + Cr(s) is 0.92V at 298K. What is the standard free-energy change for this reaction?

Ox: Al Al3+ + 3 e-

Red: Cr3+ + 3 e- Cr

ΔGΘ = -nF EΘ = -(3)(96500)(0.92) = - 2.7 x 105 J = -270 kJ mol-

31. (a) The apparatus shown above may be used to carry out a redox reaction.

(i) State the function of the salt bridge.Completes circuitIonic conductorMaintains potential difference

(ii) Write a half-equation for the oxidation reaction.

Zn Zn2+ + 2 e-

(iii) The above reactions are carried out under standard conditions. State what the standard conditions are for the cell.

T = 298 K concentration = 1 mol dm-3 electrode – pure metal

(iv) Using the Data Booklet, calculate the cell potential for the above cell.

Zn2+ + 2 e- Zn EΘ = -0.76 V

Cu2+ + 2 e- Cu EΘ = +0.34V

EΘcell = EΘ

Cu - EΘZn = +0.34 – (-0.76) = +1.10 V

(v) State and explain what happens to the concentration of the copper(II) ions when the cell is producing an electric current.

Decreases as Cu2+ ions are being deposited onto Cu mental as a precipitate

(vi) State two observations that could be made if the zinc rod were placed in a solution of copper(II) ions.

Blue solution gets clear as Cu2+ disappears

A

D irec tio n o f e le c tro n flo w

S a lt b rid g e

C u (s)Z n (s)

Z n (a q )2+

2+C u (aq )

Cu gets deposited onto Zn metal (pink brown color)Exothermic reaction gets hotter

(b) The standard electrode potentials for three electrode systems are given below.

Ti3+(aq) + e– Ti2+(aq) Eο = –0.37 V

Fe3+(aq) + e– Fe2+(aq) Eο = +0.77 V

Ce4+(aq) + e– Ce3+(aq) Eο = +1.45 V

(i) Using the data above, deduce which species is the best reducing agent, giving a reason in terms of electrons for your answer.

Ti2+ best reducing agent as stronger reactor, more willing to remove an e- than take one. (Ti3+ least willing to gain e- hence low negative numbers)

(ii) Write an equation, including state symbols, for the overall reaction with the greatest cell potential.

Ce4+(aq) + Ti2+(aq) Ce3+(aq) + Ti3+(aq)

(iii) State and explain the sign of ∆Gο for the reaction in (b) (ii).

∆GΘ negative because positive cell potential = spontaneous reaction

EΘcell = +1.45 – (-0.37) = +1.82 V

(c) (i) State the name of a solution that would produce only hydrogen and oxygen when electrolyzed using platinum electrodes.

Sodium hydroxide

(ii) Draw a diagram of apparatus that would allow the gases produced in the reaction in (c) (i) to be collected separately. Annotate your diagram to show the polarity of each electrode and the names and relative volumes of each gas.

Electrolytic Cells (SL)

32. Draw a diagram of the electrolysis of molten sodium chloride.a. Draw and label the parts of the electrolytic cellb. On the diagram, identify the anode and the cathodec. On the diagram, indicate the direction of electron flowd. On the diagram, indicate the direction of flow of sodium and chloride ions

33. Describe how electric current is carried in a molten salt.

melted ions are free to move when voltage applied. The positive ions move to negative electrode (cathode) and negative ions move to positive electrode (anode). Electrons are released to positive electrode by negative ions (where reduction occurs) and electrons are accepted at negative electrode (where oxidation occurs)

Electrolysis (HL)

34. Consider the passage of electricity through sodium chloride.a. Is the bonding in sodium chloride metallic, ionic, or covalent?

Ionic

b. State whether the following would conduct electricity:1. solid sodium chloride No2. molten sodium chloride yes 3. aqueous sodium chloride yes

c. Give names for the following:1. passing an electric current through a solution containing ions - electrolysis2. the liquid containing ions - electrolyte3. the solid conductors put into the solution - electrodes

35. Aqueous solutions of AgNO3, Cu(NO3)2 and Cr(NO3)3 are electrolyzed using the same quantity of electricity. How do the number of moles of metal formed compare?

A. Ag = Cu = CrB. Ag > Cu > Cr charge of ion equate to the amount of electricity neededC. Ag < Cu < CrD. Cu > Ag > Cr

36. Metallic tin can be produced by the electrolysis of a molten salt containing Sn2+ ions. Which change(s) would double the amount of tin produced?

