ELECTROCHEMISTRY

INTRODUCTION

The branch of science which deals with the relationship between chemical and electric energy is known as electrochemistry.Major contents of Electro chemistry(i) Ionic theory(ii) Faraday’s law of electrolysis(iii) Electrolytic conductance(iv) Electrochemical cell

IONIC THEORY

Chemical substance which dissolve in water and furnishes ions are called electrolytes. The phenomenon of the production of ions in solution is called dissociation or ionisation.

Electrolytes:(i) Strong:

Concentration of dissociated ions largely dominate on the concentration of undissociated molecule.

(ii) Weak:Lesser concentration of dissociated ions

Degree of dissociation :

Factors affecting of degree of dissociation:

(i) Nature of electrolytes(ii) Nature of solvent(iii) Presence of other solute (common ion effect)(iv) Dilution(v) Temperature

ELECTROLYSIS Process in which electrolyte is decomposed into its constituents by passing electricity through its aqueous solution or fused (molten) state.

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Reaction occurring at cathode Reduction reaction (Gain of e )

Reaction occurring at anode Oxidation reaction (Loss of e )

(Go to anode) (Primary reaction)

PREFERENTIAL DISCHARGING THEORYIf more than one type of ions are available during electrolysis, then that ion is discharged first at respective electrodes which requires least energy i.e. discharging potential.(i) Electrolysis of sodium chloride solution:

Ions produced during electrolysis:

At cathode, ions are discharged in preference to ions as the discharge potential of ions is lower than ions. Similarly at anode, ions are discharged in preference to

ions.At cathode:

At anode:

Thus remain in solution.

(i) Electrolysis of copper sulphate solution using platinum electrodes:

At cathode:

At anode:

2 212OH H O O 2e2

(iii) Electrolysis of sodium sulphate solution using inert electrodes:

At cathode:

At anode:

(iv) Electrolysis of copper sulphate solution using copper electrodes :

At cathode, copper is deposited.

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ELECTROCHEMISTRY

At anode, Cu – electrode oxidised to ions which dissolve equivalent amount of copper at the anode.

During electrolysis, copper is transferred from anode to cathode.Illustration 1. The pH of a solution of NaCl is 7. This solution is electrolysed. What would be

the pH of the solution after electrolysis?

Solution: Greater than 7, due to the formation of OH ions.

Exercise 1. Predict the products of electrolysis in each of the following case(i) AgNO3 using Pt electrodes (ii) CuSO4 using Cu electrodes(iii) Molten NaBr (iv) Aqueous NaCl

FARADAY’S LAW OF ELECTROLYSIS(i) Faraday’s first law:

The amount of substance deposited on respective electrodes is directly proportional to the quantity of electricity passed.i.e. or, where, w = weight of deposited substance i = amp t = time in secondsIf i = 1 amp and t = 1 secw = z, Where z is electrochemical equivalent.

Illustration 2. A vanadium electrode is oxidized electrically. If the mass of the electrode decreases by 114 mg during the passage of 650. coulombs, what is the oxidation state of the vanadium product? (A) +1 (B) +2(C) +3 (D) +4

Solution: n = 3

Illustration 3. For how long a current of three amperes has to be passed through a solution of AgNO3 to coat a metal surface of 80cm2 area with 0.005mm thick layers? Density of silver is 10.5 g / cc and atomic weight of Ag is 108 gm/mol.

Solution: Assuming that the coating has to be done on only side of the surface, volume of

Ag required for coating = = 0.04cc

Mass of silver = 10.5 0.04 = 0.42gm

Moles of silver = = 0.00389

Moles of electrons = 0.00389 (since Ag+ + e– Ag)Charge passed = 0.00389 96500 = 375.385 coulombs

Time = = 125.1212 sec.

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ELECTROCHEMICAL EQUIVALENT (Z)

It is the mass of substance deposited by one coulomb of charge.(ii) Faraday’s second law: When same amount of electricity passed through different electrolytes, then deposited mass of respective electrodes will be in the ratio of their equivalent masses.

Illustration 4. Find the charge in coulombs of 1 g ion of N3.

Solution. One nitride ion carries three ve charge Charge of one ion of N3 = 3 1.6 1019 coulombOne g ion consist = 6.02 1023 ions Charge on one g – ion of N3

= 2.89 105 CIllustration 5. How many moles of iron metal will be produced by passage of 4A of current

through 1L of 0.1 M Fe3+ solution for 1 hour? Assume only iron Fe+3 + 3e– Fe is reduced.

Solution: Total charge passed = (4 1 60 60) C = 14400 C = 0.149 FFirst of all ferric ion changes to ferrous and than ferrous changes to Fe.Fe+3 + e– Fe+2, Fe+2 + 2e– FeSo, charge used up in formation of Fe (II) = 0.1 FTherefore, remaining charge left = (0.149 – 0.1) F = 0.049 FWhich is used in conversion of Fe+2 to Fe

So, number of moles of Fe produced = = 0.0245

Illustration 6. The same amount of electricity was passed through molten cryolite containing Al2O3 and NaCl. If 1.8 g. of Al were liberated in one cell, the amount of Na liberated in the other cell is (A) 4.6 g (B) 2.3 g(C) 6.4 g (D) 3.2 g

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ELECTROCHEMISTRY

Solution:

GRAM EQUIVALENT WEIGHT OF SUBSTANCE

It is the mass in grams of deposited substance deposited by one Faraday (96500 coulomb) of electricity.1 coulomb = 1 ampere – second1 Faraday = Charge carried by 1 mole of

= 96500 coulombIf w = z (deposited weight by one coulomb)W = E (deposited weight by one Faraday)

Faraday’s law for gaseous electrolytic product.We know, w = zQ

z I t

For gas, Volume of evolved gas at S.T.P.

Illustration 7. The volume of gas measured at NTP, liberated at anode from the electrolysis of Na2SO4 solution by a current of 5.0A passed for 3 min 13 s:(A) 56 mL (B) 112 mL(C) 224 mL (D) None of these

Solution: Quantity of charge passed = = 0.01F

Mass of oxygen evolved = 0.01 equiv= 0.08 gm

= = 0.0025 mole

Volume of O2 at NTP = 22400 0.0025 = 56 ml

Illustration 8. A current of 0.20A is passed for 482s through 50.0 mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis? (A) 0.0159 M (B) 0.0199M(C) 0.10M (D) 0.030M

Solution: The quantity of charge passed = =

The reaction at cathode is,

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1F = 1 mol of

Concentration of in solution

After electrolysis =

Illustration 9. A constant current was flown for 2 hour through a solution of KI. At the end of experiment liberated iodine consumed 21.75 mL of 0.0831 M solution of sodium thiosulphate following the reaction: I2 + 2S2O3

2– 2I– + S4O62–. What was the

average rate of current flow in ampere?

Solution: 2S2O32– S4O6

2– + 2en-factor of sodium thiosulphate = no. of electrons lost per molecule = 1 Molarity of Na2S2O3 solution = Normalitymeq. of I2 liberated = m equiv. of Na2S2O3 = 21.75 0.0831 = 1.807Thus,

= 1.807

= 1.87 10–3 =

i = 0.0242A

Applications of electrolysis:(1) Determination of equivalent masses of element(2) Electrometallurgy(3) Manufacturing of non – metals (By electrolysis process)(4) Electro – refining of metals(5) Manufacturing of compounds(6) ElectroplatingCalculation of thickness of coated layer during electrolysis:

If volume of coated layer 3a b c cm

Mass of the deposited substance

or, I t Ea b c density96500

EFFICIENCY OF CURRENT DURING ELECTROLYSIS

% current efficiency

Illustration 10. A solution containing one mole per litre of each Cu(NO3)2; AgNO3; Hg2(NO3)2 ; Mg(NO3)2 is being electrolysed using inert electrodes. The values of standard electrode potentials (reduction potentials in volts are Ag/Ag+ =0.80 V, 2Hg/Hg2

++

= 0.79V, Cu/Cu++ = + 0.24 V, Mg/Mg++ = –2.37 V. With increasing voltage, the sequence of deposition of metals on the cathode will be (A) Ag, Hg, Cu (B) Cu, Hg, Ag(C) Ag, Hg, Cu, Mg (D) Mg, Cu, Hg, Ag

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ELECTROCHEMISTRY

Solution: Greater the value of standard reduction potential, greater will be it’s tendency to undergo reduction. So the sequence of deposition of metals on cathode will be Ag, Hg, Cu. Here, magnesium will not be deposited because it’s standard reduction potential is negative. So it has strong tendency to undergo oxidation. Therefore, on electrolysis of Mg(NO3)2 solution, H2 gas will be evolved at cathode. Hence, (A) is correct.

