electrochem tutorial solutions

2011 DHS Year 6 H2 Chemistry

Anode (+) Cathode (–)

+ – Electron Flow

Electrolyte

Anions

Cations

TUTORIAL: ELECTROCHEMISTRY

Differences between an Electrochemical and an Electrolytic cell

Electrochemical cell Electrolytic cell

Diagram Light Bulb or Voltmeter Present

No Light Bulb or Voltmeter Present

Battery No Battery Battery required

Emf Generated by the cell Supplied to the cell by battery

Energy conversion

Chemical → Electrical Electrical → Chemical

Cathode Reduction occur Positive (+)

Reduction occur Negative (–) [due to connection to battery]

Anode Oxidation occur Negative (–)

Oxidation occur Positive (+)

Electron flow From anode to cathode From anode to cathode

Cathode (+)

V

Salt Bridge

Anode (–)

Voltmeter


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Part I : Electrochemical Cell 1. Method to write cell notation:

� Determine the which half–cell undergoes reduction and oxidation

� Draw the salt bridge,

� Reduction half–cell to be written on the right

� Oxidation half–cell to be written on the left

� The species with the higher oxidation number to be written next to the salt bridge

� Include phase boundary, or use comma ‘,’

� Electrode always at the extreme ends

� Include state symbols for all species, including the electrodes

2. Calculate Eθθθθ

cell:

� Eθθθθcell

= Eθθθθred

– Eθθθθox

3. Equation Arrows:

“Anode/cathode/oxidation/reduction reactions”, “half–equation”, “Overall reaction” or

“Overall chemical equation”

Use “ →→→→ ” , include state symbols

Use will be penalised (0 marks)

4. Draw labelled diagram to measure standard electrode potential using SHE

– “Standard qn”!

• Labels: anode, cathode, polarity, ions/metal/gas, electrode

• Standard conditions: Aqueous solution – 1 mol dm–3

Gas – 298 K or 25°C, 1 atm

• Salt bridge: open end, not closed

• Gas arrow always points into “ →→→→ ” gas column not out!

• Direction of electron flow: only if asked for

• Use ruler!


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1. Given the following sets of half equations, for the spontaneous reaction that will occur: (i) Write the cell diagram (ii) Write the overall cell reaction

(iii) Calculate Eθ cell [+1.10V, +2.51V, +0.98V, +0.70V]

(a) Given

Zn2+ + 2e � Zn Eθ = –0.76 V

Cu2+ + 2e � Cu Eθ = +0.34 V

/Cu)(CuθE +2 is more positive, Cu2+ is more readily reduced.

Therefore: Eөcell = Eө red – Eөoxid = +0.34 – (– 0.76) = + 1.10 v

Reduction: Cu2+ + 2e → Cu –––(1)

Oxidation: Zn → Zn2+ + 2e –––(2)

Cell diagram: Zn (s) Zn2+ (aq) Cu2+ (aq) Cu (s) Overall Cell equation: (1) + (2)

Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq)

(b) Given

Mn2+ + 2e � Mn Eθ = –1.18 V

Cr2O72– + 14H+ + 6e � 2Cr3+ + H2O Eθ = +1.33 V

)/CrO(Crθ

3-272E +

is more positive, Cr2O72– is more readily reduced.

Eөcell = Eө red – Eөoxid = +1.33 – (–1.18) = +2.51 V Therefore

Reduction: Cr2O72– + 14H+ + 6e → 2Cr3+ + H2O –––(1)

Oxidation: Mn → Mn2+ + 2e –––(2)

Cell diagram: Mn (s) Mn2+ (aq) Cr2O72– (aq), Cr3+ (aq) Pt (s)

Overall Cell equation: (1) + (2)x 3

3Mn (s) + Cr2O72– (aq) + 14H+ (aq) → 3Mn2+ (aq) + 2Cr3+ (aq) + 7H2O (l)

Ered

Eox

Ered

Eox


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(c) Given

Sn4+ + 2e � Sn2+ Eθ = +0.15 V

2H2O + 2e � H2 + 2OH– Eθ = –0.83 V

)/Sn(Snθ

24E ++is more positive, Sn4+ is more readily reduced.

Eөcell = Eө red – Eөoxid = +0.15 – (–0.83) = +0.98 V ⇒ Eөcell >0 reaction is feasible

Reduction: Sn4+ + 2e → Sn2+ –––(1)

Oxidation: 2OH– + H2 → 2H2O + 2e –––(2)

Cell diagram: Pt (s) H2 (g) H2O (aq) Sn4+ (aq), Sn2+ (aq) Pt (s)

Overall Cell equation: (1) + (2)

Sn4+ (aq) + 2OH– (aq) + H2 (g) → Sn2+ (aq) + 2H2O (l) (d) Given

Fe3+ + e� Fe2+ Eθ = +0.77 V

PbO2 + 4H+ + 2e � Pb2+ + 2H2O Eθ = +1.47 V

)/Pb(PbOθ

22E +

is more positive, PbO2 is more readily reduced.

Eөcell = Eө red – Eөoxid = +1.47 – 0.77 = +0.70 V ⇒ Eөcell >0 reaction is feasible

Reduction: PbO2 + 4H+ + 2e → Pb2+ + 2H2O –––(1)

Oxidation: Fe2+ → Fe3+ + e ––– (2)

Cell diagram: Pt (s) Fe2+ (aq), Fe3+ (aq) PbO2 (s) Pb2+ (aq) Pt (s) Overall Cell equation: (1) + (2) x 2

PbO2 (s) + 4H+ (aq) + 2Fe2+ (aq) → Pb2+ (aq) + 2H2O (l) + 2Fe3+ (aq)

Ered

Eox

Ered

Eox


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2. N99/I/3 (a) Draw a labeled diagram of the standard hydrogen electrode and describe how it may be used to measure the standard electrode potential for Fe3+ (aq)/ Fe2+ (aq). The standard hydrogen electrode potential for Fe3+/Fe2+ is measured with reference to the standard hydrogen electrode which has a potential of 0.00 V The standard Fe3+/Fe2+ half–cell is connected to the standard hydrogen electrode by a salt bridge, which completes the circuit by allowing the passage of electricity between electrodes but physically separating the two half–cells.

