Revision on Redox & Electrochemistry (Lecture Notes) (Teacher&Apos;s Copy)

7/25/2019 Revision on Redox & Electrochemistry (Lecture Notes) (Teacher&Apos;s Copy)

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H2 Chemistry 9647 Physical Chemistry (Revision)

2015/JC2/Term3/Chem Dept/Physical Chemistry Revision(Modified from 2011 & 2014 Physical Chemistry Revision (Redox & Electrochemistry)

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Anglo-Chinese Junior CollegeDepartment of Chemistry

REDOX & ELECTROCHEMISTRY(Revision)

Lectured by: Miss Jasmine Zheng

REDOXA redox reaction consists of two half reactions, reduction and oxidation occurringsimultaneously, i.e. electrons are transferred from the reducing agent to the oxidizing agent.Hence, in a redox reaction, as one species is oxidised, another species is always reduced.

REDOX REACTION IN TERMS OF ELECTRON TRANSFER

Some examples:

Zn(s) Zn2+ + 2e

Cu2+ + 2e Cu(s)

RED = Reducing agent is Electron Donor.

REDOX REACTION IN TERMS OF OXIDATION NUMBER (STATE)

Oxidation number or state is the charge that an element would have if all the bonds in thesubstance were ionic.

Some basic rules:

Uncombined elements are zero.

F is always -1, Group I are +1, Group II +2, oxygen is -2 (except in peroxides), H is +1(except in metallic hydrides).

Oxidation numbers in a neutral compound add up to zero and in a polyatomic ion they addup to the charge on that ion.

Some examples:

Zn(s) Zn2+ + 2e Oxidation state of Zn increases from 0 to +2.

Cu2+ + 2e Cu(s) Oxidation state of Cu decreases from +2 to 0

Reduction is defined as the gain of electrons. Oxidation is defined as the loss of electrons.

Reductionresults ina decrease in the oxidation number (state) of an atom. Oxidation results inan increase in the oxidation number (state) of an atom.

TIP:You must be able todescribe and explain redoxprocesses in terms ofelectron transfer and/or ofchanges in oxidationnumber (oxidation state)


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BALANCING REDOX EQUATIONS

Overall redox equations can be obtained by adding half equations. You must check that: The number of electrons on the left of one half-equation equals the number of electrons on

the right of the other half-equation. In the overall equation, the total change in oxidation number of the substance being reduced

equals the total change in the substance being oxidized.

Using the Half-Equation Method

General steps to balance a redox reaction (in acidic medium):

1. Identify the two incomplete half-reactions, one for oxidation and the other for reduction.

2. Balance both the half-equations using the following steps:

a. Balance the elements other than H and O

b. Balance oxygen atoms by adding H2O

c. Balance hydrogen atoms by adding H+

d. Balance charges by adding electrons to the side with the greater overall positivecharge.

3. Multiply the balanced half-equations by appropriate integers such that the number of

electrons in both half-equations are equal (electrons gained = electrons lost)

4. Add the resulting half-equations together, and eliminate any common species on both sides

to obtain the overall balanced equation.

Additional steps to balance a redox reaction (in alkaline medium):

5. Use the Steps 1-4 in acidic medium.

6. Add sufficient OH-to both sides of the equation to neutralize any H+present.

(H++ OH-H2O)

7. Eliminate electrons and H2O molecules that appear on both sides of the equation.

Using the Half-Equations from the Data Booklet

The overall equation for the oxidation of I- ions by MnO4- ions is obtained from the two half

equations:MnO4

-+ 8H++5e-Mn2+ + 4H2O2I-I2+ 2e

-

The MnO4- half equation has 5 electrons, but the I -equation only 2, so the MnO4

-equation has

to be multiplied by 2 and the I-equation by 5, so that both have 10 electrons. They are thenadded to give:

2MnO4-+ 16H++10I-2Mn2++ 8H2O + 5I2


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Self-Check Assessment (5 minutes)

Q1. Complete the following half-equations:(a) Cl-Cl2

(b) Fe3+Fe2+(c) MnO4

-Mn2+(d) MnO4

-MnO2(e) C2O4

-CO2(f) S2O3

-S4O6-

Q2. Combine the following two half-equations:

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

2I-I2+ 2e-

DISCUSSION QUESTIONSQ1. The propellant used in the solid rocket booster of a space shuttle is a mixture of aluminium

and compound X. Compound X contains chlorine in an oxidation state of +7. Which of thefollowing could be compound X?