Illustration 11. One coulomb of charge passes through solution of AgNO3 and CuSO4 connected in series and the conc. of two solutions being in the ratio 1:2. The ratio of weight of Ag and Cu deposited on Pt electrode is (A) 107.9 : 63.54 (B) 54 : 31.77(C) 107.9 : 31.77 (D) 54 : 63.54

Solution: Faraday’s IInd Law = constant

So, (Ag+ + e– Ag, EAg = )

(Cu+2 + 2e– Cu, ECu = )

=

Hence, (C) is correct.

Illustration 12. 0.3605 g of a metal is deposited on the electrode by passing 1.2 ampere current for 15 minutes through its salt. Atomic weight of the metal is 96. What will be its valency?

Solution:

or,

Illustration 13. A current of 4 amp was passed for 1.5 hours through a solution of copper sulphate when 3.2 g of copper was deposited. Calculate the current efficiency.

Solution: (According to first law of electrolysis)

Illustration 14. How many gm of silver could be plated out on a serving tray by electrolysis of a solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? Density of silver is 10.5 g / cm3.

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Solution: Total electricity passed

Deposited mass = 271.65 g

Volume of Ag

Area of tray

Illustration 15. A current of 40 microamperes is passed through a solution of AgNO3 for 32 minutes using Pt electrodes. An uniform thick layer of Ag is deposited covering 43% cathode surface. What is the total surface area of cathode, if each Ag atom covers 5.4 10-16 cm2?

Solution: Total charge passed = (4 10–6 32 60) C = 0.0768 CAg+ + e– Ag

So, weight of Ag deposited at cathode surface = gm

= 8.6 10-5 gm

Now, total no. of Ag atoms deposited =

= 4.8 1017

So, surface area of cathode which is covered by Ag atoms = (4.8 1017 5.4 10–16)= 259.2 cm2

Total surface area = cm2 = 602.8 cm2

Exercise 2.(i) A current of 1.70 ampere is passed through 300 ml of 0.160 M solution of zinc sulphate

for 30 seconds with a current efficiency of 90%. Find out the molarity of Zn++ ions after the deposition of zinc. Assume the volume of the remaining solution is constant during electrolysis.

(ii) Calculate the volume of Cl2 at NTP produced during electrolysis of MgCl2 which produces 6.50 g Mg.

ELECTROLYTIC CONDUCTANCE

Property of electrolytic conductor which facilitates the flow of electricity through it. It is equal to the reciprocal of resistance i.e.

Electrolysis conductance:(i) Specific conductance(ii) Equivalent conductance(iii) Molar conductance

Specific conductance (k):

It is the reciprocal of specific resistance or it is the conductance produced by one centimetre cube of an electrolytic conductor.

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ELECTROCHEMISTRY

If

where length of conductor a = cross – sectional area of conductor

or,

or, Representation of specific conductance:

Unit at specific conductance:

Equivalent conductance Conductance of all ions which are produced by the ionisation of one equivalent of an electrolyte in a given solution.i.e. Equivalent conductance or,

or,

Unit of equivalent conductance is Molar conductance Conductance of all ions which are produced by the ionisation of one mole of an electrolyte in given solution.i.e. Molar conductance

or,

Unit of molar conductance is

Illustration16. The equivalent conductance of monobasic acid at infinite dilution is 438 If the resistivity of the solution containing 15g of acid (Molecular

weight = 49) in 1L is 18.5 ohm cm. What is the degree of dissociation of acid?(A) 45.9% (B) 40.2%(C) 60.4% (D) 50.7%

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Solution: Resistivity =

=

= 0.507

= 50.7%

Illustration 17. 0.5 N solution of a salt placed between two platinum electrode 2.0 cm apart and of area of cross-section 4.0 sq. cm has a resistance of 25 ohms. Calculate the equivalent conductance of solution.

Solution: Specific conductance (k) = Conductance Cell constant

= = = 0.02 ohm–1 cm–1

Ùc = = = 40 ohm–1 cm2 eq–1

EFFECT OF DILUTION ON CONDUCTANCE

(i) When we increase the dilution of electrolyte, magnitude of specific conductance decreases, because number of ions in 1 cm3 volume of electrolyte decreases.

(ii) When dilution of electrolyte increases, magnitude of molar conductance and equivalent conductance increases, because number of ions increases after dilution.

(iii) The molar conductivity of strong electrolytes is found to vary with concentration according to the equation.

where equivalent conductivity at fixed concentration equivalent conductivity of infinite dilution

(iv) Degree of dissociation

KOHLRAUSCH’S LAW OF INDEPENDENT MIGRATION OF IONS

At infinite dilution, conductance of any electrolyte is the sum of contribution of its constituents ions i.e. anions and cations.For an electrolyte of the type of AxBy, we have

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ELECTROCHEMISTRY

Application of Kohlrausch’s Law(i) Determination of of a weak electrolyte:

In order to calculate of a weak electrolyte say CH3COOH, we determine experimentally

values of the following three strong electrolytes:

(a) A strong electrolyte containing same cation as in the test electrolyte, say HCl(b) A strong electrolyte containing same anion as in the test electrolyte, say CH3COONa(c) A strong electrolyte containing same anion of (a) and cation of (b) i.e. NaCl.

of CH3COOH is then given as:

(CH3COOH) = (HCl) + (CH3COONa) – (NaCl)

Proof: (HCl) = …I

(CH3COONa) = …II

(NaCl) = …III

Adding equation (I) and equation (II) and subtracting (III) from them:

(ii) Determination of degree of dissociation ():

(iv) Determination of ionic product of water:From Kohlrausch’s law, we determine of H2O where is the molar conductance of water at infinite dilution when one mole of water is completely ionised to give one mole of H+ and one mole of OH– ions i.e.

(H2O) = + Again using the following equation,

, where C = molar concentration i.e. mol L–1 or mol dm–3

, where C = concentration in mol m–3

Assuming that differs very little from

=

C =

Specific conductance (k) of pure water is determined experimentally. Thereafter, molar concentration of dissociated water is determined using the above equation.Kw is then calculated as:Kw = C2

Illustration 18. The values for NaCl and KCl are 126.5 and 149.9 respectively. The ionic conductance of at infinite dilution is

Calculate the ionic conductance at infinite dilution for

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Solution:

= 149.9 – 126.5 = 23.4or,

= 23.4 + 50.1 = 73.5

Illustration 19. Calculate the limiting value of equivalent conductance of AgCl. Given that

Solution: According to Kohlrausch law

= =

= 133.4 + 149.2 145.9 = 138.2 ohm1 cm2 eq1

Illustration 20. In each of the following pairs which will allow greater conduction of electricity and why? (a) Silver wire at 20C, same silver wire at 50C. (b) NH4OH solution at 20C, same NH4OH solution at 50C. (c) NaCl solution at 20C, same solution at 50C. (d) 0.1 M acetic acid solution, 1 M acetic acid solution.