From the voltmeter reading at 25oC, Eθ

cell is found to be 0.77 V. Since reduction occurs at the Fe3+/Fe2+ half–cell, the standard electrode potential for Fe3+/Fe2+ is therefore +0.77V. (b) By calculating the relevant oxidation number, determine which elements undergo oxidation,

and which undergo reduction, during the following reactions.

(i) 2HNO2 + 2HI → 2NO + I2+ 2H2O

Nitrogen undergoes reduction (O.N. decreases from +3 in HNO2 to +2 in NO)

Iodine undergoes oxidation (O.N. increases from –1 in HI to 0 in I2)

(ii) NO2 + NaClO3 → NaNO3 + ClO2

Nitrogen undergoes oxidation (O.N. increases from +4 in NO2 to +5 in NO3–)

Chlorine undergoes reduction (O.N. decreases from +5 in ClO3

– to +4 in ClO2)

+3 –1 +2 0

+4 +5 +5 +4

platinised Pt (s)


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(c) The oxidation number of vanadium in a complex ion was determined as follows. A 0.013 mol sample of the complex was dissolved in water and the solution made up to 100 cm3. A 10 cm3 portion of this solution required 20.8cm3 of 0.025 mol dm–3 KMnO4 to oxidise all the vanadium to the +5 state.

Use these data to calculate the original oxidation number of the vanadium. [+3]

Concentration of vanadium complex =

1000

100

013.0 = 0.130 mol dm–3

No of mole of complex in 10 cm3 = 130.01000

10× = 1.30 x 10–3

No of mole of KMnO4 in 20.8 cm3 = 025.0

1000

8.20×

= 5.20 x 10–4

Ratio of complex : KMnO4 ⇒ 1.30 x 10–3 : 5.20 x 10–4

⇒ 1 : 0.4 ⇒ 5 : 2 No of electrons gained by KMnO4 = 2 x 5 = 10

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

∴ no. of electrons lost by 5 mol of complex = 10

⇒ no of electrons lost by 1 mol of complex = 25

10=

∴original oxidation state of V = +5–2 = +3


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Reduction

3. Predict which of the following reactions is expected to proceed under standard conditions.

(a) NO3– (aq) + 3H+(aq) + Cu(s) → HNO2 (aq) + H2O (l) + Cu2+ (aq) [+0.60V]

From Data Booklet:

NO3– + 3H+ + 2e � HNO2 + H2O Eθ = +0.94 V

Cu2+ + 2e � Cu Eθ = +0.34 V

Since question state that reduction have to occur from NO3– to HNO2, E

θ

red = +0.94 V

Thus, Eθ

oxd = +0.34 V

Eθ

cell = Eθ

red – Eθ

oxd = +0.94 – (+0.34) = +0.60 V ⇒ Eөcell >0 reaction is feasible

(b) I2 (s) + 2Fe2+ (aq) → 2I

– (aq) + 2Fe3+ (aq) [–0.23V]

From Data Booklet:

Fe3+ + e � Fe2+ Eθ = +0.77 V

I2 + 2e � 2I– Eθ = +0.54 V

Since question state that reduction have to occur from I2 to I–, Eθ

red = +0.54 V

Thus, Eθ

oxd = +0.77 V

Eθ

cell = Eθ

red – Eθ

oxd = +0.54 – 0.77 = –0.23V ⇒ Eөcell<0 reaction is not feasible

∴reaction is not feasible

Steps for Prediction of the Feasibility of a Chemical Reaction:

Step 1: Identify the reactants and its corresponding products.

Step 2: For the reactant undergoing reduction, assign its Eθ value from the Data

Booklet as Eθθθθ

red.

Step 3: For the reactant undergoing oxidation, assign its Eθ value from the Data

Booklet as Eθθθθ

ox.

Step 4: Calculate the overall Eθ

cell value using formula: Eθθθθ

cell = Eθθθθ

red - Eθθθθ

ox

Step 5: Check the sign of Eθ

cell. If Eθ

cell > 0, then the reaction is feasible. If Eθ

cell < 0,

then the reaction is not feasible.

Reduction Oxidation

Oxidation


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Reduction Oxidation

Reduction

Oxidation

(c) Cr2O72– (aq) +14H+ (aq) + 6Cl– (aq) → 2Cr3+ (aq) + 7H2O (l) + 3Cl2 (g) [–0.03V]

From Data Booklet:

Cl2 + 2e � 2Cl– Eθ = +1.36 V

Cr2O72– + 14H+ + 6e � 2Cr3+ + 7H2O Eθ = +1.33 V

Since question state that reduction have to occur from Cr2O72– to Cr3+, Eθ

red = +1.33 V

Thus, Eθ

oxd = +1.36 V

Eθ

cell = Eθ

red – Eθ

oxd = +1.33 – 1.36 = –0.03 V ⇒ Eөcell <0 reaction is not feasible

(d) 2V3+ (aq) + Cu2+ (aq) + 2H2O (l) → 2VO2+ (aq) + 4H+ (aq) + Cu (s) [0.00V]

From Data Booklet:

Cu2+ + 2e � Cu Eθ = +0.34 V

VO2+ + 2H+ + e � V3+ + H2O Eθ = +0.34 V

Eθ

cell = Eθ

red – Eθ

oxd = +0.34 – 0.34 = 0.00 V

⇒ Since Eөcell = 0.00 V reaction is feasible but will proceed until equilibrium is reached.

