Home A'LEVEL TOPIC 3: ELECTROCHEMISTRY ~ CHEMISTRY FORM 6

TOPIC 3: ELECTROCHEMISTRY ~ CHEMISTRY FORM 6

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TOPIC 3: ELECTROCHEMISTRY ~ CHEMISTRY FORM 6

ELECTROCHEMISTRY:- Is a branch of science which deals with the study of chemical/physical processes in which electricity is either produced or consumed.

REDUCTION-OXIDATION (REDOX) REACTIONS

Reduction is the addition of electrons to an atom or ion.

E.g. (i)

( ii)

There is an overall decrease in oxidation state.

Oxidation is the reaction in which electrons are being lost therefore removal of electrons from the atom or ion.

E.g.

Redox reaction is the reaction in which both oxidation and reduction processes takes place at the same time. In a redox reaction, electrons are transferred between ions or atoms thus electrons are lost and gained in the same reaction.

NOTE: – In a balanced redox reaction, the number of electrons lost should be equal to the number of electrons gained.

From the above reaction (overall reaction)

The substance in a redox reaction which loose electrons is called reducing agent or reductant i.e. Zn the substance in a redox reaction which gains electrons is called oxidizing agent (oxidant) i.e.

NOTE: An oxidizing agent which cause other species to undergo oxidation but itself being reduced.

Not all reactions are redox reaction :-

(i)

(ii)

Reason

To identify a redox reaction one should look for the change in oxidation number of an element in the course of the reaction. If there is increase in oxidation number, oxidation has taken place and if there is a decrease in oxidation state, reduction has taken place.

Guidelines or rules for determination of oxidation number of an element in a compound.

1. In a free element, each atom has an oxidation number of zero.

2. For ions consisting of single atom, the oxidation number is equal to the charge of that ion.

3. The oxidation of hydrogen is except in ionic hydrides where the oxidation is . E.g. NaH, KH.

The oxidation state of oxygen is -2 in most compounds except in peroxide where the oxidation state becomes -1 and +2 in oxygen fluoride due to electronegativity of fluorine.

4. The algebraic sum of oxidation number in a neutral compound must be zero and a polyatomic atom must be equal to the ion charge.

E.g.: calculate the oxidation number of the underlined elements in the following.

(i) S

(2×1)+ S + (-2×4) = 0

2 + S – 8 = 0

S = ⁺6

(ii) Cr

2 Cr + (-2 x7) = -2

2 Cr = 12

Cr = ⁺6

(iii) N

N + (-2 x 3) = -1

N – 6 = -1

N = ⁺5

(iv) KMn

1 + Mn + (-2 x4) = 0

1 + Mn – 8 = 0

Mn – 7 = 0

Mn = ⁺7

(v) P

4P= 0

P=0

(vi) N

N+ (1x 4) = 0

N = -4

(vii) S

2S + (-2×3) = -2

2S – 6 = -2

2S = 4

S = ⁺2

Disproportion reaction

Is the reaction in which an atom undergoes both oxidation and reduction reaction simultaneously in the same reaction.

OXIDATION – REDUCTION PROCESS

(i)
(ii)

Rules for balancing redox reaction

eg.

1. By using oxidation number identify which element is oxidized or reduced.

eg.

2. Write the half reaction for each process.

eg.

3. Balance the atoms that are reduced or oxidized.

4. Balancing number of oxygen atoms

a) In acidic solution

i. Add water molecules to the side that is deficient of oxygen atoms.

ii. Add appropriate number of hydrogen ions to the other side to complete oxygen balance.

eg:-

b)In basic solution.

i. For every oxygen atom required add OH^– to the side deficient of oxygen atoms.

ii. Add water molecule to the other side to complete the oxygen balance

eg:-

5. Balancing the number of hydrogen atom

(a) In acidic solution.

Add hydrogen ions to the side in deficient of hydrogen atoms.

(b) In basic solution.

(i) To every hydrogen atom required add water molecule to the side deficient of hydrogen atom.

(ii) To the other side add hydroxyl ion to complete hydrogen balance.

6. Balance the charge by adding number of electrons to the side which is more positive.

7. Multiply oxidation/reduction by the smallest number to ensure that the numbers of electrons cost/gained are equal.

8. Add the two balanced half reactions by omitting species which appear on both side to obtain the final equation.

9. Check the final result to ensure that species and charge are balanced.

eg. Overall reaction

Example 2

(a) Balance the following redox reaction equation under acidic medium.

SOLUTION
1. Half reaction equation

2. Balance under acidic medium

3. Balanced charged.

4. Overall reaction equation

(b) Balance the following redox reaction equation under basic medium

Therefore overall reaction :-

eg 2.

1.Half reaction

2.Half reaction under basic medium

3.Half reaction under basic medium

4. Balanced of charged

Overall reaction equation

Example 3
Balance the following redox reactions according to the media given.

ANSWERS

Solution

3. Solution

Solution

∴ SnO₂ is an oxidant

CrO is an reductant

ACTION OF HYDROGEN PEROXIDE

Hydrogen peroxide (H₂O₂) has both oxidizing and reducing power. However its reaction depends on the state of the second reagent whether it has to be reduced or oxidized,

(i) H₂O₂as an oxidizing agent
Like any oxidizing agent H₂O₂ is capable of accepting electrons when treated with any electron donating species and itself being reduced to water.

eg∴

(ii)H₂0₂ as reducing agent

H₂O₂ is capable of supplying electrons when treated with electron accepting species and itself being oxidized to oxygen gas.