Solution: (a) Silver wire at 20C because with the increase in temperature, metallic conduction decreases due to vibration of kernels.

(b) NH4OH at 50C because in case of a weak electrolyte, dissociation increases with increase in temperature.

(c) NaCl solution at 50C because in case of a strong electrolyte, with increase in temperature, the ionic mobilities increase.

(d) 0.1 M acetic acid because with dilution, ionization increases.

Illustration 21. The molar conductivity of acetic acid solution at infinite dilution is Calculate the molar conductivity of 0.01 M acetic acid solution,

given that the dissociation constant of acetic acid is

Solution:

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Exercise 3.(i) The resistance of a decinormal solution of a salt occupying a volume between two

platinum electrodes 1.80 cm apart 5.4 cm2 in area was found to be 32 ohm. Calculate the specific and equivalent conductance of solution.

(ii) Calculate the dissociation constant of water at 25C from the following data. Specific conductance of H2O = .

Electrochemical cell: It is a device or arrangement by which we can convert chemical energy into electrical energy i.e. chemical energy produced as a result of redox chemical reaction of cell.Some major points regarding electrochemical cell:

(i) Electrochemical cell consists of two half – cells. The half – cell in which oxidation occurs is called oxidation half – cell and the another half cell in which reduction occurs is called reduction half cell.

(ii) Electrons flow from anode to cathode in the external circuit.(iii) Chemical energy is converted into electrical energy.(iv) The net reaction is the sum of two half – cell reactions.

Oxidation half reaction

(iv) Two half – cells are connected by the salt bridge and completes the cell circuit.(v) Salt bridge prevents transference or diffusion of the solutions from one half – cell to the other.

It also helps to maintain the electrical neutrality of half cells.(vi) Salt bridge is represented by a broken line or two parallel vertical lines in a cell reaction.

(vii) To sum up:Left

Metal / Metal ion (conc.) Metal ion (conc.) / MetalOxidation occurs Salt bridge Reduction occursAnode CathodeNegative pole Positive pole

Standard Reduction Potentials at 25° C

E°, V E°, V

F2 + 2e– 2F– 2.87 AgCl + e– Ag + Cl– 0.222

2.0 + 2e– Pd + 4I– 0.18

Co3+ + e– Co+2 1.82 Cu2+ + e– Cu+ 0.15

H2O2 + 2H+ + 2e– 2H2O 1.77 Sn4+ + 2e– Sn2+ 0.13

+ 4H+ + 3e– MnO2 + 2H2O 1.70 +e– Ag+ 0.017

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PbO2 + 4H+ + + 2e– PbSO4 +

2H2O1.70 2H+ + 2e– H2 0.000

Ce4+ + e– Ce3+ 1.70 Pb2+ + 2e– Pb – 0.126

+ 8H+ + 5e– Mn+2 + 4H2O 1.51 Sn2+ + 2e– Sn – 0.14

Au3+ + 3e– Au 1.502CuO + H2O + 2e– Cu2O + 2OH– – 0.15

Cl2 + 2e– 2Cl– 1.36 AgI + e– Ag + I– – 0.151

+ 14H+ + 6e– 2Cr+3 + 7H2O 1.33 CuI + e– Cu + I– – 0.17

Tl3+ + 2e– Tl+ 1.26 Ni2+ + 2e– Ni – 0.25

MnO2 + 4H+ + 2e– Mn2+ + 2H2O 1.23 Co2+ + 2e– Co – 0.28

O2 + 4H+ + 4e– 2H2O 1.229 PbSO4 + 2e– Pb + – 0.31

+ 12H+ + 10e– I2 + 6H2O 1.20 Tl+ + e– Tl – 0.336

Br2 + 2e– 2Br– 1.09 Cu2O+ H2O + 2e–2Cu + 2OH– – 0.34

+ 3e– Au + 4Cl– 1.00 Cd2+ + 2e– Cd – 0.403

OCl– + H2O + 2e– Cl– + 2OH– 0.94 Fe2+ + 2e– Fe – 0.44

Pd2+ + 2e– Pd 0.92 Cr3+ + 3e– Cr – 0.74

2Hg2+ + 2e– 0.92 Zn2+ + 2e– Zn– 0.7628

Cu2+ + I– + e– CuI 0.85 2H2O + 2e– H2 + 2OH– – 0.828

Ag+ + e– Ag 0.799 Mn2+ + 2e– Mn – 1.18

+ 2e– 2Hg 0.79 Al3+ + 3e– Al – 1.66

Fe3+ + e– Fe2+ 0.771 H2 + 2e– 2H– – 2.25

O2 + 2H+ + 2e– H2O2 0.69 Mg2+ + 2e– Mg – 2.37

Cu2+ + Cl– + e– CuCl 0.566 Ce3+ + 3e– Ce – 2.48

I2 + 2e–- 2I– 0.535 Na+ + e– Na – 2.713

Cu+ + e– — Cu 0.52 Ca2+ + 2e– Ca – 2.87

+ e– 0.370 Ba2+ + 2e– Ba – 2.90

+ e– 0.355 Cs+ + e–- Cs –2.92

Cu2+ + 2e– Cu 0.34 K+ + e– K – 2.93

Hg2Cl2 + 2e– 2Hg + 2Cl– 0.270 Li+ + e– Li –3.03

ELECTRODE POTENTIAL

(a) Oxidation potential: Potential developed on anode electrode due to oxidation process is called oxidation potential.

(b) Reduction potential:

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ELECTROCHEMISTRY

Potential developed on cathode electrode due to reduction process is called reduction potential.

Emf of the cell = oxidation potential of anode + Reduction potential of cathode

Reference electrode (Standard Hydrogen Electrode):

By this device, we measure, the standard electrode potential of half cell. It consists of a small platinum strip coated with platinum black so as to absorb H2 gas. A platinum wire is welded to the platinum strip and sealed in a glass tube. The entire arrangement dipped into 1 M HCl solution at 298 K and specified pressure of H2 gas.

In this arrangement oxidation and reduction process occurs at same electrode and resultant potential becomes equal to zero.

(i) Measurement of electrode potential of electrode, which placed above than hydrogen

in electro – chemical series.

(ii) Measurement of electrode potential of electrode, which placed below than hydrogen

in electro – chemical series.

Some other reference electrode:(i) Calomel electrode – Hg, Hg2Cl2 /

(ii) Silver – silver chloride electrode – Ag, AgCl /

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Various types of half cells:Type Representation

(1.) Metal – metal ion half cell

(2.) Gas – ion half cell

(3.) Metal insoluble salt anion half cell

(4.) Oxidation – Reduction half cell

(5.) Metal – metal – oxide – hydroxide half cell

(6.) Quinhydrone half cell

Illustration 22. Hydrogen gas will reduces (A) heated cupric oxide (B) heated ferric oxide (C) heated stannic oxide (D) heated aluminium oxide

Solution: The standard reduction potential of Copper is greater than the standard reduction potential of hydrogen. So hydrogen gas reduces cupric oxide asCuO + H2 Cu + H2OSimilarly, hydrogen reduces ferric oxide to ferous oxide Fe2O3 + 3H2 2FeO + 3H2OHence, (A) & (B) is correct.

Illustration 23. Which are true for a standard hydrogen electrode? (A) The hydrogen ion concentration is 1 M. (B) Temperature is 25C.(C) Pressure of hydrogen is 1 atmosphere.(D) It contains a metallic conductor which does not adsorb hydrogen.

Solution: (A), (B) & (C)

Illustration 24. Which species in each pair is a better reducing agent under standard conditions?(A) Na & Li (B) H2 & I2

(B) Fe+2 & Ag (D) & Co+2

Solution: Lesser the reduction potential, more is the reducing power. Hence, (A) is correct.

