FORM SIX CHEMISTRY STUDY NOTES ELECTROCHEMISTRY

ELECTROCHEMISTRY

ELECTROCHEMISTRY

ELECTROCHEMISTRY

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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)

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( ii)

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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.

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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.

 

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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.

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

Not all reactions are redox reaction  :-

(i)

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(ii)

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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.

 

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

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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.

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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.

 

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

(i)   S

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(2×1)+ S + (-2×4) = 0

2 + S – 8 = 0

S = +6

 

(ii) Cr

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2 Cr + (-2 x7) = -2

2 Cr = 12

Cr = +6

 

(iii) N

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N + (-2 x 3) = -1

N – 6 = -1

N = +5

(iv) KMn

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1 + Mn + (-2 x4) = 0

1 + Mn – 8 = 0

Mn – 7 = 0

Mn = +7

(v) P

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4P= 0

P=0

 

(vi)  N

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N+   (1x 4) = 0

N = -4

(vii)  S

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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)

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Rules for balancing redox reaction

eg.

 

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

eg.

 

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2.     Write the half reaction for each process.

eg.

 

 

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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:-

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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:-

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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.

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SOLUTION
1. Half reaction equation

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2. Balance under acidic medium

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3. Balanced charged.

4. Overall reaction equation

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

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Therefore overall reaction :-

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eg 2.

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1.Half reaction
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2.Half reaction under basic medium

 

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3.Half reaction under basic medium

 

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4. Balanced of charged

 

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Overall reaction equation

 

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

Solution

 

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Solution

3. Solution

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Solution

 

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Solution

∴ SnO2 is an oxidant

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CrO is an reductant

ACTION OF HYDROGEN PEROXIDE
Hydrogen peroxide (H2O) 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,

 

 

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(i)  H2O2 as 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∴

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(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.

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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.

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b)     In acidic medium

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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

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eg.

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

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c)     In strong alkali medium.

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1.

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n = number of electrons transferred per mole.

 

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

 

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Examples

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

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

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

 

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Concentration = 0.032gL-1

 

202g = 1 mol

0.16g = x

 

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X = 7.92 × 10-4 moles

 

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Concentration = 0.032 g dm-3

 

 

Molarity= 2.038 x 10-4

(ii) To find molarity of KMno 4 from reaction equation :-

 

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2:5

 

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= 7.92 x 10-4 moles

 

From the reaction equation :-

2 moles of KMnO4 = 5 moles of

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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]

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Solution:

……………………………………. (1)

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Reaction of  and

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=

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s

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1 mole of Fe →1 mole of FeSO4

? x mole →0.0249 moles

X = 0.0249 moles of Fe

 

 

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n = 0.025 moles

 

 

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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.

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Solution:

 

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Ratio is 5: 2

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= 0.0956M

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0.0956 moles → 1000cm3

X → 100cm3

X = 9.5 x 10-3 moles

 

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MD = 0.0956

VD = 1000cm3

Mc =?

VC = 100

 

 

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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 Na2C2O4 in gl-1

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

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

 

Answer

  

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=

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Conc. = x M.M

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=

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b)

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x = 1.149 x 10-3 moles of C2O4

 

 

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(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.

 

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a)If the oxidant is acidified KMnO4

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       ……………… (1)

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

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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.

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Reaction with Na2S2O3

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When equation (1) and (2) are combined

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1: 5

 

 

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1: 5

 

b) If the oxidant is acidified K2Cr2O7

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Mole ratio will be

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1: 6

c) If the oxidant is acidified KIO3

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The mole ratio is

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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)

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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”

 

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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.

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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.

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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.

 

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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.

 

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III.Metal – insoluble salt electrode

This consists of metal dipped into its insoluble salts at

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

 

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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.

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

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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.

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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.          =

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The zinc electrode is negative.

Cell reaction:

 

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II.Copper

 

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

Cell reaction:

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Overall:

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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

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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

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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:

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Cathodic reaction:

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Cell reaction:

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

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Anode                    cathode

I.e. for Daniel cell

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Questions

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

a)

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b)

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2.    From the cell notation give the cell reactions

a)

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b)

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Answers

1.

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                  \
           Cu(aq) + 2Ag+(aq)→Cu2+(aq) +2Ag(s)
2.

ELECTROMOTIVE (EMF) FORCE OF A CELL

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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.

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For Daniel cell

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= 0.34 – (-0.76)

= 1.1v

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Alternatively

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

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= 0.76 + 0.34

 = 1.1v

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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

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

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Solution

Cell notation

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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.

a)

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b)

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c)

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Given for    .