E.g.

APPLICATION OF REDOX REACTIONS

(i) PERMANGANOMETRY TITRATION

It is applied on permanganometry titration. These are titration in which potassium permanganate is the tit rants and indicators are used. This is because permanganate itself acts as an indicator. It changes its colour due to the formation of manganese (Mn ²⁺) from manganese (Mn ⁷⁺) [in acidic media]

NOTE
Oxidizing power of permanganate depends on the media used.

a) In neutral media or weakly alkali medium.

b) In acidic medium

NOTE
Dilute. Sulphuric acid (H₂SO₄)is only used HCl cannot be used because will oxidize it to chlorine gas thus is interfering the power of permanganate

eg.

NOTE :- Concentration. and are also not used because they are also oxidizing agents.

c) In strong alkali medium.

n = number of electrons transferred per mole.

Basicity is the number of H⁺ per 1 molecule of acid when dissolved in H₂O.

Examples

1.A standardization of in acidic solution gave the following data:

0.16g of potassium salt i.e. needed solution. What is the molarity of the solution
(

Solution:
(i) To find molarity of K₂ C₂ 0₄ .2H₂0

Concentration = 0.032gL^-1

202g = 1 mol

0.16g = x

X = 7.92 × 10^-4 moles

Concentration = 0.032 g dm^-3

Molarity= 2.038 x 10^-4

(ii) To find molarity of KMno₄ from reaction equation :-

2:5

= 7.92 x 10^-4 moles

From the reaction equation :-

2 moles of KMnO₄ = 5 moles of

2. Calculate the percentage of iron from the following data in a sample of iron wire:

1.4g of the wire was dissolved in excess dilute sulphuric acid and the solution was made up to 250cm³. 25cm³ of this solution required for oxidation. [If it was concentration. , it could have oxidized Fe to Fe³⁺ since it is a STRONG oxidizing agent]

Solution:

……………………………………. (1)

Reaction of and

1 mole of Fe →1 mole of FeSO₄

? x mole →0.0249 moles

X = 0.0249 moles of Fe

n = 0.025 moles

3. 100cm³ of H₂O₂ solution was diluted to 1 dm³ of solution. 25cm³ of this solution, when acidified with dilute H₂SO₄ reacted with 47.8cm³ of solution. Calculate the concentration of the original H₂O₂ solution.

Solution:

Ratio is 5: 2

= 0.0956M

0.0956 moles → 1000cm³

X → 100cm³

X = 9.5 x 10^-3 moles

M_D = 0.0956

V_D = 1000cm³

M_c =?

V_C = 100

4. 25cm³ of sodium oxalate (Na₂C₂O₄) solution acidified with dilute sulphuric acid and heated to 80^°c is titrated with standard KMnO₄ solution of concentration 3gl^-1. 26.4cm³ of KMnO₄ solution was required for complete reaction.

a.Calculate the concentration of oxalate ions in Na₂C₂O₄in gl^-1

b.Hot acidified Na₂C₂O₄ reacts with MnO₂ according to the equation.

What volume of sodium oxalate (Na₂C₂O₄) will be required to react with 0.1g of MnO₂.

Answer

Conc. = x M.M

x = 1.149 x 10^-3 moles of C₂O₄

(ii) Iodometry titration

These are titrations in which iodine is produced by using of starch as an indicator. All iodometry
titrations use KI as source of Iodine.

a)If the oxidant is acidified KMnO₄

……………… (1)

The iodine produced is titrated with to pale yellow, then add starch and continue titrating until the blue-black colour is discharged.

Starch is not added at the beginning because the concentration of iodine is large. Iodine react with starch to form the blue black complex hence more volume of is required to discharge the blue black complex.

Reaction with Na₂S₂O₃

When equation (1) and (2) are combined

1: 5

b) If the oxidant is acidified K₂Cr₂O₇

Mole ratio will be

1: 6

c) If the oxidant is acidified KIO₃

The mole ratio is

1: 6

ELECTRODE POTENTIAL AND ELECTROCHEMICAL SERIES

ELECTRODE POTENTIAL

Metals have a small tendency to dissolve in solution of their ions producing cations leaving their valency electrons on the metal rod. The metal acquires a negative potential which prevents further release of cations and equilibrium is established

ne^– ⇒ number of electron (s)

As a result the region of solution very close to the rod suffers an increase in charge while the rod carries a layer of negative charge (electrons). Then an electric double layer is set up and this layer is known as “Helmholtz double layer”

Whenever there is a separation of negative and positive charges we should be able to measure the voltage i.e. voltage between the electrode and surrounding solution and this is called Electrode potential.

Definition:

Electrode potential – is the potential difference formed between an electrode and its hydrated ions.

Magnitude of electrode potential

This depends on the position of equilibrium of reversible reaction. The further to the right the greater is the electron density on the surface of metal and larger is the Potential Difference (P.D) between metal and solution. The opposite is true.

For a given metal the position of equilibrium forward or backward depends on the concentration of solution into which the electrode is dipped

i.e.

If the concentration of the solution is high the equilibrium will lie towards the left i.e. tendency of zinc rod to dissolve decrease and vice versa.