Emf of cell:The Nernst Equation:

For a reversible reaction,

Reaction Quotient / Equilibrium constant

During cell reaction, change in Gibbs free energy which is equal to the available work done

by cell.According to thermodynamic consideration,

where, Change in Gibbs free energy

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ELECTROCHEMISTRY

Standard change in Gibbs free energy

Q = Reaction quotientor,

or,

or,

If R = 8.314 J

T = 298 K F = 96500

or,

where (For same electrode)

Illustration 25. For the cell Tl Tl+ (0.001M) Cu2+ (0.1M) Cu. ECell at 25C is. 83V which can be increased (A) by increasing [Cu2+] (B) by increasing [TI+](C) by decreasing [Cu2+] (D) by decreasing [TI+]

Solution: 2Tl + Cu+2 2Tl+ + Cu

Hence, (A) & (D) is correct.

Illustration 26. The logarithm of the equilibrium constant, log K, of the cell reaction of the cell

(A) (B)

(C) (D)

Solution: Use the Nernst equation

Hence, (A) is correct.

Applications of Nernst equation:

(a) Determination of equilibrium constant for cell reaction:

If (Q = Reaction Quotient)

At equilibrium,

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(b) Determination of solubility product:If cell reaction is

for concentration cell

Value of can be calculated and

Illustration 27. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298K if the emf of the cell:Ag | Ag+ (sat Ag2CrO4 solution) || Ag+ (0.1 M) | Ag is 0.164 at 298K

Solution: For cell Ag | Ag+ (saturated solution of Ag2CrO4) || Ag+ (0.1 M) | AgEcell = 0.164V

Ecell = E0cell –

0.164 =

[Ag+]LHS = 1.66 10–4MKsp for Ag2CrO4 2Ag+ + CrO4

2–

Ksp = [Ag+]2 [CrO42–] = [1.66 10–4]2

Ksp = 2.287 10–12 mol3 liltre–3

Ionisation constant and degree of dissociation of weak acid or weak base:If cell reaction is

By this expression, we can calculate the value of Ka for acid.

If value of Ka and C are known, then we can calculate the value of

Determination of pH of a solution:If cell reaction is:

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or,

or,

Illustration28. By how much is the oxidizing power of the MnO4–/Mn2+ couple be decreased if

the H+ concentration is decreased from 1 M to 10–4 M at 25C? Assume other species have no change in concentration.

Solution: MnO4– (in acidic medium) is oxidizing agent.

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

E – E0= = = – 0.38 V

Thus MnO4+/Mn2+couple will move to position of less oxidizing power by

0.38 V.

Illustration 29. Graph between Ecell R log was linear with intercept on Ecell axis 1.1 V.

Calculate Ecell for Zn | Zn2+(0.1 M) || Cu2+(0.01 M) +| Cu.

Solution: Cell reaction is Zn + Cu2+  Zn2+ + Cu

Ecell= E0cell log …(1)

We can compare equation (1) with y = mx + c C = E0

cell = 1.1 V

Ecell = E0cell – = 1.1 – log10

= 1.1 – 0.295 = 1.0705 V

Illustration 30. What is the emf when the electrolyte is saturated with Pb?The Edison storage cell is represented asFe(s) | FeO(s) | KOH(aq) | Ni2O3(s) | Ni(s)The half cell reactions areNi2O3(s) + H2O(l) + 2e– 2NiO(s) + 2OH (aq) Eo = 0.4 VFeO(s) + H2O (l) + 2e– Fe(s) + 2OH (aq) Eo = – 0.87 V(i) What is the cell reaction?(iii) What is the cell emf? How does it depends on the concentration of KOH?(iii) What is the maximum amount of electrical energy that can be obtained from

one mole of Ni2O3.

Solution: (i) Given cell is Fe(s) | FeO | Ni2O3(s) | Ni(s)Cell Reaction Eo

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L.H.S. Fe(s) + 2OH (aq) FeO(s) + H2O(l) + 2e– Eoox = 0.87 V

R.H.S. Ni2O3(s) + H2O(l) + 2e– 2NiO(s) + 2OH (aq) Eored= 0.4 V

––––––––––––––––––––––––––––––––––––––––––––––Fe(s) + Ni2O3(s) FeO(s) + 2NiO)(s) Eo

cell = 1.27 V(ii) Eo

cell = EoOX + Eo

red = 1.27 VSince net reaction does not contain OH, hence cell emf does not depend on the concentration of KOH.

(iii) Maximum amount of electrical energy Go = – nFEocell

= – 2 96500 1.27 = 2.45 105 J mole = 2.45 102 kJ mole–1

= 2.45 102 kJ mole–1

Illustration 31. The standard reduction potential at 25°C for the reaction 2H2O + 2e- H2 + 2OH-

is –0.8277 V. Calculate the equilibrium constant for the reaction 2H2O H3O+ + OH- at 25°C.

Solution: Consider an electrode H as2H+ + 2e– H2 E0

= 0.0 VGiven electrode is2H2O + 2e– H2 + 2OH– E0 = –0.8277 VCell reactions are At anode: H2 + 2OH– 2H2O + 2e–

At cathode: 2H+ + 2e– H2

————————————————————Net cell reaction is 2H+ + 2OH– 2H2O

Ecell =

= (0.8277 + 0) - = 0.8277 – 0.591

At equilibrium, Ecell = 0

0.8277 = 0.0591

= 14.0051 log Kw = –14.0051

Kw = 9.88 10–15

Thus for 2H2O H3O+ + OH–

Kw = [H3O+] [OH–]Kw = 9.88 10–15

DETERMINATION OF THERMODYNAMICAL DATA

(Gibbs – Helmholtz equation)

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or,

or,

(Temperature coefficient of the emf of the cell)

Illustration 32. Calculate the standard free energy change in kJ for the reaction:

Given:

(A) (B) 135.1(C) 1.78V (D) 1.75V

Solution:

The for the reaction:

= = =

Type of Electrochemical cell:

(i) Electrode cell(ii) Concentration cell

Electrode cell:

Different electrolytes are used in both half cell.

Resultant cell emf determined by

Concentration cell:

Same type of electrolytes used in both half cell but concentration of these electrolytes are different.

Resultant cell emf determined by

Here,

Illustration 33. Stronger the oxidizing agent, greater is the(A) Standard reduction potential (B) Standard oxidation potential(C) Ionic nature (D) None

21

Solution: (A)

Illustration 34. ERP for Fe+2 / Fe and Sn+2 / Sn are – 0.44 and 0.14 volts respectively. The standard emf for cell Fe+2 + Sn Sn+2 + Fe is (A) + 0.30 V (B) –0.58 V(C) + 0.58 V (D) –0.30 V

Solution: Fe+2 + Sn Sn+2 + Fe

= – 0.44 – 0.14 = – 0.58 VHence, (B) is correct.

Illustration 35. EMF of the cellCd | CdSO4 | H2SO4 (0.01 M) | H2 (1 atm) | Ptis + 0.362 Volt at 25° C. The is –0.403 Volt at same temperature.

Calculate the solubility product of CdSO4.

Solution: Anode half cell: Cd Cd+2 (aq) + 2e–

Cathode half cell: 2H+ (aq) + 2e– H2

————————————————————Net cell reaction: Cd + 2H+ (0.02 M) H2 (1 atm) + Cd+2 (aq)

Ecell =

+ 0.362 = + 0.403 –

–0.041 = –

1.3875 = log =24.41

[Cd+2] = 9.764 10-3 MKsp of Cd SO4 = [Cd+2]

= (9.764 10–3) (0.01) = 9.764 10–5 M2

Illustration 36. The normal oxidation potential of zinc referred to the standard hydrogen electrode is 0.76 Volt and that of copper is -0.34 Volt at When excess of zinc is added to a solution of copper sulphate, the zinc displaces copper till equilibrium is reached. What is the ratio of concentration of ions

at equilibrium?