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ANSWERS
I.

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=-0.14 – (-0.76)

= 0.62V

II.  Cell notation:

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=1.09 – (1.36)

= -0.27V

III.  Cell notation:

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=-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

 

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Example
Given the following  values

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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

 

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(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.

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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)

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= 5.91v

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For minimum emf, we choose two electrodes with closest reactivity

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= -0.4 – (-0.44)

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= 0.04v

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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.

 

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Where;  – emf of a cell at any conditions

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– emf of a cell under standard conditions

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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)

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OR

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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.

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Given

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Alternative Solution 1

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= -0.13 – (-2.56)

= 2.43v

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Alternative Solution 2

 

 

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n = 6

 

Now;

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Example 2

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

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Alternative Solution 1

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= 0.34 – (-0.76)

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 = 1.1v

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Alternative Solution 2

Half reactions

(Oxidation)

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(Reduction)

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Here n = 2

Now using

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Ecell =1.08v

Example 3

Calculate the emf for the following voltaic cells

a)

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(b)Zn/Zn2+(10-3M)//Ag+(10-3M)/Ag

c)

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Given

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Alternative Solution 1

 

 

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n = 2

Now;

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= 0.8 – (-0.76)

= 1.56v

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Alternative Solution 2

From

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= 1.47v

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d)  Solution

Alternative Solution 1

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= 0.8 – (0.77)

= 0.03v

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Alternative Solution 2

Half reactions

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n = 1

From

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Ecell =0.0122v

 

c)  Solution

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= -0.14 – (-2.56)

= 2.42v

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Half reactions

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2Al + 3Sn2+ ® 2Al3+ + 3Sn

n = 6

Now;

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word image 2513

Ecell =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.

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Applying the Nernst equation to a Daniel’s cell

 

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Since the reaction is at equilibrium, the ratio

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KC = 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;

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Example

Calculate KC for the following voltaic cell

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Given

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Cell reaction

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From

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= -0.12 – (-0.46)

= 0.34v

Now,

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KC = 3.20 x 1013

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

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Solution

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Cu + 2Ag+ ® Cu2+ + 2Ag

n = 2

From
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When

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Now,

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= 0.8 – (0.338)

= 0.462v

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

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Solution

 

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= 0

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Cu + 2Cu2+ ® 2Cu+

n = 2

Using

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= 0.53 – 0.34

 = -0.187v

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word image 2561

= -6.3282

word image 2562

 

KC = 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.

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(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

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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.)

 

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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

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Solution

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Using Nernst equation

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= 0.0591v

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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 = –

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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

 

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Reaction:

H2(g) + 2 H+(aq) +2e                       E 0 = 0.00v

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2Ag+ (aq)+ 2e  Ag(s)                   E = 0.8v

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H2(g) + 2Ag+(aq) 2H+(aq0+ Ag(s)        E0   = 0.8V

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Applying Nernst equation at standard temperature

E cell =  cell

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= 0.8 –

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= 0.8 + 2(-

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E cell = 0.8 + 0.0591

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pH=   =

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

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Zn/Zn2+ // H+ /H2, pt

Zn2+ /Zn = -0.76v

E cell = 0.115v

Solution

= 0 – (-0.760)

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= 0.76v

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=  –

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Zn(s)   →  Z(aq)  +  2

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2H+(aq)  +  2  →  (g)

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Zn(s) + 2H+(aq)  Zn2+(aq)+ (g)

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 =  –

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 =  +  x 2

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0.115 = 0.76 + 0.0591

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-0.0591) = 0.645

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– = 10.913

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= 10.91

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            pH DEPENDENCE OF REDOX POTENTIALS

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

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Example

(aq)+  (aq)+    (aq)+ 4O(l)      E = 1.52V

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The Nernst equation at standard temperature is

=  –

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=  –

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Under standard conditions, all effective concentrations are 1M

= 1.52 –

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= 1.52 – 0

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= 1.52v

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

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= 1.52 – )-8

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= 1.52 +

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= 1.04 v

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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

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Question1.
Calculate  of the following cell;-

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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

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

(aq)+  (aq)+    (aq)+ 4H2O(l)

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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

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= 0.8v

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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)

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2Ag+(aq)+ 2  2Ag(s)

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Mg (s) + 2Ag+(aq) Mg2+(aq)+ 2Ag (s)

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Mg(s)/Mg2+(aq)//Ag+(aq)/Ag(s)

=  –

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= 0.8 – (-2.37)

= 3.17v

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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)  ZnCl2(s) + H2(g)

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ii.     Cu(s) + HCl(aq)  No reaction

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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.