For different metals placed in solution contain same concentration of their ions at the same temperature i.e. the positions of equilibrium is governed by the overall energy change forming hydrate ions from the metal.

Electrode potential involves the following stages

(i)Atomization of electrode

(ii) Ionization of gaseous atoms

(iii)hydration of ions

Atomization ionization hydration

Zn_(s) → Zn_{(g) →}Zn²⁺_→Zn_(aq)²⁺ Energy Energy Energy

If the total energy is low, electrode dissolve more easily and its equilibrium will move forward hence large potential (electrode potential) value

Metals with large electrode potential release electrons easily and are good reducing agents

TYPES OF ELECTRODES FOUND IN ALL GALVANIC CELLS

I. Metal – metal ion electrodes

This consists of metal dipped into its soluble salts.

II. Gaseous electrode

This consists of platinum to which a gas at 1 atm, and 25^°c is bubbled and dipped in ions of gas at a given concentration.

III. Metal – insoluble salt electrode

This consists of metal dipped into its insoluble salts at

IV. Redox electrodes
It consists of platinum dipped in cations having different oxidation states at a given concentration at 25^oc.

Measurements of electrode potential

We can always measure the electrode potential for an electrode in combination with standard electrode which is standard hydrogen electrode (SHE)

It was conventionally agreed that electrode potential for hydrogen is zero.

i.e.

When electrode potential of any element is measured against hydrogen electrode at 1 atm and 1M concentration is called standard electrode potential.

Salt Bridge

Is an inverted U-tube that contains on electrolyte (e.g. ) which connects the solution of two half cells to balance the charge and to complete the circuit.

NOTE
The standard redox potential is actually reduction potential. All elements below hydrogen have negative reduction potential and positive oxidation potential hence strong reducing agents.

All elements above hydrogen have positive reduction potential and negative oxidation potential hence strong oxidizing agents.

When writing the cell description the hydrogen electrode is placed on the left conventional to determine the polarity of the right hand electrode.

Functions of salt bridge
I.To complete the circuit

II.To balance the charge

Example

Consider the following arrangements for determine electrode potential fo
I. =

The zinc electrode is negative.

Cell reaction:

II.Copper

The copper electrode is positive. (It can easily be reduced)

Cell reaction:

Overall:

Application of standard electrode potential

I. Construction of electrochemical cells.

II. Prediction of occurrence of chemical reaction

III. Determination of of a solution without meter

IV. Replacement of elements in the electrochemical series.

Electrolytic cell = produce electrical power through chemical reaction in electrochemical cells.

I. CONSTRUCTION OF ELECTROCHEMICAL CELLS

Electrochemical cells are devices that use chemical reactions to produce electrical power. These are sometimes known as Galvanic or Voltaic cells.

e.g. dry cells, car batteries.

It contains two half cells connected together by external circuits.

Half cell is an arrangement which consists of an electrode dipped into a solution containing its ions. When the two half cells are connected, the resulting component is called electrochemical power. A good example is Daniel cell. It is constructed from Zinc and copper electrode.

PROCEDURE FOR CONSTRUCTION OF ELECTROCHEMICAL CELL

1. Identify between the electrodes, which electrode is supplying or gaining electrons by studying their electrode potential.

II. An electrode with negative standard electrode potential is more reactive (supplies electrodes) than the positive one.

III. If both are negative the one which has more negative electrode potential is more reactive.

E.g. Sn = -0.16

Mg = -0.25 (more reactive)

If both electrodes are positive, the one which has less positive value is more reactive.

E.g. 0.07v (more reactive)

0.2v

The electrode which supplies electrons should be placed on the left and electrode which gains electrons should be on the right hand side.

E.g. construct a Daniel cell and shows the direction of flow of electrons and current given that

The electrode which supplies electrons oxidation takes place and the electrode is oxidized. This electrode is called ANODE. The electrode which receives electrons reduction takes place and the electrode is REDUCED. It is called CATHODE.

Zn – anode

Cu – cathode

Overall reaction is obtained by adding the two half reactions and is called cell reaction.

Anodic reaction:

Cathodic reaction:

Cell reaction:

Electrochemical cell can also be represented in an abbreviation way known as cell notation.

Anode cathode

I.e. for Daniel cell

Questions

1. Given the overall reactions, write their corresponding cell notations

2. From the cell notation give the cell reactions

Answers

_\_{Cu(aq) + 2Ag}⁺_(aq)→Cu²⁺_{(aq) +2Ag(s)}_2.
ELECTROMOTIVE (EMF) FORCE OF A CELL

The difference between electrode potential of the two electrodes constituting an electrochemical cell is known as electromotive (emf) or cell potential.

This acts as a driving force for a cell reaction and it is expressed in volts.

The emf of a cell is calculated by subtracting the standard electrode potential of the left electrode from that of the right electrode.

For Daniel cell

= 0.34 – (-0.76)

= 1.1v

Alternatively

Since standard electrode potentials are in reduced form, for oxidation half reaction the sign of electrode potential should be reversed.

= 0.76 + 0.34

= 1.1v

1.Prediction of the occurrence of chemical reaction.

If the emf of the cell calculated is negative the reaction is non spontaneous i.e. the reaction does not occur in the way it is written unless external forces apply but the reaction is spontaneous in the opposite direction.