Solution: The reaction is:

At equilibrium,

22

ELECTROCHEMISTRY

or,

or,

Illustration 37. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of hydrogen ions. The emf of the cell is 0.118 Volt at. Calculate the concentration of hydrogen ions at the positive electrode.

Solution: ,

or,

or,

Illustration 38. Calculate the equilibrium constant for the reaction:Fe2+ + Ce4+ Ce3+ + Fe3+

Given that, Fe3+ + e- Fe2+; E0 = 0.771 VCe4+ + e- Ce3+; E0 = 1.61 V

Solution: The equilibrium reaction is Fe2+ + Ce4+ Ce3+ + Fe3+

Since the reaction is at equilibrium, Ecell = 0

=

= 0.0591 log K

– 0.771 + 1.61 = 0.0591 log K

Log K = = 14.1963

K = 1.57 1014

Illustration 39. Iron is corroded by atmospheric oxygen under acidic condition to product Fe2+

(aq) ions initially. The standard reduction potential

E0Fe2+/Fe = 0.44 V and for the reaction H2O ( ) 2H+ (g) +

E0 = 1.23 V Find whether the formation of Fe2+ (aq) is thermodynamically favorable or not.

Solution: The reactions are

(i) Fe (s) Fe+2 (aq) + 2e–,

(ii) 2H+ + (g) + 2e H2O (l), E0 = +1.23V

Adding equation (i) and (ii), we get

23

Fe (s) + 2H+ + (g) Fe+ (aq) + H2O (l), = 1.67V

= positive = 1.67V G0 is negative. So, the reaction is thermodynamically favourable or spontaneous.

Illustration 40. The emf of the cell reaction,

Calculate the entropy change. Given that enthalpy of the reaction is 216.7 KJ mol1 And and = + 0.34V.

Solution: For the cell reaction,

= 0.76 + 0.34= 1.1V

G0 = nF

= 2 96,500 1.1 = 212.3 KJ

= 14.76 J K1 mol1.

Exercise 4.(i) The standard oxidation potential of Ni/Ni2+ electrode is 0.236 V. If this is combined

with a hydrogen electrode in acid solution, at what pH of the solution will be measured emf be zero at 25C? Assume [Ni22+] = 1 M and .

(ii) The voltage of the cell given below is 0.46 V

Also

Calculate the value of K2, where K2 =

Common cell:

(i) Leclanche dry cell:

Construction:

24

ELECTROCHEMISTRY

Zinc (Using anode filled with a mixture of ammonium chloride and MnO2, filled with a graphite cathode in the center.

Cell reaction:At cathode:

These gases are removed by depolariser MnO2.

Ammonia is absorbed by zinc ions.

At anode:

Cell reaction:

Voltage Disadvantage: (i) Rapid withdrawl of current causes gases to collect at cathode and voltage suddenly

drops.(ii) Slow but spontaneous reaction of zinc with ammonium ions results in poor shell life.

Alkaline Battery:

No gases are formed, hence it is an improvement over a previous model.

Cell reaction:

At cathode:

At anode:

Voltage = 1.54 V

Buffer cells:

Construction: Anode is metallic zinc while cathode is HgO. Both in the form at tightly compacted powders separated by a moist paper of HgO containing some NaOH / KOH. Moistened paper serves as salt bridge.

Cell reaction:

At cathode:

25

At anode:

Storage Batteries: (discharging)A rechargeable array of many cells.

(i) Lead storage Battery:

Construction: Alternate plates of porous lead and lead deposited with PbO2. Electrolyte is aqueous H2SO4.Cell reaction:At anode:

At cathode:

Overall reaction:

Cell reaction (charging): For this cell is connected to an external source of current. At anode:

(From electrical source)

At cathode:

Over all reaction:

Reactions for discharging are enthalpy opposite to that of charging.Density of solution decreasing (due to consumption of H2SO4) when it is discharged and density increases (due to formation of H2SO4) when it is charged.

Fuel cells

The main disadvantages of a primary cell is that it can deliver current for a short period of time due to limited amounts of oxidising agents. A fuel cell is a device which converts chemical energy into electrical energy as long as the outside supply fuel is maintained and example is H2 O2 fuel cell. The over all reactions are

The fuel cells are efficient (upto 95%) and are pollution free.

26

ELECTROCHEMISTRY

The main disadvantage of fuel cell is high cost of catalyst and difficult storage of H2 and O2 gas.

Corrosion:

The conversion of a metal into undesirable compounds by the action of air, moisture is called corrosion, In case of iron corrosion it is called rusting. Corrosion is a redox reaction by which metal gets oxidised by air in presence of moisture. Small cathodic and anodic cells are set up on metal. The area of metal in contact with water acts as anode. Metal loses electrons to form cations.

The electrons flow along the metal into the cathode area. Where O2 is reduced to hydroxyl ions in presence of H2O.

Cathode reaction

The over all reaction is

Fe(OH)2 may dehydrate to form its oxide FeO, which than further gets oxidised to Fe2O3. Which is further hydrated to form rust?

Corrosion of a metal weakness the metal and hence is undesirable. The following methods are used to protect a metal from corrosion.(i) Applying a protective film such as metal oxide. (ii) Plating the metal with a more electropositive metal, which loses electrons in preference to the

less electropositive metal e.g. Zn is coated on steel. (Galvanization) (iii) By connecting the metal with more electropositive metal (Electrical protection)

Exercise 5.

27

(i) Which of the following metals can be used for galvanisation of iron?Cu, Zn, Mn, Au

(ii) Sea water promotes corrosion. Why?

28

ELECTROCHEMISTRY

ANSWER TO EXERCISES

Exercise 1:(i) Ag at cathode and O2 at anode(ii) Cu at cathode and Cu will dissolve from anode(iii) Na+ at cathode and Br2 at anode(iv) H2 at cathode and Cl2 at anode

Exercise 2:(i) Molarity of Zn++ ions after deposition of zinc

=

(ii) Equivalent of Mg formed at cathode = Equivalent of Cl2 formed at anode

Moles of Cl2 =

Exercise 3:

(i) Specific conductance (K) = = 0.01041 mho cm1

Also = K V(ml) = 0.01041 10.000

= 104.1 mho cm2

(ii)

= 350 + 198 = 548 Molarity of water = 55.6

Equivalent conductance of water =

Keq = = 2.0 1016 mol lit1

29

Exercise 4:(i) Ni Ni2+ + 2e

2H+ + 2e H2

0 =

or

pH = 4

(ii)

[H+] = 4.0 106

The dissociation of is suppressed in pressure of due to common

ion effect.

Thus

Exercise 5: (i) Zn and Mn can be used as these are more reactive than iron.(ii) Sea water contains salts and impurities which increase the corrision.

30

ELECTROCHEMISTRY

MISCELLANEOUS EXERCISES

Exercises 1: What is cell constant? What are its units?

Exercises 2: Why the variations of equivalent conductance on dilution of a strong electrolyte differ from that of a weak electrolyte?

Exercises 3: Why do acetate ions have lower ionic conductance than Cl ions?

Exercises 4: Through Li+ is smaller in size than Cs+, the equivalent conductance of Li+ is lower than that of Cs+. Why

Exercises 5: Iron may be protected from rusting by coating with zinc or tin. By referring to the electrode potentials given explain why zinc protects iron effectively than tin once the protective coating has been scratched.

ORWhy galvinisation of iron is preferred over tinning?

Exercises 6: What are concentration cells?

Exercises 7: Although aluminium is above hydrogen in the electrochemical series, it is stable in air and water. Explain.

Exercises 8: How many hours does it take to reduce 3 mol of Fe3+ to Fe2+ with 2.00 amp current.

Exercises 9: The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is Calculate the resistivity, conductivity and molar conductivity.