⇒ Overall eqn: 2V3+ (aq) + 2H2O (l) + Cu2+ (aq) � 2VO2+ (aq) + 4H+ (aq) + Cu (s)

4. The standard redox potential for hydrazine in acidic solution

N2 (g) + 5H+(aq) + 4e → N2H5+ (aq)

is –0.17V. By using this, and any other necessary information from your Data Booklet, predict

the outcome of the reaction of chlorine with hydrazine in acidic solution and write a balanced equation for the overall reaction. [+1.53V] From Data Booklet:

Cl2 (g) + 2e � 2Cl– (aq) Eθ = +1.36 V

Given in question:

N2 (g) + 5H+(aq) + 4e → N2H5+ (aq) Eθ = –0.17 V

Since question mentioned chlorine reacting with hydrazine

∴ Eθ

red = +1.36 V

Eθ

oxd = –0.17 V

Eθ

cell = Eθ

red – Eθ

oxd = +1.36 – (–0.17) = +1.53 V ⇒ Eөcell >0 reaction is feasible


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Reduction: Cl2 + 2e → 2Cl– –––(1)

Oxidation: N2H5+ (aq) → N2 (g) + 5H+ (aq) + 4e –––(2)

Overall equation: (1) x 2 + (2)

2Cl2 (g) + N2H5+ (aq) → 4Cl– (aq) + N2 (g) + 5H+ (aq)

5. A cell consisting of a Cu2+ (aq) Cu(s) half cell and a Fe3+ (aq), Fe2+ (aq) Pt (s) half–cell is shown below, using the conventional notation:

Cu (s) Cu2+ (aq) Fe3+ (aq), Fe2+ (aq) Pt (s)

(a) Predict the effect on the standard potential, Eθ of this cell if aqueous sodium hydroxide is

added to the Cu2+ (aq) Cu (s) half–cell.

Cu2+(aq) + 2e � Cu (s) Eθ

oxid –––(1)

• Addition of OH– causes precipitation of Cu2+ as Cu(OH)2 hence decreasing [Cu2+]

• By Le Chatelier’s Principle, the equilibrium in (1) shift left to increase the [Cu2+]

• Eθθθθ

oxid will become less positive/decrease, oxidation is favoured

• Since Eθ

cell = Eθ

red – Eθ

oxd

Overall Eθθθθ

cell will increase / become more positive.

(a)** Predict the effect on the standard potential, Eθ of this cell if aqueous potassium iodide is added to the Fe3+ (aq) , Fe2+ (aq) half–cell.

(b) It is not possible to use Eθ values reliably to decide whether a chemical reaction will occur: Suggest why it is nevertheless probable that the reaction for which you have written an in (a) will proceed if performed in a test–tube.

Two reasons why the reaction will proceed:

1. Reaction is kinetically feasible though not thermodynamically feasible.

2. Reaction takes place under non–standard conditions

(c) Suggest a replacement half–cell for Cu2+ (aq) Cu (s) which would reverse the direction of the

electron flow in the Fe3+ (aq), Fe2+ (aq) Pt (s) half–cell. Your answer needs to state both the electrode and reagents of your new half–cell.

Ag electrode dipped in AgNO3 solution Reason: From Data Booklet:

Ag+ + e � Ag (+0.80V)

Fe3+ + e � Fe2+ (+0.77V)

Cu2+ + e � Cu (+0.34V)

From the Eθ value, to reverse the direction of electron

flow, there is a need to cause oxidation to take place

at the Fe3+ (aq), Fe2+ (aq) Pt (s) half–cell.

⇒ Choose half–cell with Eθ

value > +0.77 V, so that

Eθ

cell will be > 0.

oxidation

reduction


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6. A simple rechargeable cell may be constructed by dipping two lead electrodes into aqueous lead (II) nitrate and passing a current for a few minutes. During the process, lead (IV) oxide is deposited on one of the electrode. When the power source is disconnected and a bulb is connected across the two electrodes, the bulb lights for a time as the cell discharges.

(a) By reference to the Data Booklet, choose two half–cell equations to construct the full equation

for the reaction that occurs during discharge. Calculate the value of Eθ for this cell reaction. [+1.60V]

During charging the Pb2+ is being oxidised to PbO2. When the power source is disconnected, the cell will discharge and PbO2 will be reduced to Pb2+

From the Data Booklet For reduction:

PbO2 + 4H+ + 2e � Pb2+ + 2H2O +1.47 V ––– (1)

Pb4+ + 2e � Pb2+ +1.69 V

(The first half–equation is selected since the question mentioned lead (IV) oxide) For oxidation:

Pb2+ + 2e � Pb – 0.13 V ––– (2)

During discharge PbO2 will be reduced to Pb2+ (NB: reduction of NO3

– will not take place.)

∴∴∴∴ Eθ

cell = Eθ

red – Eθ

oxd = +1.47 – (–0.13) = +1.60 V

(1) + (2):

PbO2 + 4H+ + 2e → Pb2+ + 2H2O +1.47 V

Pb → Pb2+ + 2e –0.13 V

PbO2 (s) + 4H+ (aq) Pb (s) → 2Pb2+ (aq) + 2H2O (l) (b) In a lead–acid battery, similar reactions takes place but the electrolytes of dilute sulfuric acid

causes the lead (II) ions to be precipitated as PbSO4 (s), which coat the electrode. The e.m.f. of the cell is 2.0V.

Explain the difference between this e.m.f. and the Eθ calculated in (a) by reference to the concentrations of relevant aqueous ions.

• When Pb2+ is precipitated out as PbSO4, the [Pb2+] decreases

• Hence the equilibrium of the reaction will shift right according to Le Chatelier's principle to increase [Pb2+]

PbO2 + 4H+ + 2e � Pb2+ + 2H2O Eθ

red

• In addition, the increase in [H+] due to presence of H2SO4 further shifts the equilibrium to the right.

• Eθθθθ

red will increase and since Eθθθθ

cell = Eθ

red – Eθ

oxd, new Eθ

cell value will increase


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7. (a) Describe the components in any one fuel cell and the reactions taking place at the electrodes. Suggest one advantage of fuel cells as source of energy.