If the emf of the cell is positive the reaction is spontaneous

Example: given the following reaction

Predict the direction of the reaction given that reduction potential of nickel (Ni²⁺/Ni) E^o = -0.25v of mercury Hg⁺/Hg = 0.14v

Solution

Cell notation

The reaction is non spontaneous since emf is negative hence backward reaction is favored.

Example
For each of the following reactions at standard condition, decide if it occurs spontaneously in the direction written.

Given for .

ANSWERS
I.

=-0.14 – (-0.76)

= 0.62V

II. Cell notation:

=1.09 – (1.36)

= -0.27V

III. Cell notation:

=-0.4 – 0.34

=-0.74V

Therefore reaction (1) is spontaneous while reaction (2) and (3) are non-spontaneous

Example
Briefly explain what happen when

I.Fe is dipped in CuSO₄ solution

II.Cu is dipped in FeSO₄ solution

Example
Given the following values

a) State which species are the strongest oxidant and which oxidant and weakest reductant.

b)State which species is the strongest oxidant and weakest reductant.

c)The lead rods are placed in a solution of each CuSO₄, FeSO₄, AgNO₃ and ZnSO₄. In which solution do you except coating of another metal on lead rod. Explain.

ANSWERS

2. (i) since Fe has a negative reduction potential then it has a positive oxidation potential and hence will displace copper metal from CuSO₄ solution

(ii)Since Cu has positive reduction potential, then it has a negative oxidation potential and hence it will NOTE displace Fe from FeSO₄solution therefore no reaction will be possible.

3. (a) strongest oxidant

Weakest reductant

Maximum and minimum emf of a cell

It is possible to decide which cell has to be constructed either of great or smallest emf for a given electrodes. A cell with greatest emf is obtained by

choosing two electrodes of greatest cell reactivity difference. A cell with minimum emf, choose two electrodes with closest reactivity.

Example

Study the following electrodes.

4. Explain how you can construct a cell that will yield

i. Maximum emf

ii. Minimum emf

ANSWER

For maximum emf, we choose two electrodes of greatest cell reactivity difference.

= 2.87 – (-3.04)

= 5.91v

For minimum emf, we choose two electrodes with closest reactivity

= -0.4 – (-0.44)

= 0.04v

EFFECTS OF CONCENTRATION AND TEMPERATURE ON CELL POTENTIALS

We have been considering electrode potential under standard conditions of molar solution, pressure of 1 atom and 298K. When the conditions are altered he values of electrode potential changes thus we have to define the potential of the cell under non-standard conditions.

The Nernst equation shows the relationship between emf of the cell at standard conditions and emf under non-standard conditions.

Where; – emf of a cell at any conditions

– emf of a cell under standard conditions

R – Universal gas constant (8.314 Jmol^-1K^-1)

n – Number of moles electrons being transferred

F – Faraday’s constant (96500c)

At standard temperature (298K)

The above equation can be applied to half reactions and overall reactions

NOTE
Always write a balanced cell reaction to obtain ‘n’ and position of ions, either reactants or products together with their stoichiometry

Example

Calculate the emf of the given cell at standard temperature.

Given

Alternative Solution 1

= -0.13 – (-2.56)

= 2.43v

Alternative Solution 2

n = 6

Now;

Example 2

Calculate the emf of Daniel cell at using 2M ZnSO4 solutions and 0.5M CuSO₄ solution

Alternative Solution 1

= 0.34 – (-0.76)

= 1.1v

Alternative Solution 2

Half reactions

(Oxidation)

(Reduction)

Here n = 2

Now using

E_cell =1.08v

Example 3

Calculate the emf for the following voltaic cells

(b)Zn/Zn²⁺(10^-3M)//Ag⁺(10^-3M)/Ag

Given

Alternative Solution 1

n = 2

Now;

= 0.8 – (-0.76)

= 1.56v

Alternative Solution 2

From

= 1.47v

d) Solution

Alternative Solution 1

= 0.8 – (0.77)

= 0.03v

Alternative Solution 2

Half reactions

n = 1

From

E_cell =0.0122v

c) Solution

= -0.14 – (-2.56)

= 2.42v

Half reactions

2Al + 3Sn²⁺ ® 2Al³⁺ + 3Sn

n = 6

Now;

E_cell =2.418v

EQUILIBRIUM CONSTANT OF GALVANIC CELL

Consider to what happens to a Daniel’s cell if we use it to do some electrical work i.e. it can be connected to a small electric motor.

After sometimes the motor will stop. The cell will run down. When this happens and there is no overall transfer of electricity from one half cells to the other.

When there is no overall change taking place in a chemical reaction the equilibrium has been established. At equilibrium, electron density of both electrodes is equal and there is no transfer of electricity between the two half cells.

Applying the Nernst equation to a Daniel’s cell

Since the reaction is at equilibrium, the ratio

K_C = 1.67 x 10³⁷

The large value tells us that, the equilibrium lies almost entirely in favour of copper metal and zinc ions.