Exercises 10: The molar conductance of NaOH, NaCl and BaCl2 at infinite dilution are respectively.

Calculate molar conductivity of Ba(OH)2.

31

ANSWERS TO MISCELLANEOUS EXERCISES

Exercises 1: Cell constant = l/a, where l is the distance of separation of two electrodes (in cm) of the area of cross section ‘a’ cm2. Its units are cm1 or m1.

Exercises 2: In strong electrolytes, the increase in equivalent conductance is due to weakening of the ion-ion interaction on dilution which increases ionic mobility as a result of solvation effect. On the other hand in weak electrolytes the increase in equivalent conductance is also due to increases in degree of dissociation of weak electrolyte with dilution.

Exercises 3: Acetate ions being larger in size are less mobile than chloride ions and because of this the ionic conductance of acetate ions have lesser value.

Exercises 4: Being smaller in size Li+ ions in solution are highly hydrated as compared with larger Cs ions in solution. Consequently Li+ (H2O)n is bigger in size than Cs+

(H2O)n and hence are less mobile. Equivalent conductance of Li+ ions in solution is therefore less than as compared with that of Cs+ ions.

Exercises 5: Standard reduction potential of zinc is less than that of iron. Therefore, even if some surface of iron gets exposed by the scratching of protective surface, zinc acts as anode and iron acts as cathode. As such iron cannot get rusted. But in case of tin its standard reduction potential is more than that of iron. As such one the tin coating gets scratched, iron begins to acts as anode while tin acts as cathode. Consequently electrochemical cell is set up and iron begins to get rusted rapidly.

Exercises 6: These are those cells in which there is no resultant chemical reaction due to the fact that the reaction occurring in one half cell is exactly reversed in other cell.

The electrical energy arises because of transfer of a substance from the solution of higher concentration around one electrode to that of lower concentration around the other electrode.

Exercises 7: This is because of the formation of a thin protective layer of aluminium oxide on the aluminium surface.

Exercises 8: 40.2 hours

Exercises 9: 87.135 r cm, 0.01148 S .

Exercises 10:

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ELECTROCHEMISTRY

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1. Can we use direct current for experimental determination of resistance of a solution?

Sol. No, it will cause electrolysis of solution. The concentration and resistance of solution will change with time.

Prob 2. Why is a salt bridge not necessary in lead storage cell?

Sol. In lead storage cell, the oxidant (PbO2), reduction (Pb) and their redox product (PbSO4) are solids. Thus half cells need not to be in separate vessels; also they have common electrolyte solution of 40% of H2SO4.

Prob 3. Colour of KI solution containing starch turns blue when Cl2 water is added. Explain.

Sol. Chlorine is placed below iodine in electrochemical series having higher and thus undergoes reduction whereas I undergoes oxidation.

The I2 so formed gets absorbed in starch to give blue colour.

Prob 4. Why will Ag not react with dil. H2SO4 whereas zinc reacts?

Sol. Ag is placed lower than hydrogen in electrochemical series having higher and thus reduction of Ag+ occurs in place of H+ whereas Zn being above H in the series will be oxidised to Zn2+ and H+ will be reduced to H2.

Prob 5. What is the role of ZnCl2 in a dry cell?

Sol. ZnCl2 combines with NH3 produced to form the complex salt otherwise

the pressure developed due to NH3 would crack the seal of the cell.

IIT Level Questions

Prob 6. How much charge is required to reduce?(a)1 mole of Al3+ to Al and(b)1 mole of

Sol. (a)Reduction reaction is:

Thus, 3 moles of are needed to reduce 1 mole of Al+3.

33

(b)The reduction reaction is :

Prob 7. An electric current of 100 Amp is passed through a molten liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at N.T.P.

Sol. The reaction taking place at anode is:-

Number of eq. of chloride

Number of moles of chlorine

Volume of Cl2 liberated at NTP

Prob 8. An ammeter and a copper voltameter are connected in series through which a constant current flows. The ammeter shows 0.52 amp. If 0.635 grams of copper is deposited in one hour, what is the percentage error of the ammeter?

Sol. The electrode is

Deposited wt

or,

Difference in current = 0.536 – 0.52

% error

Prob 9. Cadmium amalgam is prepared by electrolysis of a solution of CdCl2 using a mercury cathode. Find how long a current of 5 amp. should be passed in order to prepare 12% Cd – Hg amalgam on a cathode of 2 g mercury. [Atomic mass of Cd = 112.40].

Sol. 2 g Hg require Cd to prepare 12% amalgam

Charge required to deposit 0.273 g of Cd

34

ELECTROCHEMISTRY

Prob 10. The equivalent conductivity of solution of acetic acid at

Calculate the degree of dissociation of CH3COOH if is 390.71 ohm1cm2eq1.

Sol.

Degree of dissociation,

= 0.0366 i.e. 36% dissociation

Prob 11. Calculate the minimum amount of NaOH required to be added in R.H.S. to consume all the present in R.H.S. of the cell of emf +0.701 volt at before its use. Also report the emf of the cell after addition of NaOH.

Sol. Cell reaction is:-

Applying Nernst equation:

Thus 0.0316 mol / lit of NaOH is required to neutralise Mass of NaOH After addition of NaOH, the solution becomes neutral i.e. the concentration of ions

in cathodic solution becomes

Applying again Nernst equation

= 0.3759 volts

35

Prob 12. For the galvanic cell

Calculate the emf generated and assign correct polarity to each electrode for the spontaneous process after taking into account the cell reaction at

Given

Sol.

or

= -0.0371 Volts

Prob 13. The observed emf of the cellis 0.154 V.

Calculate the value of M1 and pH of cathodic solution.

Sol.

or,

or,

Prob 14. In a fuel cell H2 and O2 react to produce electricity. In the process H2 gas is oxidised at the anode and O2 is reduced at the cathode. If 67.2 Lit. of H2 at NTP reacts in 15 minutes, what is the average current produced? If the entire current is used for electro – deposition of Cu from how many gram of Cu are deposited.

Sol.

67.2 Lit of H2 correspond

Average current

Mass of copper deposited by

36

ELECTROCHEMISTRY

Prob 15. Neglecting the liquid – liquid junction potential. Calculate the emf of the following cell at . Ka for HCOOH and

CH3COOH are respectively.

Sol.

37

Objective:

Prob 1. Dissociation of H3PO4 occurs in ………… stages(A)1 (B) 2(C) 3 (D) 4

Sol. H3PO4 is a tribasic acid.Hence, (C) is correct.

Prob 2. Which one is the correct equation that represents the first law of electrolysis?(A) mz = ct (B) m = cit(C) mc = zt (D) c = mzt

Sol. m = zitHence, (B) is correct.

Prob 3. Copper sulphate solution is electrolysed between two platinum electrodes. A current is passed until 1.6 g of oxygen is liberated at anode. The amount of copper deposited at the cathode during the same period is:(A)6.36 g (B) 63.6 g(C) 12.7 g (D) 3.2 g

Sol. For O2,

For Cu,

Hence, (A) is correct.

Prob 4. When electricity is passed through a solution of AlCl3, 13.5 g of Al is discharged. The amount of charge passed is:(A)1.5 F (B) 0.5 F(C) 1.0 F (D) 2.0 F

Sol.

Hence, (A) is correct.

Prob 5. The equivalent conductivity of 0.1 N CH3COOH at is 80 and at infinite dilution 400 The degree of dissociation of CH3COOH is(A)1 (B) 0.2(C) 0.1 (D) 0.5

Sol.

Hence, (B) is correct.

Prob 6. A solution of sodium sulphate in water is electrolysed using platinum electrodes. The products at cathode and anode are respectively.(A)H2, O2 (B) O2, H2

(C) O2, Na (D) O2, SO2

Sol. Discharge potential of H+ is less than Na+ and OH is less than .Hence, (A) is correct.