Anode: Graphite impregnated with finely powdered Pd catalyst Cathode: Graphite impregnated with finely divided cobalt (II) oxide and Pt as catalyst

Electrolyte Aqueous KOH

Anode reaction 2H2(g) + 4OH–(aq) → 4H2O (l) + 4e

Cathode reaction O2 (g) + 2H2O (l) + 4e → 4OH– (aq)

Overall reaction 2H2 (g) + O2 (g) → 2H2O (l)

Advantages: – Pollution free – High power to mass ratio – Highly efficient (75% efficiency) – Easy maintenance

Any One

Hydrogen Fuel Cell used in rockets to propel space craft into orbit

Futuristic MP3 player from Toshiba which make use of Fuel Cell Technology

Fuel Cell Technology used in hybrid cars

Industrial use of Fuel


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(b) Hydrogen and oxygen can be used in a fuel cell. By choosing suitable electrode reactions from Data Booklet, calculate the e.m.f. of a cell consisting of oxygen electrode and a hydrogen electrode, each under acidic conditions. [+1.23V]

From the Data Booklet choose:

O2 + 4H+ + 4e � 2H2O +1.23 V

2H+ + 2e � H2 0.00 V

In the case of hydrogen and oxygen fuel cell, from the Eθ values, reduction of O2 and oxidation of hydrogen will occur

Thus, Eθ

cell = Eθ

red – Eθ

oxd = +1.23 – 0.00 = +1.23 V

(c) One type of rechargeable battery makes use of the nickel–cadmium cell, in which nickel and

cadmium electrodes, coated with their respective hydroxides, are immersed in potassium hydroxide solution. In normal use, the cadmium electrode is the one that releases electrons to the external circuit.

The relevant electrode reactions are:

Cd(OH)2 (s) + 2e �� Cd (s) + 2OH– (aq)

NiO(OH) (s) + H2O (l) + e �� Ni(OH)2 (s) + OH– (aq)

Write a cell diagram for this arrangement, showing the polarity of the electrodes, and construct a balanced equation for the reaction that occurs during the discharge.

Since cadmium electrode is the one that releases electrons to the external circuit oxidation must occur at the cadmium electrode and it must be the anode (–) Nickel electrode must be the cathode (+) where reduction occurs.

Cell Diagram: Cd (s), Cd(OH)2 (s) NiO(OH) (s), Ni(OH)2 (s), Ni (s)

Reduction: NiO(OH) (s) + H2O (l) + e → Ni(OH)2 (s) + OH– (aq)

Oxidation: Cd (s) + 2OH– (aq) → Cd(OH)2 (s) + 2e

Overall equation: Cd (s) + 2NiO(OH) (s) + 2H2O (l) → Cd(OH)2 (s) + 2Ni(OH)2 (s)


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Self–Attempt Questions 8. Use the standard electrode potential data in your Data booklet, to calculate the standard e.m.f of the cell: [+0.48V]

Sn (s) Sn2+ (aq) Cu2+ (aq) Cu (s) and write a balanced equation for the reaction. From the Data Booklet:

Cu2+ + 2e � Cu Eθ = +0.34 V

Sn2+ + 2e � Sn Eθ = –0.14 V

Eөcell = Eө red – Eөoxid = +0.34 – (–0.14) = +0.48 v ⇒ Eөcell >0 reaction is feasible From the cell diagram

Reduction: Cu2+ + 2e → Cu –––(1)

Oxidation: Sn → Sn2+ + 2e –––(2) Overall Cell equation: (1) + (2)

Cu2+ (aq) + Sn (s) → Cu (s) + Sn2+ (aq)

9. Use the half equations in the Data Booklet to construct a balanced equation, under conditions

of low pH, for each of the reactions between (a) manganate (VII) and iron (II) ions [+0.75V]

From Data Booklet:

MnO4– + 8H+ + 5e � Mn2+ + 4H2O Eθ = +1.52 V

Fe3+ + e � Fe2+ Eθ = +0.77 V

Eθ

red = +1.52 V

Eθ

oxd = +0.77 V

Eθ

cell = Eθ

red – Eθ


Reduction: MnO4– + 8H+ + 5e → Mn2+ + 4H2O –––(1)

Oxidation: Fe2+ → Fe3+ + e –––(2) Overall reaction (1) + [(2) x 5]

MnO4– (aq) + 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq)+ 4H2O (l) + 5Fe3+ (aq)

Ered

Eox


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(b) dichromate (VI) and iron (II) ions [+0.56V]

From Data Booklet:

Cr2O72– + 14H+ + 6e � 2Cr3+ + 7H2O Eθ = +1.33 V

Fe3+ + e � Fe2+ Eθ = +0.77 V

Eθ

red = +1.33 V

Eθ

oxd = +0.77 V

Eθ

cell = Eθ

red – Eθ


Reduction: Cr2O72– + 14H+ + 6e → 2Cr3+ + 7H2O –––(1)

Oxidation: Fe2+ → Fe3+ + e –––(2) Overall reaction (1) + [(2) x 6]

Cr2O72– (aq) + 14H+ (aq) + 6Fe2+ (aq) → 2Cr3+ (aq) + 7H2O (l) + 6Fe3+ (aq)

Suggest why potassium manganate (VII) rather than potassium dichromate (VI) is used to titrate Fe (II) in volumetric analysis. [i.e. consider the titration procedure]

If MnO4– is used, a distinct end–point of yellow (due to mostly Fe3+) → pale orange (due to mixing

one drop of purple MnO4– in excess with yellow Fe3+) is seen.

If Cr2O72– is used, the end–point of yellow–green (due to mostly Fe3+ + Cr3+) → orange–green is

not distinct. Hence MnO4

– is used.


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10. (a) (i) Describe the experimental procedure you would use to measure the standard electrode potential of I2 (aq) /I– (aq) system.

A standard I2 (aq) / I– (aq) half–cell (which consists of a Pt electrode dipping into a solution of I– of concentration 1 mol dm–3) is connected to a standard hydrogen electrode via a salt bridge. The electrode potential is measured using a voltmeter.

(Need not draw diagram to illustrate since question specifically mention “describe”).

(ii) What effect would an increase in the concentration of iodide ions have in the Eθ

cell of this system?

I2 + 2e �� 2I– Ered

θ = +0.54 V

• When [I–] increases, equilibrium shift to the left by Le–Chaterlier’s Principle to decrease [I–]

• oxidation process is favoured, Eredθθθθ becomes less positive

• Ecellθθθθ becomes decreases

(b) Use the Data Booklet to predict the reaction that would occur when the following pairs of

solutions are mixed. Calculate the Eθ

cell and write an equation for each reaction. State the observation if reaction occurs.