A NH4Cl B N2H5Cl C NH4ClO3 D NH4ClO4

Q2. Disproportionation occurs when an element is both oxidised and reduced in a reaction.Which named element does not disproportionate in the reaction shown?

Element ReactionA Carbon H2C2O4H2O + CO + CO2

B Nitrogen H2O + 2NO2HNO3+ HNO2C Sulphur 2FeSO4Fe2O3+ SO2+ SO3D Chlorine 3ClO-ClO3

-+ 2Cl-

Q3. In an experiment, 25.0 cm3of a 0.10 mol dm-3solution metallic salt reacted exactly with12.5 cm3of 0.10 mol dm-3aqueous sodium sulphite. The half equation for the oxidation ofthe sulphite ion is shown below.

SO32- (aq) + H2O (l) SO4

2- (aq) + 2H+ (aq) + 2e-

If the original oxidation number of the metal in the salt was 3, what would be the newoxidation number of the metal?

A 0 B 1 C 2 D 4

Q4. 25.0 cm3 of a solution of M2O5 of concentration 0.100 mol dm-3 is reduced by sulphur

dioxide to a lower oxidation state. To reoxidise Mto its original oxidation number required50.0 cm3 of 0.0200 mol dm-3 potassium manganate (VII) solution. To what oxidationnumber was Mreduced by sulphur dioxide?

A +2 B +3 C +4 D +5


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Q5. The Winkler method is used to determine the amount of dissolved oxygen in a sample. Inthis procedure, oxygen reacts with Mn2+under alkaline conditions to produce a precipitateof MnO(OH)2.

2Mn2+ (aq) + O2 (aq) + 4OH (aq) 2MnO(OH)2 (s)

The precipitate is then dissolved in acid and reacted with iodide, forming iodine and Mn 2+.MnO(OH)2 (s) + 2

(aq) + 4H+ (aq) I2 (aq) + Mn2+ (aq) + 3H2O (aq)

Finally, the amount of iodine produced is determined by reaction with thiosulphate.I2(aq) + 2 S2O3

2-(aq) 2I-(aq) + S4O62-(aq)

When a sample of water was analysed using the Winkler method, a total of 0.60 mol ofthiosulphate was used in the reaction. What was the mass of oxygen present in theoriginal sample?

A 4.8 g B 9.6 g C 19.2 g D 38.4 g

Q6. A sample of 10.0 cm3of 0.10 mol dm-3 iron(II) sulfate is titrated against 0.025 mol dm-3potassium manganate(VII) in the presence of an excess of fluoride ions. It is found that10.0 cm3 of the manganate(VII) solution is required to reach end-point. What is the

oxidation number of the manganate at the end-point?

Q7. Sodium thiosulfate is used in the textile industry to remove an excess of chlorine from thebleaching processes by reducing it to chloride ions:

S2O32- + 4Cl2+ 5H2O 2HSO4

-+ 8H++ 8Cl-

In this reaction, how many moles of electrons are supplied per mole of thiosulphate?


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ELECTROCHEMICAL CELL

Standard electrode potential is defined as the potential difference between a standard

hydrogen electrode and a metal (the electrode) which is immersed in a solution containingmetal ions at 1 moldm-3concentration at 25 oC and 1 atmospheric pressure.

Standard cell potential is a measure of the tendency of electrons to flow through the externalcircuit under standard conditions of 25 oC, 1 atmospheric pressure and 1.00 moldm-3concentration. It is the maximum potential difference between the electrodes.

Standard hydrogen electrode consists of a platinised electrode in a solution of 1.00 moldm-3H+ions (e.g. aqueous HCl)

H2 gas at a pressure of 1 atmospheric pressure isbubbled over the Pt electrode. On the Pt surface,

equilibrium is set up between the adsorbed layer of H2(g) and the H+(aq) ions in solution at 25 oC.