Generally, the cell at equilibrium at standard temperature is given by the following expression;

Example

Calculate K_C for the following voltaic cell

Given

Cell reaction

From

= -0.12 – (-0.46)

= 0.34v

Now,

K_C = 3.20 x 10¹³

Example
What concentration of will emf of the cell be zero at if concentration of is

Solution

Cu + 2Ag⁺ ® Cu²⁺ + 2Ag

n = 2

From

When

Now,

= 0.8 – (0.338)

= 0.462v

3. Given and . Calculate the equilibrium constant for the reaction State the significance of equilibrium constant.

Solution

= 0

Cu + 2Cu²⁺ ® 2Cu⁺

n = 2

Using

= 0.53 – 0.34

= -0.187v

= -6.3282

K_C = 4.69 x 10^-7

Question
(a) Given the which is thermodynamically feasible, the reduction of Cu²⁺ by Ag or reduction of Ag⁺ by Copper.

(b) Calculate the standard emf of the cell and the equilibrium constant.

CONCENTRATION CELL

A half concentration cell which consists of two half cells with identified electrode that differs in ion concentration because the electrodes are identified. for oxidation is numerically equal and opposite in sign to for reduction. As a result

The reaction in the cell takes place in order to reduce the difference in concentration until (when the two concentrations are equal. The higher concentration is reduced and the lower concentration is increased.)

Oxidation reduction

Cu ® Cu²⁺ + 2e^–Cu²⁺ + 2e^– ® Cu

Electrochemical cell: Anode (-)

Cathode (+)

Electrolytic cell: Anode (+)

Cathode (-)

Question

Calculate the emf of the following concentration cell

Solution

Using Nernst equation

= 0.0591v

3. MEASUREMENT OF pH OF A SOLUTION USING STANDARD ELECTRODE POTENTIAL

pH is the degree of alkalinity, or acidity of a substance. It is obtained by using the hydrogen ion concentration pH = –

The pH of a solution is determined by using any electrode provided its standard electrode potential is known and the concentration of ions in that electrode should be 1M. This will be one of the half cells another half cell is made up hydrogen electrode dipped into the solution whose pH is to be determined

Reaction:

H_2(g) +2 H⁺_(aq) +2e^– E ⁰= 0.00v

2Ag⁺_(aq)+ 2e^– Ag_(s) E = 0.8v

H_2(g)+ 2Ag⁺_(aq)2H⁺_(aq0+ Ag_(s) E⁰ = 0.8V

Applying Nernst equation at standard temperature

E _cell= _cell–

= 0.8 –

= 0.8 + 2(-

E _cell= 0.8 + 0.0591

pH= =

Question: Calculate the of the following cell and hydrogen ion concentration.

Zn/Zn²⁺// H⁺/H₂, pt

Zn²⁺/Zn = -0.76v

E _cell= 0.115v

Solution

= 0 – (-0.760)

= 0.76v

= –

Zn_(s) → Z_(aq) + 2

2H⁺_(aq) + 2 → _(g)

Zn_(s) + 2H⁺_(aq)Zn²⁺_(aq)+ _(g)

= –

= + x 2

0.115 = 0.76 + 0.0591

-0.0591) = 0.645

– = 10.913

= 10.91

pH DEPENDENCE OF REDOX POTENTIALS

Whenever a redox half equation involves H⁺ or OH ion, its redox potential depends on the of the solution.

Example

_(aq)+ _(aq)+ _(aq)+ 4O_(l) E = 1.52V

The Nernst equation at standard temperature is

= –

Under standard conditions, all effective concentrations are 1M

= 1.52 –

= 1.52 – 0

= 1.52v

Suppose the is changed from O to 5 i.e. the concentration changes from 1M to 1 x 10^-5M

= 1.52 – )^-8

= 1.52 +

= 1.04 v

Due to change of from O to 5, the electrode potential decreases from 1.52 to 1.047v.
That means, permanganate (vii) becomes a less powerful oxidizing agent when PH increases

Question1.
Calculate of the following cell;-

Zn/Zn²⁺//H⁺/H₂, pt

Zn²⁺/Zn = -0.76v

E _cell= 0.115v

Question2.
If E for Zn /Zn²⁺//Cn²⁺/Cn is 1.1v

i.Calculate the E _cell when concentration of Zn²⁺ is 2M and Cu²⁺ is 0.5M

ii.What is E _cell when concentration of Zn is 0.4M, = 0.1M

Question3.
Write down the expression for the cell emf for the following reaction;

_(aq)+ _(aq)+ _(aq)+ 4H₂O_(l)

Briefly explain why the oxidizing power of permanganate (Vii) ion is quite sensitive to the concentration of H⁺ in the solution.

ELECTROCHEMICAL SERIES

Is the series of standard electrode potential with respect to their elements from more negative standard electrode potential to the more positive standard electrode potential.

Is the arrangement of electrodes of elements in order of reducing power

Uses

It is a good guide for predicting reaction that takes place in solution especially displacement reaction.

Displacement reaction

Is a types of reaction in which an atom or element displace another element or atom in a compound

Example
What will happen when magnesium ribbon is added to a solution of AgNo₃?

= -2.37v

= 0.8v

The more negative the electrode potential the greater is the reducing power of that element i.e. the more likely it is to give out the electrons and acts as a reducing agent therefore Mg will reduce Ag⁺ions to Ag _(s).