(i) acidified manganate (VII) and aqueous sulfur dioxide. [+1.35V] From the Data Booklet:

MnO4– + 8H+ + 5e � Mn2+ + 4H2O Eθ = +1.52 V

SO42– + 4H+ + 2e � SO2 + 2H2O Eθ = +0.17 V

Eθ

cell = Eθ

red – Eθ


Reduction: MnO4– + 8H+ + 5e → Mn2+ + 4H2O –––(1)

Oxidation: SO2 + 2H2O → SO42– + 4H+ + 2e –––(2)

Overall reaction: [(1) x 2] + [(2) x 5]

2MnO4– (aq) + 2H2O (l) + 5SO2 (g) → 2Mn2+ (aq) + 5SO4

2– (aq) + 4H+ (aq)

Eθθθθ

red

Eθθθθ

oxd


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Part II : Electrolytic Cell Important Concept: Selective discharge of ions With an aqueous solution as electrolyte, there may be more than one type of ions E.g. NaCl (aq) contains Na+ and Cl– from NaCl and H+ and OH– from auto–ionisation of water. The orders in which ions are discharged at the electrodes depend on three factors:

(1) Nature of the electrode (reactive or inert e.g. platinum or graphite) (2) Concentration of the ions in the electrolyte

(3) Nature of aqueous electrolyte (Comparing Eθ values)

General Steps for Calculations involving Electrolysis Step 1 : Write the half equation for the reaction occurring at the cathode or anode

Step 2 : Determine the no. of mol of electrons required Step 3 : Calculate the quantity of electricity, Q

Step 4 : Calculate the no. of moles of substance (= Faradays of no.

Q)

Step 5 : Calculate the mass / volume of substance 1(a) A current of 3.20 A passed through dilute aqueous sodium sulfate for 50 minutes. Calculate

the volume of gas (measured at room temperature and pressure) liberated at the anode.

Analysis: Why not SO42–?

From Data Booklet

SO42– + 4H+ + 2e �� SO2 + 2H2O +0.17

S2O82– + 2e �� 2SO4

2– +2.01 O2 + 4H+ + 4e– 2 H2O +1.23V

Since oxidation must occur at anode, ion concerned must be on the right hand side. Thus first equation is out, since it shows SO4

2– being reduced.

Since oxidation have to occur, the half equation with a less positive Eθ value will be favoured. That’s why SO4

2– did not under go oxidation

Step 1: Analysis, at anode: 2 H2O (l) → O2 (g) + 4H+ (aq) + 4e Step 2: 2F is required to liberate one mole of O2 gas Step 3: Quantity of electricity passed Q = I x t = 3.20 x (50 x 60) = 9600 C

H2O Na+ SO4

2–


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Step 4: No of moles of O2 gas liberated = Faradays of no.

Q

= 965004

9600

4F

9600

×=

= 0.02487 mol

Step 5: Volume of gas liberated at r.t.p. = n x Molar volume = 0.02487 x 24.0 = 0.597 dm3

(b) A layer of chromium is electroplated on an automobile bumper by a passing constant current of 250 A through a cell that contains Cr3+ (aq). Calculate the time taken in minutes to deposit 250 g of chromium.

Step 1: Analysis, at Cathode: Cr3+ + 3e → Cr Step 2: 3F is required to deposit 1 mol of Cr Step 3: Quantity of electricity passed Q = I x t = 250 x t

Step 4: No of moles Cr deposited = Faradays of no.

Q

= 965003

t250

×

×

Step 5: Determining time to deposit 250 g of Cr

Since n=m/Mr 965003

t250

52

250

×

×= ⇒ t = 5567.3 s ≈ 92.8 min

(c) In an experiment, 0.80 g of element Europium was discharged at the cathode by the passage

of 0.50 A of current for 2000 seconds. Europium has relative atomic mass of 152. Calculate the charge on the europium ion.

Using Fn

Q

Mr

m

×= ⇒

96500n

20000.50

152

0.80

×

×= ⇒ n = 1.96 ≈ 2

Since 2F is required to discharge 1 mol of Europium ion, Therefore, charge on Europium ion = 2 2. The following pairs of electrolysis experiments were carried out. In each case, determine which

experiment produces the bigger increase in the mass of the cathode (Show all relevant working). (a) Exp I : 3.00 A for 60 minutes through 1.0 mol dm–3 Na2SO4 (aq) using platinum electrodes. Exp II : 5.00 A for 10 minutes through 1.0 mol dm–3 CuSO4 (aq) using copper electrodes. For Expt 1: Analysis: Current 3.0 A, time 60 min, given Na2SO4 (aq)

(1) At the cathode: 2 H2O (l) + 2e– → H2 (g) + 2OH– (aq) (2) 2 Faraday is required to discharge 1 mol of H2 (g) (3) Q = I x t = 3 x 60 x 60 = 10800 C

Na+ SO42–

H2O


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(4) No of mole of H2 liberated = 965002

10800

2F

Q

×= = 0.05595

(5) Volume of H2 liberated = 0.05595 x 24 = 1.34 dm3 Thus, reduction of Na+ to Na will not occur at the cathode.