H2(g) 2H+(aq) + 2e- 0.00 V

To measure the standard electrode potentials of:(i) metals or non-metals in contact with their ions in aqueous solution

e.g. for metals e.g. for non-metals

TIP:You should be able to define the terms:(i) standard electrode (redox) potential(ii) standard cell potentialand describe the standard hydrogen electrode.

TIP:You should be able to describe methods used tomeasure the standard electrode potentials of:(iii) metals or non-metals in contact with their

ions in aqueous solution(iv) ions of the same element in different

oxidation states


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(ii) ions of the same element in different oxidation states

Electrochemical cells (Galvanic cells)

To calculate a standard cell potential by combining two standard electrode potentials:

**ALWAYS:Cathode: reductionAnode: oxidation

Note:i) Eredvalues substituted are as given in the Data Booklet.ii) Positive Ecell

valuesuggestsa thermodynamicallyfeasible reaction. Ecellvalues give no

information on the reaction rates and apply to standard conditions only.iii) Electrons flow from the oxidation half-cell to the reduction half-cell in the external circuit.

TIP:You should be able tocalculate a standard cellpotential by combining two

standard electrodepotentials

Ecell= Ered(reduction half-cell) - E

red(oxidation half-cell)

In this topic, you must:a) understand the limitations in the use of standard cell potentials to predict the feasibility of a

reactionb) construct redox equations using the relevant half-equationsc) predict qualitatively how the value of an electrode potential varies with the concentration of the

aqueous iond) state the possible advantages of developing other types of cell, e.g. the H2/O2fuel cell and

improved batteries (as in electric vehicles) in terms of smaller size, lower mass and higher voltage


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Electrochemical cell: combination of two half cells to generate/produce electricity (its abattery!)


to determine the direction of electron flow, identify which half-cell loses/gains electrons (lookat the half equation involved)

to determine the polarity of the electrode: "electrons exit from the negative terminal of thecell and enter into the positive terminal of the cell"

salt bridge- completes the circuit (by allowing charge to move from one half cell to another)and maintains electrical neutrality (prevents the build up of charges in one half cell)

Anode : H2(g) 2H+(aq) + 2e- Cathode : Cu2+(aq) + 2e-Cu(s)

E(Cu2+|Cu) = +0.34V

Daniel cell:oxidation reduction

e.g. Pt(s) | H2(g) | H+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s) cell diagram

H2(g) H+(aq); Fe3+(aq) Fe2+(aq)

H2(g) + 2Fe3+(aq) 2H+(aq) + 2Fe2+(aq)

"|" is used to separate different phases, a comma is used if they are in the same phase"||" = salt bridge

Ecell> 0V energetically feasible;

Ecell< 0V energetically non-feasible

e.g. Predict if the following reaction is feasible:

Fe(s) + Zn2+(aq) Fe2+(aq) + Zn(s)

Zn2+(aq) + 2e Zn(s) 0.76 V

Fe2+(aq) + 2e Fe(s) -0.44 V

Fe(s) + Zn2+(aq) Fe2+(aq) + Zn(s) Ecell= 0.32 V < 0energetically non-feasible

H2(g) at 1 atm, 25 C

1 mol dm3H+(aq)

e

Pt(s) anode

Cu(s) cathode

1 mol dm3Cu2+(aq)

salt bridge (filter paper soaked in

NaNO3(aq) KNO3(aq))

(Read from left to right)


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e.g. Which of the following statements are true for a standard cell set up using the half-cellsbelow?

Ni2+(aq) | Ni(s) E= 0.25 V

Sn2+(aq) | Sn(s) E= 0.14 V

1 Electrons flow in the external circuit from Ni to Sn.

2 The concentration of Sn2+(aq) will decrease. 3 Oxidation occurs at the Ni terminal.

Consider: Ni2++ 2e Ni E= 0.25V

Sn2++ 2e Sn E= 0.14V (more favourable reduction, therefore the Ni2+/Ni half-cell undergoes oxidation)

Therefore,

Ni Ni2++ 2e

Sn2++ 2e Sn

Ni + Sn2+ Ni2++ Sn Ecell= +0.11 V > 0energetically feasible

e.g. Use the Data Booklet to predict the outcome of mixing acidified potassiummanganate(VII) and aqueous hydrogen peroxide.