Mg_(s)

2Ag⁺_(aq)+ 2 2Ag_(s)

Mg _(s)+ 2Ag⁺_(aq) Mg²⁺_(aq)+ 2Ag _(s)

Mg_(s)/Mg²⁺_(aq)//Ag⁺_(aq)/Ag_(s)

= –

= 0.8 – (-2.37)

= 3.17v

Thus, the element higher in electrochemical series will displace the one lower in the series.

2. Displacement of hydrogen from mineral acids

Metals which are higher in the electrochemical series than hydrogen reacts with acids and replaces hydrogen but metals below hydrogen have no action with mineral acids.

Complete the following reaction:-

i. Zn_(s) + 2HCl _(aq) ZnCl_2(s)+ H_2(g)

ii. Cu_(s) + HCl_(aq) No reaction

3. The knowledge of electrochemical series helps as on choosing method for extraction of metals.

Example

Higher most metal can’t be extracted from the oxides by chemical reduction process. This is because they are strong reducing agent hence they can be reduced easily from their oxide by electrolysis.

Questions

For electrolysis, fused or molten metal should be used and not aqueous solution. Why?

Answer:-

In aqueous solution, there are H⁺ ions. Hence metals prefer to react with element in lower electrochemical series than the metal itself, hence for electrolysis we use fused or molten metal and not aqueous solution.

CORROSION AND ITS PREVENTION

Corrosion is the deterioration of the metals due to the chemical reactions taking place on the surface. Usually, the process is due to the loss of metal to a solution in some forms by a redox reaction (unwanted redox reactions)

For corrosion to occur on the surface of a metal there must be anodic area where a metal can be oxidized to metal ions as electrons are produced.

Anode area:-
M_(s) Mⁿ⁺+ ne⁺ And cathodes area where electrons are consumed by any of all of several half reactions.

Cathodic reaction:-

2H⁺_(aq) + 2e^– H_{2 (g)}

2H₂O_(l)+ 2e^– 2OH^–_(aq)+ H _2(g)

4e^–+ O_2(g)+H₂O_(l) 4OH^–_(aq)

Anodic reactions occur at cracks or around the area with some impurities.

RUSTING OF IRON

The most common corrosion process is rusting of iron

Rust is hydrated iron (III) Oxide (Fe₂O_3.XH₂O) which appears as a reddish brown substance on the surface of the iron bar.

Both water and air (oxygen) are required for rusting to occur.

The presence of dissolved salts and acid in water increases its conductivity and speed up the process of rusting.

If the iron object has free access to oxygen and H₂O as in flowing water, a reddish brown ion (III) oxide will be formed which is a rust.

For iron:

Anode: Fe_(s) Fe²⁺ _(aq)+ 2e^–

Cathode:O_2(g)+ 2H₂O_(l) + 4e^– 4OH^–_(aq)

2Fe_(s) + O_2(g)+ 2H₂O(l) 2Fe²⁺_(aq)+ 40H^–_(aq)

(rust)

4Fe(OH)_2(s)+ O_2(g)2H₂O_(l) + 2Fe₂O₃.XH₂O

If oxygen is not freely available: The farther oxidation of iron (II) hydroxide is limited to formation of magnetic iron oxide.

6Fe(OH)_2(s)+ O_2(g)2Fe₃O₃.H₂O + 4H₂O_(l)

Fe₃O₄ + H₂O_(l)Fe₃O₄+ 4H₂O

Black magnetite

Prevention of rusting

Corrosion can’t be made non-spontaneous but it can be prevented by making the rate of the reaction negligible. This can be done by covering the metal surface with protective coating or by providing alternative redox pathways (oxidation-reduction pathways).

Protective coatings are usually of 3 types:

i.Painting: – is the simplest and most common method where a metal surface is properly cleaned and then applied with several layers of rust – proofing paint.

ii.Corrosion inhibitors: – these interfere with flow of charges needed for corrosion to take place i.e. phosphate (II) coating on a surface of iron of steel. Using phosphoric acid serves that purpose.

iii.Galvanization: (sacrificial protection) a metal can also be protected by coating with a thin film of second metal where the second metal is oxidised instead of the 1^st metal.

Often iron is coated with another metal like zinc, tin or chromium for protection on the surface.

Note

zinc is preferred to tin because zinc protects iron against rusting when its coating has broken down. This is because it has a more negative reduction potential than iron: it acts as a cathode hence it is not changed. This is called cathodic protection.

Tin protects iron only as long as coating is intact. Once the coating is broken down, tin actually promotes corrosion of iron as iron has more negative reduction potential than tin. Thus iron acts as anode and dissolves while tin acts as cathode and does not change.

Factors which affect corrosion:

i. Position of metal in the electrochemical series(E.C.S). the reactivity of metal depends upon its position in the electrochemical series. More the reactivity, the more likely it is to be corroded.

ii. Presence of impurities in the metal. The impurities help to set up the voltaic cells which increase the speed of corrosion.

iii. Presence of electrolytes: Presences of electrolytes in water also increase the rate of corrosion. E.g. Corrosion of iron in sea water takes place to a large extent than in distilled water.

iv.Presence of CO₂in water: water containing CO₂ acts as an electrolyte and increases the flow of electrons from one place to another. (CO₂+ H₂O) form carbonic acid which dissolves into ions and hence acts as an electrolyte).