38

ELECTROCHEMISTRY

Prob 7. A certain current liberated 0.504 g of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in CuSO4 solution?(A)12.7 g (B) 15.9 g(C)31.8 g (D) 63.5 g

Sol. According to Faraday’s second law

m2 = 15.9 g.Hence, (B) is correct.

Prob 8. If the specific resistance of a solution of concentration C g equivalent is R, then its equivalent conductance is:-

(A) (B)

(C) (D)

Sol.

Since,

Hence, (C) is correct.

Prob 9. The reaction, occurs in the galvanic

cell.(A)(B)(C)(D)

Sol. (C) is correct.

Prob 10. Cu+ is not stable and undergoes disproportionation. E for Cu+ disproportionation ………………………….

( = + 0.153 V, = 0.53 V)(A) + 0.683 V (B) – 0.367(C) + 0.3415 V (D) + 0.367

Sol. 2Cu+ Cu+2 + Cu

= –0.153 + 0.53 = + 0.367 VHence, (D) is correct.

Fill in the Blanks

Prob 11. In electrolytic cell, electrical energy is ………………….

Sol. Consumed

Prob 12. In electrochemical cell, electrical energy is ……………… for a cell.

Sol. Generated

39

Prob 13. If for a cell is less than zero, the cell reaction is …………..

Sol. Spontaneous

Prob 14. With dilution, the conductance of strong electrolyte …………

Sol. Increases

Prob 15. Conductance of a weak electrolyte is …………. than the conductance of strong electrolytic of same concentration.

Sol. less

True/False

Prob 16. If a cell have one electrode made up of Cu and the other made up of iron, Cu electrode acts as anode.

Sol. False

Prob 17. In the electrolysis of (aq)AgNO3 solution, using Ag electrode, Ag acts as anode.

Sol. True

Prob 18. If salt bridge is removed from an electrochemical cell, the current produced by the cell increases.

Sol. False

Prob 19. In an electrochemical cell current flows from cathode to anode in outer circuit.

Sol. True

Prob 20. Dry cell is an example of secondary cell.

Sol. False

40

ELECTROCHEMISTRY

ASSIGNMENT PROBLEMS

Subjective:

Level - O

1. Define electrochemical equivalent. How is it related to the equivalent weight of the element?

2. What is the effect of dilution on specific conductivity and equivalent conductivity? Explain briefly giving reasons.

3. What are primary cells? How does a dry cell function?

4. What is understood by a normal hydrogen electrode? Give its significance?

5. Explain the function of salt bridge in an electrochemical cell.

6. What is an electrochemical series? How does it help us in predicting whether a redox reaction is feasible in a given direction or not?

7. What does positive value of in case of galvanic cell like indicate?

8. Write Nernst equation for the general electrochemical change of the following type at

9. What is rusting? Indicate the chemical changes in rust formation.

10. How does electrochemical series help in?(i) Comparing the relative oxidising or reducing powers of different elements.(ii) Predicting whether a metal will react with the acid to give hydrogen gas or not.

11. Why fluorine cannot be obtained by electrolysis of aqueous HF solution, though it is a good conductor of electricity?

12. What would happen if no salt bridge were used in an electrochemical cell (like Zn – Cu) cell?

13. Explain the terms, equivalent conductivity and molar conductivity. How are they inter – related?

14. Give the construction of a lead battery. Explain giving reactions how it can be recharged.

15. Give a short account of corrosion and its mechanism.

41

Level-I

1. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5.0 amp for 30 minutes. How much of nickel will be produced at the cathode.

2. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of hydrogen ions. The emf of the cell is 0.118 V at Calculate the concentration of hydrogen ions at the positive electrodes.

3. The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms. Calculate the equivalent conductivity of the solution if the electrodes in the cell were 2 cm apart and each has an area of 3.5 square cm.

4. A 100 watt, 110 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hrs.

5. Calculate the emf of the following cell

a = 0.6 a = 0.2

6. A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at

298 K. Calculate the electrode potential .

7. Calculate the potential (emf) of the cell

8. The standard electrode potentials are given as

Calculate the cell potential (E) for the cell containing 0.100 M

9. For the reaction; , enthalpy and entropy changes are -92.4 kJ and

respectively. Calculate the equilibrium constant of the reaction

(R = 8.314

10. Calculate the cell e.m.f. at for the following cell:

Given

Calculate the maximum work that can be accomplished by the operation of this cell.

42

ELECTROCHEMISTRY

Level-II

1. Conductivity of two electrolyte solutions A and B each having a concentration of 0.1 M are respectively. Which of the two offers less resistance to the

flow of current.

2. The electrolysis of a metal salt solution was carried out by passing a current of 4 amp for 45 minutes. It resulted in the deposition of 2.977 g of the metal. If atomic mass of the metal is 106.4 g calculate the charge on the metal cation.

3. Iodine (I2) and bromine (Br2) are added to a solution containing iodide ions. What reaction would occur if the concentration of each species is 1 M? The electrode potential for the reactions are:

4. The atomic mass of a metal M is 52.01. When the metal chloride of the metal ‘M’ is electrolysed, metal is deposited at the cathode. Chlorine gas is evolved at anode. For every gram of metal deposited, 725 ml of dry Cl2 gas is evolved at and 740 mm Hg. Find the formula of the chloride of the metal.

5. The standard reduction potential at of the reaction is -0.8277 V. Calculate the equilibrium constant for the reaction

6. The emf of the cell is -0.03V at Calculate the value

of x.

7. The mass of copper deposited from a solution of copper sulphate by a uniform current of 1 amp flowing for 33 minutes is 0.65 g. Find the electro – chemical equivalent of hydrogen from the result.

8. An electrode is prepared by dipping a silver rod in a saturated solution of AgCl and another electrode is prepared by dipping a silver rod in a saturated solution of Ag2CrO4. These two electrodes are used to construct a galvanic cell. Calculate the emf of such a galvanic cell at

It is given that

9. The standard reduction potential for is +0.34 Volt. Calculate the reduction potential

at pH = 14 for the above couple.

10. Calculate the equilibrium constant for the reaction

Given

43

Objective:

Level – I

1. A redox reaction is spontaneous in a given direction, if:(A) EMF is zero (B) EMF is negative(C) EMF is positive (D) None of these

2. Which of the following is a secondary cell?(A) Dry cell (B) Mercury cell(C) Ni – Cd cell (D) H2 – O2 cell

3. The amount of electricity required to deposit 1 mol of Aluminium from a solution of AlCl3 will be(A) 0.33 F (B) 1 F(C) 3 F (D) 1 amp

4. The standard EMF for the cell reactionis 1.1 Volt at . The emf of the cell reaction when 0.1 M Cu2+

and 0.1 M Zn2+ solutions are used, at (A) 1.10 V (B) 0.10 V(C) -1.10 V (D) -11.0 V

5. The specific conductance of a solution of KCl at is 0.002765 mho. If the resistance

of a cell containing this solution is 400 ohms. What is cell constant?(A) 2 (B) 1.106(C) 3 (D) 3.2

6. A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (Atomic weight = 177). The oxidation state of the metal in the metal salt is:(A) +1 (B) +2(C) +3 (D) +4

7. The conductivity of a saturated solution of BaSO4 is and its

equivalent conductance is 1.53 The Ksp for BaSO4 will be(A) (B)(C) (D)

8. for the cell

is 1.10 V at The equilibrium constant for the reaction

is of the order of

(A) (B)(C) (D)

9. For the cell reaction, of an electrochemical cell, the

change in free energy, at a given temperature is a function of

(A) (B)

(C) (D)

44

ELECTROCHEMISTRY

10. Which of the following graphs will correctly correlate Ecell as a function of concentrations for the cell (for different M and M¢)?Zn(s) + Cu+2(M) Zn+2(M¢) + Cu(s), = 1.10 V

X-axis: , Y-axis: Ecell

(A) (B)

(C) (D)

11. is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction.