MnO4-+ 8H++ 5e- Mn2++ 4H2O +1.52V (+7,highest o.s. for Mn-reduction)

O2+ 2H++ 2e H2O2 +0.68 (H2O2-oxidation)

_______________________________________________________________5H2O2+ 2MnO4

-+ 6H+ 5O2+ 2Mn2++ 8H2O E

cell= +0.84 V > 0

energetically feasible

e.g. Use the Data Booklet to predict the outcome of mixing aqueous tin(II) chloride andacidified aqueous hydrogen peroxide.

Sn2++ 2e Sn 0.14 (reduction)

Sn4++ 2e Sn2+ +0.15 (oxidation) - most fav ox

O2+ 2H++ 2e H2O2 +0.68 (oxidation)

H2O2+ 2H++ 2e 2H2O +1.77 (reduction) - most fav red

H2O2+ 2H++ Sn2+ 2H2O + Sn

4+ Ecell= +1.62 V > 0energetically feasible

V

Sn(s)Ni(s)

Sn2+

(aq)Ni2+

(aq)


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e.g. For the following circuit, will the voltage of the cell increase, decrease, or remain thesame if the concentration of KI is increased.

2Fe3+(aq) + 2I(aq) 2Fe2+(aq) + I2(aq)

If [I] is increased, by Le Chatelier's Principle, the position of equilibrium of half equation (2)

shifts to the left. i.e. more favourable to form I2.

Fe3+(aq) + e Fe2+(aq) E1 -------- (1)

I2(aq) + 2e 2I(aq) E2 -------- (2)

Ecell= E1- E2Increases.____________________________________________________________________________

e.g. Fuel cells are electrochemical cells that were developed as portable energy sources forthe Apollo space program. The overall equation for the reaction occurring in themethane-oxygen fuel cell is

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

a) Write the two half-equations for the reaction occurring at the electrodes in the fuel cell.

CH4(g) + 2H2O(g) CO2(g) + 8H+(aq) + 8e-

O2(g) + 4H+(aq) + 4e- 2H2O(g)

b) At which electrode (positive or negative) does the oxygen react?

Oxygen reacts at the cathode where reduction takes place. In an electrochemicalcell, it is the positive electrode

c) What is the advantage of oxidizing methane in a fuel cell rather than producing energyby the complete combustion of methane in oxygen?

The combustion of methane results in heat losses to the atmosphere hence isless efficient than fuel cell. In the fuel cell, there is direct conversion of chemicalto electrical energy.


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ELECTROLYSIS

The number of Faradaysrequired to discharge one mole of an ion at an electrode equals thenumber of charges on the ion. The Faraday is the charge carried by one mole of electrons.

Factors affecting selective discharge during electrolysis:

1) Position in the Redox Series. Cation with the most positive Eredis discharged first. Anion with the most negative Eredis discharged first.

2) Relative concentration of ions. Ions of higher concentration will be selectively discharged

3) Nature of electrodes.

The quantity of charge(Q)passed during electrolysis can be calculated from

Q = I x t units: coulombs, C.

where I = current in amperes, A.

t = time in seconds, s

Electrolytic cell:electrolysis = decomposition of a molten or aqueous compound using electricity


Direction of electron flow: electrons flow out of the negative terminal and into the positiveterminal of the battery

Polarity of the electrode depends on which terminal (+/) of the battery it is connected to Inert electrodes are used, e.g. Pt, graphite, steel Electrolyte can be a molten salt or an aqueous solution of a salt (ions are free to move to

conduct electricity)

F= Le where L= the Avogadro constante = charge on an electron (1.60 x 10-19C)

1 F= 96 500 Cmol-1

TIP:You should be able to state

the relationship, F = Le,between the Faradayconstant, the Avogadroconstantand the charge onthe electron.