⇒⇒⇒⇒ Mass of cathode remains unchanged (since no Na metal is deposited) For Expt 2: Analysis: Current 5.0 A, time 10 min, given CuSO4 (aq)

(1) At the cathode: Cu2+ (aq) + 2e → Cu (s) (2) 2 Faraday is required to deposit 1 mol of Cu (3) Q = I x t = 5 x 10 x 60 = 3000 C

(4) No of mole Cu deposited = Q / 2F 965002

3000

×= = 0.0155

(5) Mass of Cu deposited = 0.0155 x 63.5 = 0.987 g Expt 2 will show the greater increase in the mass of the cathode. (b) Exp I : 0.2 A for 30 minutes through PbBr2 (aq) using carbon electrodes. Exp II : 0.2 A for 30 minutes through AgNO3 (aq) using silver electrodes

For Expt 1: Analysis: Current 0.2 A, time 30 min, given PbBr2 (aq)

(1) At the cathode: Pb2+ (aq) + 2e → Pb(s)

(2) 2 Faraday is required to discharge 1 mol of Pb2+(aq) (3) Q = I x t = 0.2 x 30 x 60 = 360 C

(4) No of mole of H2 liberated = 965002

360

2F

Q

×= = 1.18 x10–3

(5) mass of Pb deposited = 1.18 x 10–3 x 207.2 = 0.244 g For Expt 2: Analysis: Current 0.2 A, time 30 min, given AgNO3 (aq)

(1) At the cathode: Ag+ (aq) + e → Ag (s) (2) 1 Faraday is required to deposit 1 mol of Ag

Cu2+ SO42–

H2O

Pb2+ Br – H2O

Ag+ NO3– H2O


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(3) Q = I x t = 0.2 x 30 x 60 = 360 C

(4) No of mole Ag deposited = Q / F = 96500

360

(5) Mass of Ag deposited = 96500

360x 108 = 0.403 g

Expt 2 will show the greater increase in the mass of the cathode. (c) When a current if 0.25 A was passed through dilute aqueous NaCl for 32 minutes and 10 seconds at s.t.p., 56 cm3 of gas was evolved at the cathode. Calculate the Faraday constant from these data and hence derive a value for the Avogadro’s constant, obtaining any other data from the Data Booklet. Analysis: I = 0.25 A t = 32 min 10 sec

Given NaCl (aq) → Na+, Cl– , H2O are present

(1) Cathode reaction: 2 H2O (l) + 2e– → H2 (g) + 2OH– (aq) (2) 2 Faraday is required to discharge 1 mol of H2 (g) (3) Q = I x t = 0.25 x [(32 x 60) + 10] = 482.5 C

(4) no. of mole of H2 evolved at s.t.p. = 3102.5

22.4

0.056 −×=

⇒ 2.5 x 10–3 mol of H2 required 482.5 C of electricity to discharge.

Using the formula: no. of mole of H2 = F

Q

2

)102(2.5

482.5

n2

QF

3H2

−×

=×

= = 96500 C mol–1

Since F = L x e Where L is Avogadro’s Constant and e is charge of electron and from the Data Booklet e = 1.60x10–19 C

L = 19101.6

96500

e

F−

×= =

23106.03×

Proven

Proven

H2O discharged due to

more +ve Eθ value compared to Na+


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3. (a) Draw and label the apparatus used to electrolyse molten lead (II) bromide in an evaporating basin using graphite electrodes and including an ammeter in the circuit.

(b) Give the ion–electron equations for the reaction occuring at the (i) anode (ii) cathode Suggest how you would detect the product formed at the anode.

Anode: 2Br – (l) → Br2 (l) + 2e

Cathode: Pb2+ (l) + 2e → Pb (s) [grey solid] The electrolyte turns reddish–brown to show the formation of the Br2 at the anode.

Label: –Electrode –Electrolyte –State symbols –cathode/anode – +/ –

A

Anode (+) (Oxidation)

Cathode (–) (Reduction)

Graphite electrode Graphite electrode

Electrolyte PbBr2 (l)

2Br– (l) � Br2 (l) + 2e Pb2+ (l) + 2e � Pb (s)

Ammeter


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(c) When a current of 4.0 A was passed for 30 minutes, a bead of lead of mass 6.90 g was obtained.

(i) How many coulombs were passed? (ii) In this experiment, how many coulombs would be required to deposit one mole of lead? (i) Q = I x t = 4 x 30 x 60 = 7200 C (ii) mass of Pb deposited = 6.90 g No. of mole of Pb deposited = 6.90/ 207 = 0.0333 Therefore quantity of electricity required to deposit 1 mol of Pb = 7200 ÷ 0.0333 = 2.16 x 105 C

4. J03/IV/1 Because of its increased scarcity, cheaper copper ornaments are no longer made from the

solid metal, but from iron that has been copper plated. (a) Draw a diagram to show the set up for a copper electroplating process. Show clearly the

polarity (+/–) of the power source, and suggest a suitable electrolyte.

NB: The object to be plated must always be the cathode i.e. reduction take place

⇒ Cu2+ (aq) + 2e → Cu (s)

(b) A current of 0.500 A is passed through the electroplating cell. Calculate the time required to deposit a mass of 0.500 g of copper on to the ornament. [50.6 min]

Cu2+ (aq) + 2e → Cu (s)

No of mole of Cu deposited =3107.874

63.5

0.500 −×=

No of Faraday to produce 1 mol of Cu = 2F

Therefore, using no. of mole = F

Q

2 ⇒ Q = no. of mol x 2F = 7.874 x 10–3 x 2 x 96500 = 1519.6 C

Since Q = I x t ⇒ t = seconds 30390.500

1519.6

I

Q== ≈ 50.7 min

+ –

A ı ı ı

Electrolyte: CuSO4 (aq)

Object to be plated

Cu (s) electrode


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5.(a) When a current is passed through a cell containing aqueous silver nitrate (using inert electrodes ), silver is deposited at the cathode and oxygen is evolved at the anode. What volume of oxygen, at s.t.p, would be produced if the current were passed for a sufficient time to deposit 0.60 g of silver?

At Cathode: Ag+ + e → Ag

At Anode: 4OH– (aq) → 2H2O (l) + O2 (g) + 4e

No. of mole of silver deposited = 108

60.0 = 5.555 x 10–3 = no. of mole of electron

Since the two electrodes are connected in series, the same no. of mole of electron must pass through both electrodes at the same time. 4e Ξ O2

No. of mole of O2 evolved = 4

10555.53−

× = 1.388 x 10–3

Volume of O2 evolved at s.t.p = 1.388 x 10–3 x 22.4 = 0.0311 dm3 (b) Calculate the ratio of the mass of silver to the mass of nickel deposited on the cathodes when

the current is passed through electrolytic cells containing silver (I) nitrate and aqueous nickel (II) sulfate, connected in series.