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Cathodic reaction:-

2H+(aq) + 2e H2 (g)

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2H2 O(l) + 2e  2OH(aq)+ H 2(g)

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4e+ O2(g) +H2 O(l)  4OH(aq)

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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 (Fe2O3. 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.

 

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For iron:

Anode:   Fe(s)    Fe2+ (aq)+ 2e

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Cathode:O2(g) + 2H2O(l) + 4e 4OH(aq)

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2Fe(s) + O2(g) + 2H2 O(l)  2Fe2+(aq)+ 40H(aq)

word image 2660

(rust)

4Fe(OH)2(s) + O2(g)  2H2O(l) + 2Fe2 O3.XH2 O

word image 2661

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) + O2(g)  2Fe3 O3.H2O + 4H2O(l)

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Fe3O +   H2O(l)  Fe3O4 + 4H2O

word image 2663

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 1st 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 CO2 in 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

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R = ρx

word image 2666

ρ=Resistivity

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

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. ()

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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)

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=  (

word image 2669

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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)

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= K

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R = ρ   =

word image 2675

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= , =                     =

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=

word image 2685

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Molar conductance (λm)

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

Or

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.

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

V =

word image 2688

=

word image 2689

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Where C is the concentration in mol/dm3 and  is m-1

word image 2692

word image 2693

= Ѕm2mol-1 or m2Ὡ -1 mol-1

word image 2694

Example

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

Solution:

Given l = 0.2mol/dm3

V =

word image 2695

V = 5dm3mol-1

2.0.0055M silver nitrate has a molar conductivity of 2.98 x 10-3m2 Ω-1 mol-1 .  Calculate conductivity of that solution.

Solution

=

word image 2696

word image 2697

=  x c

word image 2698

word image 2699

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

= 2.98 x 10-3 m2 Ω-1 x 5.5cm-3

= 0.01529 Ὡ-1 m-1

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3. Calculate the molar conductivity of 0.3M, KOH solution which has a conductivity of 391Ὡ1-1 m-1.

=

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=

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=

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=

word image 2705

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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 10-4 Ð…

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i.  What is the conductivity of the solution?

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

Solution:

i) R = ρ

word image 2708

=  =

word image 2709

word image 2710
word image 2711

K = 0.215cm-1

= 6.6 x 10-4 s

word image 2712

= K

word image 2713

word image 2714

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

                                                     = 1.419 x 10-4 Ð…cm-1

ii)   =

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=

word image 2717

= 14.195

word image 2718

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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 31.6Ὡ 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 747.5Ὡ at 250c,when the same cell was filled with aqueous solution of 0.05M CaC, the resistance was 876Ὡ. Calculate.

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i.Conductivity of the solution

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

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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 .

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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.

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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.

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= K ……………. (i)

word image 2726

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At infinity dilution, the weak electrolyte is completely ionized.

= 1 or 100%

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= K (Constant)……….. (ii)

word image 2729

 

Taking ratio of equation (i) and (ii)

 

word image 2730

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 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/).

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AB     A+ (aq)   +    B(aq)

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1mole     0               0       start

At equilibrium

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But V =

word image 2738

AB  A+ (aq) + B(aq)

word image 2739

K =

word image 2740

But AB = CH3 COOH

=

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

1 –   1

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Ka =

word image 2745

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K a = C

word image 2747

 

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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)

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2.  A 0.001 moldm-3 solution 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 :-

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

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

have a resistance of 550Ω

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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.

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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.  =  +

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E.g. =  + (C)

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λ(Al2(S) = 2(AL3+) + 3(S)

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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)

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CH3COOH + KCl  CH3COOK + HCl

word image 2772

λ(CH3COOH) = (CH3COOK) + (HCl) – KCl)

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

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λ(CH3 COOH) = (CH3COOH) + (H+)

REVIEW QUESTIONS

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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.

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2.The molar conductivity of 0.093 CH3COOH solution at 298k is 536 x 10-4 Sm2mol-1 .The molar conductivity at infinity dilution of H+ and CH3COO are 3.5 x 10-2 and 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-1 at 298k.At the same temperature (NaF),   (NaoH) and (H2O) are 493360 and 162Sm2mol-1 respectively. 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.

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