+

+

Pt electrode

(anode)Pt electrode

(cathode)

electrolyte


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Factors affecting substance liberated during electrolysis (for electrolytes containing morethan one salt or aqueous salts):

i. Eredcations/anions with a more positive E

redwill be discharged first

Note that for aqueous solutions, water can also be discharged at either electrodes.

2H2O(l) + 2e H2(g) + 2OH

(aq)

2H2O(l) 4H+(aq) + O2(g) + 4e

Note that the equations 2H++ 2eH2and 2OH

2H+(aq) + O2(g) + 4e

are notusedas [H+] and [OH] are low

compared to H2O in aqueous solutions.

ii. Relative concentration of ionsif a cation/anion is in very high concentration, it maybe discharged in preference to one with a more favourable Ered(if their E

redare closeto each other)

Note that electrolysis of dilute aqueous solutions produces H2and O2(electrolysis of water).

iii. Nature of electrodesometimes the electrodes are deliberately chosen to take part inthe redox reactions

e.g.

(a) A current of 11 A is passed through an aqueous solutions of CrCl3for 52 min. Calculatethe mass of Cr deposited on the cathode and the volume of C l2at stp released at theanode.

Q = It = 11(52 x 60) = 34320 C

n(e) =96500

34320= 0.3556 mol

Cr3++ 3e Cr Cl Cl2+ e

n(Cr) = 1/3(0.3556) = 0.1185 n(Cl2) = (0.3556) = 0.1778mass(Cr) = 0.1185(52.0) = 6.16 g V(Cl2) = 0.1778(22.4) = 3.98 dm

3

(b) How long would it take to produce 25.0 g of Cr from a solution of CrCl3by a current of2.75 A?

n(Cr) =0.52

0.25= 0.4808

Cr3++ 3e Cr

n(e) = 3 x 0.4808 = 1.442

Q = 1.442 x 96500 = 139200 C

Q = It t =I

Q=

75.2

139200= 50610 s 14.1 h


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DISCUSSION QUESTIONS

[ACJC 2006 Prelim Q3]

Q1. (a) The diagram shows a laboratory illustration of a simple hydrogen-oxygen fuel cell.

(i) Write the half equation for the reaction occurring at the left hand (oxygen)

electrode when the cell operates and state its polarity.

O2(g) + 4H+(aq) + 4e- 2H2O (l)

Polarity of electrode: Positive

(ii) Only a very small current can be drawn from this laboratory cell. Suggest oneway in which it could be modified to enable a larger current to be drawn fromit.

Increase concentrations of ions in reduction half-cell ORIncrease in operatingpressure

(iii) A fuel cell in an orbiting satellite is required to produce a current of 0.010A for400 days. Calculate the mass of hydrogen that will be needed.

Time taken = 400 x 24 x 60 x 60 = 34560000 secondsQuantity of electricity = 0.01 x 34560000 = 345600 CAmount of e-= 345600/96500 = 3.6 mol

H2(g) 2H+(aq) + 2e-

Amount of H2= 3.6/2Mass of H2= (3.6/2) x 2 = 3.60 g

(b) A major use of molybdenum is as an alloy with iron, ferromolybdenum. Extremelyhard and strong, even at high temperatures, this alloy is used in filament supports,in heating elements for furnaces and for drill bits. The table below shows fivestandard reduction potentials for molybdenum-containing species.

E/VMo3+ + 3e-Mo -0.20MoO2

+ + 4H++ 2e-Mo3++ 2H2O 0.00MoO4

2- + 4H+ + 2e-MoO2 + 2H2O -1.40MoO3 + 6H

+ + 6e- Mo + 3H2O +0.10MoO4

2- + 4H2O + 6e-Mo + 8OH- -0.97


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(i) Using the above data and any other data from the Data Booklet, suggest areducing agent that could reduce aqueous molybdate(VI) ion, MoO4

2-(aq), to

molybdenum(IV) oxide. Calculate the Ecell for the reaction and write abalanced equation.