Choose correct for the above reaction if

(A) + 0.49 V (B) + 0.38 V(C) (D)

12. For the cell

(A) -1.20 V (B) -0.32 V(C) 1.20 V (D) 0.32 V

13. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because(A) Zn acts as oxidising agent when reacts with HNO3

(B) HNO3 is weaker acid than H2SO4 and HCl(C) In electrochemical series, Zn is above hydrogen(D) is reduced in preference to hydronium ion

14. At cathode, the electrolysis of aqueous Na2SO4 gives(A) Na (B) H2

(C) SO3 (D) SO2

15. What will be the emf of the given cell

(A) (B)

(C) (D) None of these

45

16. A current is passed through 2 voltameters connected in series. The first voltameter contains XSO4(g)(aq.) the relative atomic masses of X and Y are in the ratio of 2:1. The ratio of the mass of X liberated to the mass of Y liberated is(A) 1:1 (B) 1:2(C) 2:1 (D) None of these

17. The correct statement among the following is:(A) The EMF of a cell is an extensive property and so is the G of the cell reaction(B) The EMF of a cell is an intensive property and so is the G of the cell reaction(C) The EMF of a cell is an intensive property whereas G of the cell reaction is an extensive

property.(D) The EMF of a cell is an extensive property but G of the cell reaction is intensive.

18. Moderate electrical conductivity is shown by (A) Silica (B) Graphite(C) Diamond (D) carborundum

19. The standard reduction potentials of the elements A, B and C are + 2.73V, - 1.85 V and -1.36 respectively. The order of their reducing power is (A) B > C > A (B) A > B > C(C) C > B > A (D) B > A > C

20. In electrolysis of alkaline water, a total of 1 mole of gases is evolved. The amount of water decomposed would be(A) 1 mol (B) 2 mol(C) 1/3 mol (D) 2/3 mol

46

ELECTROCHEMISTRY

Level – II

1.The of the cell is 2.12 V. To increase E(A) Zn2+ concentration should be increased(B) Zn2+ concentration should be decreased(C) concentration should be increased(D) Partial pressure of Cl2 should be decreased

2.

(A) 0.54 V (B) -0.054 V(C) +0.18 V (D) -0.18 V

3. Time required to deposit one millimole of aluminium metal by the passage of 9.65 amp. through aqueous solution of aluminium ion is(A) 30 sec. (B) 10 sec.(C) 30,000 sec. (D) 10,000 sec.

4. An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108 g more at the end of the electrolysis. The volume of oxygen evolved at STP is(A) 56 ml (B) 550 ml(C) 5.6 ml (D) 11.2 ml

5. The standard reduction potentials for two reactions are given below :

The solubility product of AgCl under standard conditions of temperature (298 K) is given by :(A) (B)(C) (D)

6. On the basis of information available from the reaction

the minimum emf required to carry out

electrolysis of Al2O3 is :(A) 2.14 V (B) 4.28 V(C) 6.42 V (D) 8.56 V

7. A Chemist found that the standard electrode potential , of the zinc electrode is –

0.763 Volt. Which cell will give this value of EMF?(A) Pt, H2 (1 atm) | H+ (10M) || Zn+2 (10M) | Zn(B) Pt, H2 (1 atm) | H+ (1M) || Zn+2 (1M) | Zn(C) Zn | Zn+2 (1M) || H+ (1M) | H2 (1 atm), Pt(D) Zn | Zn+2 (10M) || H+ (10M) | H2 (1 atm), Pt

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8. The standard reduction potentials at 298 K for the half reactions are given against each

Which is strongest reducing agent?(A) Zn(s) (B) Cr(s)

(C) H2(g) (D) Fe2+(aq)

9. When a lead storage battery is discharged(A) SO2 is evolved (B) Lead sulphate is consumed(C) Lead is formed (D) Sulphuric acid is consumed

10. A dilute aqueous solution of Na2SO4 is electrolysed using Platinum electrodes. The products at the anode and cathode are(A) O2, H2 (B)

(C) O2, Na (D)

11. The standard reduction potential of Cu2+ / Cu and Cu2+ / Cu+ are 0.337 and 0.153 V respectively. The standard electrode potential of Cu2+ / Cu half cell is(A) 0.184 V (B) 0.827 V(C) 0.521 V (D) 0.490 V

12. A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M 1 M

If the reduction potential of z > y > x, then(A) y will be oxidise x and not z (B) y will be oxidise z and not x(C) y will be oxidise both x and z (B) y will reduce both x and z

13. The correct order of equivalent conductance of infinite dilution of LiCl, NaCl and KCl is(A) LiCl > NaCl > KCl (B) KCl > NaCl > LiCl(C) NaCl > KCl > LiCl (D) LiCl > KCl > NaCl

14. Saturated solution of KNO3 is used to make salt bridge because(A) Velocity of K+ is greater than that of

(B) Velocity of is greater than that of K+

(C) Velocity of both K+ and are nearly the same(D) KNO3 is highly soluble in water

15. In an electrolytic cell, flow of electron is from(A) Cathode to anode in the solution(B) Cathode to anode through external supply(C) Cathode to anode through internal supply(D) Anode to cathode through internal supply

48

ELECTROCHEMISTRY

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level - O

1. Electrochemical equivalents of any element is the deposited weight on electrode during electrolysis by the passing of 1 coulomb of electricity.Electrochemical equivalent (z) = Deposition by 1 coulombEquivalent weight (E) = Deposition by 1 Faraday

2. During dilution, specific conductivity decreases, because number of ions per ml electrolyte decreases during electrolysis. But, when electrolysed diluted, then number of ions in given volume for eq. dissolved electrolyte increases, hence, resultant equivalent conductivity increases.

3. Primary cells: Such type of cell in which redox reaction occurs only once and the cell becomes dead after some time. It cannot be again used.Dry cell: Dry cell consists of a catalysed zinc container which acts as the anode. A graphite rod placed in the center acts as cathode. Space between electrodes is packed by the paste of NH4Cl and ZnCl2. An anode:

At cathode:

4. Ans. Given in theoretical part.

5. Ans. Given in theoretical part.

6. Explanation given theory part.

7.

For positive value of

8.

9. Refer to theory.

10. Refer to theory

11. On electrolysis of aqueous HF solution, we get H2 and O2 instead of F2, because discharge potential of H2O is less than F.

12. The circuit will not be completed and the current will not flow.

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13. Given in theoretical part.

14. The electrodes are arranged alternately and separated by thin wooden or fibre glass sheets. Generally 38% by mass of H2SO4 (d = 1.30 g/ml) acts as electrolyte. When external circuit is closed, then electrode reactions occur as following:At anode:

At cathode:

15. Corrosion is the process in which atmospheric air reacts with metal surface resulting into formation of compounds such as oxides, sulphides, carbonates, etc. Rust formation can be explained on this basis.

Level – I

1. 2.74 g

2. 40.2 hours

3. 23.32

4. 19.06 gm

5. 0.566 Volt

6. -0.79021 Volts

7. 0.4 V 8. 0.383 V

9. 10.

Level – II

1. (A) 2. +4

3. 4. MCl3

50

ELECTROCHEMISTRY

5. 6. 0.009616 mol

7. Mass of hydrogen

8. 0.065 Volt

9. -0.22 V

10.

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Objective:

Level - I

1. C 2. C 3. C4. A 5. B 6. C7. D 8. B 9. B10. B 11. B 12. D13. D 14. B 15. B16. A 17. C 18. B19. A 20. D

Level – II

1. B 2. A 3. A4. C 5. B 6. A7. C 8. A 9. D10. B 11. C 12. A13. B 14. C 15. C

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