I. Doubling the current passed during electrolysisII. Doubling the time used for electrolysisIII. Using Sn4+ ions instead of Sn2+ ions

A. I onlyB. II onlyC. I and II only 2 other factors affecting product is: amount of current and

duration doubling these would double to product (I and II). III would decrease the product 2x

D. I, II and III

37. Which of the following factors affect the amount of product formed during electrolysis?I. The current usedII. The duration of electrolysisIII. The charge on the ion

A. I and II onlyB. I and III onlyC. II and III onlyD. I, II and III

38. Aqueous solutions containing different concentrations of NaCl were electrolysed using platinum electrodes. What is the major product at the positive electrode in each case?

0.001 mol dm-3 NaCl(aq) 1.0 mol dm-3 NaCl(aq)

A. H2 Na

B. H2 H2

C. O2 Cl2

D. Cl2 O2

At low concentration of Cl-, water (OH-) is going to be discharged creating O2 as a product. At high concentration of Cl-, Cl- is discharged creating Cl2 as a product

39. Which statement is correct about the electrolysis of copper(II) sulfate solution using graphite electrodes?

A. A colourless gas is produced at the negative electrode.B. The electrolyte does not change colour. – it does change color (loses blue)C. The negative electrode decreases in mass. – Cu is ppt out therefore gaining massD. A colourless gas is produced at the positive electrode.

O2 is produced when OH- is discharged

40. A metallic object is electroplated with copper using a solution of copper(II) sulfate. Which statement is correct?

A. The positive electrode increases in mass.B. The concentration of Cu2+ ions in the solution decreases.C. Reduction occurs at the positive electrode.D. The reaction occurring at the negative electrode is Cu2+ + 2e- Cu.

Negative electrode is cathode41. In the electrolysis of acidified water, if 8.4 cm3 of hydrogen gas is evolved, what volume of

oxygen gas is evolved?A. 4.2 cm3 1:2 ratioB. 8.4 cm3

C. 12.6 cm3

D. 16.8 cm3

42. If a quantity of electricity is passed through a molten sample aluminum chloride, it is found that 0.2 moles of chlorine are formed. What is the mass of aluminum formed at the other electrode?

AlCl3 Al3+ + 3 Cl-

Cathode: Al3+ + 3 e- Al

Anode: 2Cl- Cl2 + 2 e-

If 0.2 mole of Cl2 gas is formed at cathode, then this can be used to determine the total amount of electrons released in the oxidation of Cl-

n(e-) = 2 x 0.2 mol = 0.4 mol e-

n(Al) = n(e-)/3 = 0.4/3 = 0.13 mol

m(Al) = 0.13 mol x 26.98 g mol-1 = 3.5g

43. Two electrolytic cells are connected in series so that the same current flows through both cells for the same length of time.

+ - - + S n S O (aq )4 C u S O (aq )4

The amount of tin deposited is 0.01 mol. How much copper is deposited?A. 0.005 molB. 0.01 mol same current, same time lapse, same ion concentration, same amt ppt

formedC. 0.02 molD. 0.05 mol

44. (a) In one experiment involving the electrolysis of molten sodium chloride, 0.1 mol of chlorine was formed. Deduce, giving a reason, the amount of sodium formed at the same time.

0.2 mol Na

Overall equations: 2 Na+ + 2 Cl- 2 Na + Cl2 (2:1 ratio)

(b) In another experiment involving the electrolysis of molten sodium chloride, the time of the electrolysis was halved and the current increased from 1 amp to 5 amp, compared to the experiment in (a). Deduce the amount of chlorine formed, showing your working.

Charge = amount of product formed

C1 = A1 x t1 C2 = C1 x A2/A1 x t2/t10.1(1/2)(5/1) = 0.25 mols

C2 = A2 x t2

(c) If dilute aqueous sodium chloride is electrolyzed, a different product is obtained at each electrode. Identify the product formed at each electrode and write an equation showing its formation.

Negative electrode water (H+) reduced 2 H2O + 2 e- H2 + 2 OH-

Positive electrode water (OH-) oxidized because [Cl-] is low

2 H2O 4 H+ + O2 + 4 e-

45. Two copper strips X and Y are placed in an aqueous solution of copper(II) sulfate and electrolyzed for a certain time. X was then dried and weighed.

(i) State and explain what would happen to the mass of X.

A

– +

X Y

C u S O (aq )4

Mass would increase since negative electrode (cathode), reduction would occure

Cu2+ + 2 e- Cu deposited

(ii) State two ways in which the change in the mass of X could be increased.

Increase in time and current