Al (s) Al3+ (aq) + 3e Ered= -1.66V

MoO42-

+ 4H+

+ 2e MoO2 + 2H2O E

red= -1.40V

2Al (s) + 3MoO42- (aq) + 12H+ (aq) 2Al3+ (aq) + 3 MoO2(s) + 6H2O (l)

Ecell= -1.40-(-1.66) = +0.26V

Mg (s) Mg2+ (aq) + 2e Ered= -2.90V

MoO42-+ 4H++ 2e MoO2+ 2H2O E

red= -1.40V

Mg (s) + MoO42- (aq) + 4H+ (aq) Mg2+ (aq) + MoO2+ 2H2O (l)

Ecell= -1.40-(-2.90) = +1.50V

Although Ered

has to be more negative than -1.40V, Ca(s), Ba(s), K(s), Li(s)or Na(s) will react with water. Therefore it is not a good answer.

(ii) An aqueous solution of molybdenum(III) sulphate is electrolysed using inertelectrodes. Give ion-electron equations, with state symbols, for the electrodereactions at the anode and the cathode.

Anode: 2H2O (l) O2 (g) + 4H+ + 4e

Cathode: Mo3+ (aq) + 3e Mo (s)

(iii) 0.00200 moles of MoO42- (aq) solution was reduced chemically. The reduced

molybdenum species was then re-converted at an electrode of an electrolytic

cell to its original form by passage of 579 C of electricity. Calculate theoxidation state of the molybdenum in the reduced form.

(Assume that the oxidation state of oxygen remains unchanged throughoutthe process.)

96500C = 1 mol of electrons579C = 6 x10-3mol of electrons

Let reduced species of molybdenum have an oxidation state of +nAmount of Mon+ : Amount of electrons lost

2 x 10-3: 6 x 10-31 : 3

n-(-3) = +6n = +3


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Q2. (a) The electrical current needed to start an automobile engine is provided by a lead-acid storage battery. This battery contains aqueous sulfuric acid in contact with twoelectrodes. One electrode is metallic lead and the other is solid PbO2. Eachelectrode becomes coated with solid PbSO4as the battery operates.

(i) Determine the oxidation and reduction half-equations at both electrodes.

anode: Pb + SO42- PbSO4 + 2e

cathode: PbO2+ 4H++ SO4

2-+ 2e PbSO4+ 2H2OORanode: Pb + HSO4

- PbSO4 + H+ + 2e

cathode: PbO2+ 3H++ HSO4

-+ 2e PbSO4+ 2H2O

(ii) Indicate the anode of the lead-acid storage battery and state its polarity.

Pb is the anode and it is the negative electrode.

(iii) Automobile headlights typically draw 5.9 A of current. The lead-acid storagebattery consumes Pb and PbO2as it operates.

A typical electrode contains 250 g of PbO2. Assuming that the battery cansupply 5.9 A of current until all the PbO2has been consumed, how long will ittake for a battery to run down if the lights are left on after the engine is turnedoff?

Amount of PbO2= 250/239 = 1.05 mol

PbO2+ 4H++ SO4

2-+ 2e PbSO4+ 2H2OAmount of e = 1.05 x 2 = 2.10 mol

Quantity of charge = 2.10 x 96500 = 202650 CTime taken = 202650 / 5.9 x60 x60 = 9.5 hrs

(b) A 1.00 g sample of steel containing manganese was dissolved in nitric acid to givea solution containing manganese in oxidation state +2. All the manganese presentwas then oxidized to manganate(VII) by adding sodium bismuthate, NaBiO3.

After the destruction of any excess bismuthate ion, the resulting purple solutionrequired 36.00 cm3of an iron(II) sulfate solution of concentration 0.10 mol dm-3 toreach an end-point, the iron(II) being oxidized to iron(III).

(i) What is the oxidation state of bismuth in NaBiO3? +5

(ii) Given that the bismuthate ion is reduced to Bi3+, write an equation for thereaction in which it functions as an oxidizing agent. Hence write a balancedequation for the oxidation of manganese(II) to Mn(VII) by bismuthate ion inacidic solution.