Ni2+ + 2e → Ni ––––– (1)

Ag+ + e → Ag –––– (2) Since the two electrolytic cells are connected in series, the same no. of mole of electron must pass through both cells at the same time.

Eqn (2) x 2: 2Ag+ + 2e → 2Ag Hence no. of mole of Ag deposited : no. of mole of Ni deposited = 2 : 1 Mass of Ag deposited : Mass of Ni deposited = 108 x 2 : 58.7 = 3.68 : 1


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6. Integrated Question – N2006/III/1 (a) Predict the products formed at the anode, and at the cathode, when the following liquids are

electrolysed using inert electrodes. In each case explain your reasoning, using data from the Data Booklet. [6] (i) NaBr (l) Only Na+ (l) and Br– (l) are present as ions to be discharged at the cathode and anode respectively. Products are: Na (s) at the cathode and Br2 (l) at the anode (ii) NaBr (aq) At the cathode: 2H2O + 2e H2 + 2OH– –0.83V

Na+ + e Na –2.71 V

Since the )O/H(Hθ

22E is more positive / less negative, H2O is more readily reduced.

⇒ H2 (g) is produced at the cathode.

At the anode: O2 + 4H+ + 4e 2 H2O +1.23V

Br2 + 2e 2Br– +1.07 V

Since the )/Br(Brθ

2-E is less positive, Br– is more readily oxidised.

⇒ Br2 (aq) is produced at the anode.

(iii) CuF2 (aq) At the cathode: 2H2O (l) + 2e H2 (g) + 2OH– –0.83V

Cu2+ + 2e Cu +0.34 V

Since the )(Cu/Cuθ

2E +

is more positive / less negative, Cu2+ is more readily reduced.

⇒ Cu (s) is produced at the cathode.

At the anode: O2 + 4H+ + 4e 2 H2O +1.23V

F2 + 2e 2F– +2.87 V

Since the )O/O(Hθ

22E is less positive, H2O is more readily oxidised.

⇒ O2 (g) is produced at the anode.


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(b) The electrolysis of a solution of potassium fluoride, KF, in a mixed solvent of hydrogen fluoride and water produces a colourless triatomic gas A at the anode. When 0.900 g of A is introduced into an evacuated vessel of 1.00 dm3 capacity at a temperature of 282 K, it produces a pressure of 3.90 x 104 Pa.

(i) Calculate the Mr of A.

)10)(1.0010(3.90

31)(282)(0.900)(8.

PV

mRT M

34 −××

== = 54.1 g mol–1

Mr = 54.1 (answer to question!) (ii) Suggest the molecular formula and the shape of A. [4] Analysis:

KF (electrolysed) + HF + H2O → colourless triatomic (three atoms!) gas A at anode

⇒ suspect F– oxidised (since at anode) to F2

Mr of 54 ≈ 2(19) (due to F2) + 16 ⇒ remaining atom in A is oxygen (since triatomic).

Possible molecule formula of A is F2O.

Shape: Bent about O atom. [like H2O]

7. Copper often occurs in the Earth’s crust as the sulfide, Cu2S, associated with the sulfides of zinc, ZnS, and silver, Ag2S. The copper is extracted from its ores by partial air–oxidation to copper (I) oxide and sulfur dioxide, followed by the reaction between this copper (I) oxide and unchanged copper (I) sulfide to give copper metal and more sulfur dioxide.

(a) Write balanced chemical equations for these two reactions.

The first stage produces copper containing some zinc and silver as impurities.

Reaction 1: Cu2S (s) + 3/2 O2 (g) → Cu2O (s) + SO2 (g)

Reaction 2: 2Cu2O (s) + Cu2S (s) → 6Cu (s) + SO2 (g)

(b) Describe in detail with the aid of relevant equations and diagram, how this impure copper is purified electrolytically, explaining what happens to the zinc and silver impurities. The use of

relevant Eθ data from the Data Booklet is recommended.


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In the electrochemical purification of copper: Anode: Crude copper Cathode: Pure copper Electrolyte used: Copper (II) sulfate From the Data Booklet:

Ag+ + 2e � Ag +0.80V

Cu2+ + 2e � Cu +0.34V

H+ + e � H2 0.00 V

Zn2+ + 2e � Zn –0.76V

• When an electric current is applied, both zinc and copper are oxidised to Zn2+ and Cu2+ respectively.

• Silver, due to its more positive (Eθ) reduction potential will not be oxidised.

• Silver will drop to the electrolytic bed as elemental silver.

• H+, Zn2+ and Cu2+ migrate to the cathode.

• Only Cu2+ is reduced to copper due to its more positive reduction potential and higher concentration.

• H+ and Zn2+ remains in the solution.


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Self–Attempt Questions 8. An aqueous solution of copper (II) nitrate is electrolysed for 20 minutes with a constant current of

2.40 A. Calculate the mass of copper deposited at the electrode.

Step 1: Analysis, at Cathode: Cu2+ (aq) + 2e → Cu (s) Step 2: 2F is required to deposit 1 mol of Cu Step 3: Quantity of electricity passed Q = I x t = 2.40 x (20 x 60) = 2880 C

Step 4: No of moles of Copper deposited = Faradays of no.

Q

= 965002

2880

2F

2880

×= = 0.014922

Step 5: Mass of copper deposited = n x Ar = 0.014922 x 63.5 = 0.94756 = 0.948 g 9. State and explain, with the aid of the Data booklet where necessary, what would happen in the

following experiment: (a) Dilute aqueous potassium bromide using C electrodes.

At the cathode (reduction takes place): K+ and H2O migrate

From the Data Booklet:

Eθ/ V

K+ + e �� K –2.92 V

2H2O + 2e– H2 + 2OH– – 0.83V

Since the )O/H(Hθ

22E is more positive, H2O is more readily reduced.