Examples

1. Why do you think zinc on iron is sometimes called sacrificial anode?

2. Explain why blocks of Mg can be attached to wills of ship or irons pipes with the aim of preventing rusting.

3. Tin cans are made of Iron coated with thin film of tin. After a crack occurs in the film, a can corrodes much more rapidly than Zinc coated with Iron. Explain this behavior.

4. Why is it that with enough time, corrosion will always defect the protection applied to iron?

ANSWERS

Zinc or Iron is sometimes called sacrificial anode because it after it wears off, the metal can get exposed and hence starts to undergo rust.

Conductivity in solutions

Electrolytes

These are substances which allow electricity to pass through them in their molten state or in form of their aqueous solution, and undergo chemical decomposition e.g., acids, bases and salts.

Classification of electrolytes

All electrolytes do not ionize by the same extent in the solution. According to this we have strong and weak electrolytes

Strong electrolytes are those that ionize completely into ions in the solution
e.g. Salts, mineral acids, some bases.

Weak electrolytes are these that ionize partially into ions in the solution
e.g. in organic acids. HCN, Na₄OH.

Electrolytic conduction

When a voltage is applied to the electrodes dipped into an electrolytic solution, ions of the electrolyte move towards their respective electrodes and therefore electric current flows through the electrolytic cell. The process of the electrolyte to conduct electric current is termed as conductance or conductivity.

Like metallic conductors, electrolytic solution also obey ohm’s law which states that “the strength of the current flowing through a conductor is directly proportional to the potential difference applied across the conductor and inversely proportional to the resistance of the conductor
i.e. V = RI.

Resistance of any conductor is directly proportional to the length L and inversely proportional to the area of cross – section.

R = ρx

ρ=Resistivity

Resistivity is the resistance of a conductor having unit length and unit area of cross-section. SI unit Ohm-meter (á½©m).

Conductance is the measure of the ease with which the current flows through a conductor a (λ orΛ)
Conductance is the reciprocal of electric resistance. ()

From the above expression high resistance means low conductance and vice versa.

Conductivity is the reciprocal of resistivity and is also called specific resistance (kappa)

= (

For an electrolytic cell, l is the distance apart between the two electrodes and A is the total area of cross-section of the two electrodes. Therefore, for a given cell, l and A are constant. If the dimensions of the cell are not altered, the ration is referred as cell constant (K)

= K

R = ρ =

= , = =

Molar conductance (λm)

Is the conductivity of volume of a solution which contains 1 mole of the solute.

It is the conducting power of the ions produced by dissolving 1 mole of an electrolyte in a solution.

Molar conductance is given by where v = volume containing 1 mole of a solute and is called dilution.

Concentration of an electrolyte depends on the volume i.e.

V =

Where C is the concentration in mol/dm³and is m^-1

Example

1. What is the dilution of 0.2M, NaOH solution?

Solution:

Given l = 0.2mol/dm³

V =

V = 5dm³mol^-1

2.0.0055M silver nitrate has a molar conductivity of 2.98 x 10^-3m²Ω^-1mol^-1. Calculate conductivity of that solution.

Solution

= x c

= 2.98 x 10^-3m²Ω^-1mol^-1x 0.0055moldm^-3

= 2.98 x 10^-3m²Ω^-1x 5.5cm^-3

λ∞= 1.303

5.The cell constant of the conductivity cell was stated as 0.215 cm^-1.
The conductance of the 0.01 moldm^-3solution of KNO₃was found to be 6.6 x 10^-4Ð…

i. What is the conductivity of the solution?

ii.What results does this give for the molar conductivity of KNO₃

Solution:

i) R = ρ

= =

K = 0.215cm^-1

= 6.6 x 10^-4s

= K

= 6.6 x 10^-4s x 0.215cm^-1

= 1.419 x 10^-4Ð…cm^-1

ii) =

= 14.195

Questions
1. How many grams of acetic acid must be dissolved in 1dm³ of water in order to prepare a solution with a conductivity and molar conductivity of 575cm^-1 and 9255cm²mol ^{– 1} respectively?

2. 0.05M NaOH solution offered a resistance of 31.6á½© in a conductivity cell at 298K. If the cell constant of a cell is 0.367 cm^-1. Calculate the molar conductivity of NaOH solution.

3. The conductivity cell filled with 0.01M KCl has a resistance of 747.5á½© at 25⁰c,when the same cell was filled with aqueous solution of 0.05M CaC, the resistance was 876á½©. Calculate.

i.Conductivity of the solution

ii.Molar conductivity of the solution if given of 0.01M KCl is 0.4114Ð…m^-1

VARIATION OF MOLAR CONDUCTIVITY WITH CONCENTRATION

The intensity of electricity that can pass through the solution depends on

i.The number of concentration of free ions present in the solution.

ii. Speed with which ions move to their respective electrodes.

An increase in concentration gives an increase in total number of solute particles in a given volume of solution and this might well be expected to give an increased conductivity.

Conduction in strong electrolytes.

The molar conductivity is high since strong electrolytes ionize completely into free ions and the molar conductivity increases slightly in dilution. WHY?

In strong electrolytes, there are vast numbers of ions which are close to each other. These ions tend to interfere with each other as they move towards their respective electrodes. The positive ion are held back by the negative ions and vice versa which in turn interrupts their movement to the electrodes (reduce the speed with which they move)

Dilution ions get separated from each other and at an average distance they can move freely or easily.

NOTE:-

As the dilution increases, there comes a time for amount of interference become small so that further dilution to has no effect.