BiO3- (aq) + 6H+ (aq) + 2e Bi3+ (aq) + 3H2O (l)

Mn2+ 4H2O MnO4- (aq) + 8H+ (aq) + 5e

2Mn2+ (aq) + 5BiO3- (aq) + 14H+ (aq) 2MnO4

- (aq) + 5Bi3+ (aq) + 7H2O (l)


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(iii) Calculate the percentage by mass of manganese in the steel sample. [1 molMnO4

-(aq) 5 mol Fe2+(aq)]

Amount of Fe2+ = 36.00/1000 x 0.10 = 0.00360 molAmount of MnO4

-= 0.0036/5 = 0.00072 = Amount of Mn2+

Mass of Mn = 0.00072 x 54.9 = 0.0395 g% of Mn in steel = 0.0395/1.00 = 3.95%


Q3. Electrochemical cells are used as portable sources of electricity. When the chemicals getdepleted, the cell becomes flat. Consider the following two examples of electrochemicalcells being used as batteries.

(a) A heart pacemaker consists of zinc and platinum electrodes implanted into the bodytissues. These electrodes in the oxygen-containing body fluid, with a pH of 7.4,form a cell in which zinc is oxidised and oxygen is reduced.

(i) Write the equations for the reactions occurring at the anode and cathode andhence write a balanced equation for the overall reaction that takes placewhen a current flows.

anode: Zn Zn2++ 2e-

cathode: O2+ 2H2O + 4e-4OH-

overall: 2Zn + O2 + 2H2O 2Zn(OH)2

(ii) Calculate the Eocellof this cell.

Eocell= +1.16 V

(iii) How would the Eocellvalue change if the pH of the body fluid decreases below7.4? Briefly explain your answer.

pH falling below 7.4 implies fluid gets more acidic. This allows the position ofequilibrium of the overall reaction to shift right because Zn(OH)2dissolves.Therefore, e.m.f. increases.

(iv) If a current of 4 10-5A was drawn from the cell, calculate how long a zincelectrode weighing 5.0 g will last before it needs to be replaced. Correct your

answer to the nearest year. Assume that there are 365 days in a year andthat the cell e.m.f. remains unchanged throughout usage of the cell.

Amount of Zn = 5 65Complete consumption of Zn requires passage of 2 5 65 mol e-.

Charge passed 2 (5 65) 96500 = 4 10-5tt = 12 years.


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(b) A concentration cell is one that generates voltage due to a concentration difference.One example is shown below:

Cu(s) | Cu2+(aq) (0.01 mol dm-3) || Cu2+ (1.0 mol dm-3) | Cu(s).

(i) Calculate the reduction potential of the oxidation half-cell if the Ecell

generated is 0.06 V.

Ecell= Eored(Cu

+/Cu) Ered(Cu+/Cu)

Ered(Cu2+/Cu) = +0.28 V

(ii) State how the concentrations of the Cu2+ions in the two half-cells are relatedwhen the cell becomes flat, i.e. Ecell= 0.00 V.

When Ecell= 0.00 V, electrons don't flow. [Cu+] in the two half cells are equal.


Q4. In an electrolytic cell, a current of 0.250 A is passed through a concentrated solution of achloride of iron, producing iron metal and chlorine gas.

(a) Write the equation for the half-reaction taking place at the anode.

2Cl- Cl2 + 2e-

(b) When the cell operates for 2.00 hours, 0.521 g of iron is deposited at one electrode.Determine the formula of the chloride of iron in the original solution.

Let formula of Fe ion be Fex+.

Amount of Fe formed = 0.521 56

No. of faradays required = (0.521 56)x

Amount of charge passed = 96500 (0.521 56)x = 1800 C

Therefore, x = 2 & Formula of chloride is FeCl2.

(c) Write a balanced equation for the overall reaction that occurs in the cell.

Fe2+ + 2Cl- Fe + Cl2

(d) What volume of chlorine gas, measured at r.t.p., is produced when the cell operates

as described in part (b)?

Amount of Cl2evolved = 0.521 56Vol. of Cl2at r.t.p. = 0.223 dm

3

(e) Calculate the current that would produce chlorine gas from the solution at a rate of3.00 g per hour.

Amount of Cl2liberated = 3 71Amount of charge required to produce (3 71) mol of chlorine

= 2 96500 (3 71) = I60 60

I = 2.27 A.

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Revision on Redox & Electrochemistry (Lecture Notes) (Teacher&Apos;s Copy)