⇒ Cathode reaction: 2 H2O (l) + 2e– → H2 (g) + 2OH– (aq) –––––(1) At the anode (oxidation takes place): Br– and H2O migrate From the Data Booklet:

Eθ/ V

Br2 + 2e �� 2Br– +1.07 V O2 + 4H+ + 4e 2 H2O +1.23V

Since the )/Br(Brθ

2-E is less positive, Br– is more readily oxidised.

⇒ Anode reaction: 2Br– (aq) → Br2 (aq) + 2e ––––– (2)

Overall reaction (1) + (2):

2H2O (l) + 2Br– (aq) →→→→ 2H2 (g) + 2OH– (aq) + Br2 (aq) Observation: Effervescence of H2 is seen at the cathode and solution turns reddish brown.

H2O K+ Br– Inert electrode

Cu2+ NO3–


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(b) Concentrated aqueous copper (II) chloride is electrolysed between graphite electrodes.

At the cathode (reduction takes place): Cu2+ and H2O migrate From the Data Booklet:

Eθ/ V

Cu2+ + 2e �� Cu +0.34 2H2O + 2e– H2 + 2OH– – 0.83V

/Cu)(Cuθ

2E + is more positive, Cu2+ is more readily reduced.

⇒ Cathode reaction: Cu2+(aq) + 2e →→→→ Cu (s) –––––(1)

At the anode (oxidation takes place): Cl– and H2O migrate

Cl – ions are preferentially discharged instead of H2O since there is a high concentration of Cl– ions. Chlorine gas is liberated.

⇒⇒⇒⇒ Anode reaction: 2Cl – (aq)→ Cl2 (g) + 2e ––––(2) Overall reaction (1) + (2):

Cu2+ (aq) + 2Cl– (aq) → Cu (s) + Cl2 (g) Observation: Effervescence of greenish–yellow gas of Cl2 is seen at the anode and pink solid of copper metal will appear at the cathode. Decolourisation of blue solution.

(c) Molten lead (II) oxide using C electrodes. At the cathode: only Pb2+ migrate

Cathode reaction: Pb2+ (l) + 2e → Pb (s) At the anode: only O2– migrate

Anode reaction: 2O2– (l) →O2 (g) + 4e Observation: Grey solid deposited at the cathode and effervescence observed at the anode

H2O Cu2+ Cl– Inert electrode

Pb2+ O2–


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Multiple Choice Questions Q1 A weedkiller can be prepared by heating a bleach solution

3NaClO → 2NaCl + NaClO3 bleach weedkiller What are the oxidation states of chlorine in these compounds?

A –1 –1 +5 B +1 –1 +5 C +1 –1 +7 D +2 –1 +7

Q2 The equations for three reactions are given below:

Cl2 (g) + 2H2O (l) + SO2 (g) → 2HCl (aq) + H2SO4 (aq)

Cl2 (g) + H2S (g) → 2HCl (g) + S (s)

SO2 (g) + 2H2S (g) → 2H2O (l) + 3S (s) What is the correct order of strength of the three reacting gases as reducing agents? Strongest Weakest

A Chlorine Hydrogen sulfide Sulfur dioxide

B Chlorine Sulfur dioxide Hydrogen sulfide C Hydrogen sulfide Sulfur dioxide Chlorine

D Sulfur dioxide Hydrogen sulfide Chlorine Q3 Four standard electrode potentials are listed below:

Cu2+(aq) + 2e → Cu(s) +0.34V

AgCl(s) + e → Ag(s) + Cl– (aq) +0.22V

H+ (aq) + e → ½ H2 (g) 0.00V

Zn2+ (aq) + 2e → Zn (s) –0.76V Which cell potential could be obtained by combining two of these standard electrodes?

A 0.39 V B 0.42 V

C 0.54 V D 0.56 V E 0.98 V


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Q4 The use of the Data Booklet is relevant to this question.

Sir Humphrey Davy showed that the corrosion of copper hulls of sea going ships could be prevented by placing strips of “sacrificial” metals on the hulls Which of these metals is least likely to dissolve when attached to the copper hull of a sea going ship?

A Iron

B Magnesium C Tin D Zinc Q5 In the construction of heart “pacemakers”, it is possible to use a tiny magnesium

electrode which creates an electrical cell with the inhaled oxygen. The relevant half–cells are as follows

Mg2+ + 2e → Mg Eθθθθ = –2.38V

½ O2 + 2H+ + 2e → H2O Eθθθθ = +1.23V Under standard conditions, the cell emf would be 3.61 V, but in the body a potential of 3.25 V is more than usual. What is the best explanation of this lower emf?

A The small size of the magnesium electrode. B The low concentration of Mg2+ ions surrounding the magnesium electrode. C The high resistance of the body fluids surrounding the electrodes. D The pH of between 7 and 8 of the body fluid surrounding the electrodes. Q6 A current of 8A is passed for 100 minutes through molten aluminium oxide using inert

electrodes. What will be the approximate volume of gas liberated, measured at s.t.p?

A 2.8 dm3 B 5.6 dm3 C 8.4 dm3 D 11.2 dm3

E 22.4 dm3


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Q7 The standard cell potentials for the redox equilibria of aqueous vanadium–containing

ions and the colours of these ions are below.

VO2+ + 2H+ + e � H2O + VO2+ Eθθθθ = +1.00 V

Yellow blue

VO2+ + 2H+ + e � H2O + V3+ Eθθθθ = +0.34 V Blue green

V3+ + e � V2+ Eθθθθ = –0.26 V

Green purple

What is likely to the final colour when metallic tin is added to a solution containing VO2+?

(Sn2+ + 2e � Sn Eθθθθ = –0.14 V) Colourless

A Yellow B Blue

C Green D Purple Q8 Aqueous sodium chloride (brine) is electrolysed by using inert electrodes in a cell which

is stirred so that the products of electrolysis are able to react. The cell is kept cold. Which one of the following pairs of substances is among the final products?

A Hydrogen and chlorine only B Hydrogen and sodium chlorate (I)

C Hydrogen and sodium chlorate (V) D Hydrogen chloride and sodium chlorate (I) E Sodium hydroxide and chlorine only

Answers:

1. B 2. C 3. E 4. C 5. D 6. B 7. C 8. B

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