At this point the molar conductivity remains constant and it is known as molar conductivity at zero concentration .

Conduction in weak electrolyte

The molar conductivity is less because there is less number of particles as the minority of particles are dissociated into ions. On dilution the molar conductivity increases as the molecules dissociate more into ions which increases the number of free ions.

Therefore for weak electrolytes the molar conductivity depends on the degree of dissociation of molecules into ions.

Question 1

At infinity dilution, will the molar conductivity of strong and weak electrolyte of same concentration be the same?

Answer

NO, the molar conductivity will be different because it also depends on the size of the ions which would either increase or decrease the speed of ions.

Question 2

Why at infinity dilution, the molar conductivity of weak electrolyte remains constant?

Answer

Because at infinity dilution the molecule must have to dissociate into free ions.

MOLAR CONDUCTIVITY AND DEGREE OF DISSOCIATION

The molar conductivity of weak electrolyte is proportional to the degree of dissociation.

i.e.

= K ……………. (i)

At infinity dilution, the weak electrolyte is completely ionized.

= 1 or 100%

= K (Constant)……….. (ii)

Taking ratio of equation (i) and (ii)

Ostwald’s dilution law

It states that

“For a weak electrolyte the degree of dissociation is proportional to the square root of reciprocal of concentration”

Consider a weak binary electrolyte AB (i.e. ethanoic acid) in solution with concentration C (mol/).

AB A⁺_(aq)+ B^–_(aq)

1mole 0 0 start

At equilibrium

But V =

AB A⁺_(aq)+ B^–_(aq)

K =

But AB = CH₃COOH

For a weak electrolyte, degree of dissociation is very small thus the expression

1 – 1

K_a=

K_a = C

Example

1. At 25⁰c the solution of O.1M of ethanoic acid has a conductivity of 5.0791×10^-2m^-1mol^-1.
calculate the pH of the solution and dissociation constant Ka (Answer Ka=1.69 10^-2M)

2. A 0.001 moldm^-3solution of ethanoic acid was found to have molar conductivity of 14.35Sm²mol^-1. Use this value together with molar conductivity at infinity dilution of ethanoic acid (C) is 390.7Ð…to calculate :-

i. Calculate degree of dissociation of acid
ii. Equilibrium constant

3.The resistivity of M KCl solution is 361Ω and conductivity cell containing such a solution was found to

have a resistance of 550Ω

a.Calculate the cell constant

b.The same cell filled with 0.1M ZnSO₄ solution had resistance of 72Ω. What is the conductivity of this solution?

KOHLRAUSCH’S LAW OF INDEPENDENT IONIC MOBILITY

Kohlrausch notes that the difference between molar conductivity at infinity dilution λ∞ values for the two salts which were strong electrolytes and of the same cation and anion was always constant.
Using the λ∞ values in D.. cm² mol^-1

This observation shows that each type of ion (caution or anion) contribute a definite amount of molar conductivity of an electrolyte of infinity dilution independently from other ions present in the solution i.e. a fraction of current that an ion carry is always constant and it doesn’t depend on the compound in which it is contained.

Hence Kohlrausch’s law

States that;”The molar conductivity of an electrolyte at infinity dilution is equal to the sum of the molar conductivities of the caution and anion”.
OR State that the molar conductivity at infinity dilution of the solution equal to the sum of molar conductivity at infinity dilution of its components ions.

i.e. = +

E.g. = + (C)

λ_∞(Al₂(S) = 2(AL³⁺) + 3(S)

Application of Kohlrausch’s law

In direct determination of molar conductivity at infinity dilution for weak electrolytes. Thus by using strong electrolytes we can easily. Calculate the molar conductivities at infinity. Dilution for weak electrolytes.

E.g.: The (CCOOH) can be determined from of potassium ethanoate (CCOOK) hydrochloric acid (HCl) and potassium chloride (KCl)

CH₃COOH + KCl CH₃COOK + HCl

λ_∞(CH₃COOH) = (CH₃COOK) + (HCl) – KCl)

= (CH₃COO^–) + (K⁺) + (H⁺) + (Cl) – (K⁺) – (Cl)

λ_∞(CH₃COOH) = (CH₃COOH^–) + (H⁺)

REVIEW QUESTIONS

1. Calculate (NH₄OH given that of three strong electrolyte NaCl, NaOH and NH₄Cl in Ð…cm² mol^-1 are 126.4, 248.4 , 149.8 respectively.

2.The molar conductivity of 0.093 CH₃COOH solution at 298k is 536 x 10^-4Sm²mol^{-1 .}The molar conductivity at infinity dilution of H⁺ and CH₃COO^– are 3.5 x 10^-2and 0.41 x 10^-2Sm² mol^-1.what is are the dissociation constant of CH₃COOH

3. A 0.05M HF solution has a conductivity of 91.81mol^-1 m^-1at 298k.At the same temperature (NaF), (NaoH) and (H₂O) are 493360 and 162Sm²mol^-1respectively. Calculate dissociation constant.
4. The λ_∞(NaI), λ_∞(CH₃COONa) and λ_∞(CH₃COO₂Mg) are 12.69,9.10 and 18.78Sm² mol^-1 respectively at 25⁰C. What is the molar conductivity of MgI₂ at infinity dilution.