Electrolytes non electrolytes. Types of Electrolytes

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Archibald Grant
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1 Electrochemistry

2 Chemical reactions where electrons are transferred between molecules are called oxidation/reduction (redox) reactions. In general, electrochemistry deals with situations where oxidation and reduction reactions are separated in space or time, connected by an external electric circuit to understand each process. Electrolytes Substances whose solution in water conducts electric current. Conduction takes place by the movement of ions. Examples are salts, acids and bases. Substances whose aqueous solution does not conduct electricity are called non electrolytes. Examples are solutions of cane sugar, glucose, urea etc. Types of Electrolytes Strong electrolyte are highly ionized in the solution. Examples are HCl, H 2 SO 4, NaOH, KOH etc Weak electrolytes are only feebly ionized in the solution. Examples are H 2 CO 3, CH 3 COOH, NH 4 OH etc

3 Electronic conductors (1) Flow of electricity take place without the decomposition of substance. (2) Conduction is due to the flow of electron (3) Conduction decreases with increase in temperature Electrolytic conductors (1)Flow of electricity takes place by the decomposition of the substance. (2) Flow of electricity is due to the movement of ions (3) Conduction increases with increase in temperature Conductance The reciprocal of the resistance is called conductance. It is denoted by C. C=1/R Conductors allows electric current to pass through them. Examples are metals, aqueous solution of acids, bases and salts etc. Unit of conductance is ohm -1 or mho or Siemen(S)

4 Specific Conductivity Specific conductance 1 Conductance of unit volume of cell is specific conductance. a But ρ = R K x Conductance a l/a is known as cell constant Unit of specific conductance is ohm 1 cm 1.SI Unit of specific conductance is Sm 1 where S is Siemen Equivalent Conductance It is the conductance of one gram equivalent of the electrolyte dissolved in V cc of the solution. Equivalent conductance is represented by λ

5 Mathematically, k V Where, k = Specific conductivity k Molar conductance 1000 Normality V = Volume of solution in cc. containing one gram equivalent of the electrolyte. It is the conductance of a solution containing 1 mole of the electrolyte in V cc of solution. it is represented as m. Where V = volume solution in cc =k * 1000/M m Molar conductance k = Specific conductance M=molarity of the solution. Molar conductivity = n- factor x equivalent conductivity

6 Effect of Dilution on Conductivity (i) The conductivity of solution increases on dilution. (ii) The specific conductivity decreases on dilution (as number of ions decreases w.r.t. to volume). (iii) The equivalent and molar conductivities increase with dilution. (iv) The equivalent and molar conductivities tend to acquire maximum value with increasing dilution. [Maximum at dilution] For strong electrolytes, Λ = Λ o - a c but for weak electrolytes, use Kohlrausch s Law

7 Kohlrausch s law of independent ionic mobilities At time infinite dilution ( m), the molar conductivity of an electrolyte can be expressed as the sum of the contributions from its individual ions Λ m = v + λ + + v - λ - v + and v- are the number of cations and anions per formula unit of electrolyte respectively and, λ + and λ - are the molar conductivities of the cation and anion at infinite dilution respectively But for EQUIVALENT conductivity, the v terms are omitted no matter what their value is a c

8 Applications of Kohlrausch s Law Determination of Λ m for weak electrolytes eg Λ CH 3 COOH = Λ CH 3 COONa + Λ HCI - Λ NaCI Determination of the degree of ionization of a weak electrolyte Determination of the ionization constant of a weak electrolyte v Determination of the solubility of a sparingly soluble salt Example 1: The resistance of 0.01N NaCl solution at 25 0 C is 200 ohm. Cell constant of conductivity cell is unity. Calculate the equivalent conductance and molar conductance of the solution.

9 Solution: Conductance of the cell =1/resistance =1/200 =0.005 S. Specific conductance= conductance x cell constant =0.005 x 1 =0.005 S cm -1 Equivalent Conductance = Specific conductance x (1000/N) = x 1000/0.01 = 500 ohm -1 cm 2 eq -1 Molar Conductivity = Equivalent conductivity x n-factor = 500 x 1 = 500 ohm -1 mol -1 cm 2 Example 2: A decinormal solution of NaCl has specific conductivity equal to If ionic conductance of Na + and Cl ions are 43.0 and 65.0 ohm 1 respectively. Calculate the degree of dissociation of NaCl solution. Solution: Normality=0.10 N Equivalent conductance of NaCl Sp.conductivity 1000 v N

10 Example 3: Equivalent conductance of NaCl, HCl and C 2 H 5 COONa at infinite dilution are , and 91 ohm 1 cm 2 respectively. Calculate the equivalent conductance of C 2 H 5 COOH. Solution : = = ohm 1 cm 2 Example 4: Calculate (a) the degree of dissociation and (b) the dissociation constant of 0.01M CH 3 COOH solution; given the conductance of CH 3 COOH is S cm 1 and (CH 3 COOH) = S cm 2 mol 1 Solution: c K m c m 16.5 Scm mol Molarity 0.01 c m o m

11 Faraday s law 2 c 0.01 Ka Now, = = Faraday's first law: It states that, the mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed. W α Q W = Mass of ions liberated in gm, Q = Quantity of electricity passed in Coulombs = Current in Amperes ( i ) Time in second (t) W α i t W = Z i t Where, Z = constant, known as electrochemical equivalent (ECE) of the ion deposited

Faraday's second law: It states that, when the same quantity of electricity is passed through different electrolytes, the masses of different ions liberated at the

12 Faraday's second law: It states that, when the same quantity of electricity is passed through different electrolytes, the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalents (Equivalent weights). W α E W1/W2 = E1/E2 or Z1it / Z2it or Z1/Z2 = E1/E2 (W = Zit) E α Z or E = FZ or E = Z

13 Example 5: Find the total charge in coulombs on 1 g ion of N 3. Solution: No. of moles in 1g N 3- ion = 1/14 = Electronic charge on one mole ion N 3 = coulombs = coulombs Therefore, charge on 1g N 3 ion = x coulombs =2.06 x 10 4 Coulombs Example 6: On passing 0.1 Faraday of electricity through aluminium chloride what will be the amount of aluminium metal deposited on cathode (Al = 27) Solution: Example 7: What current strength in amperes will be required to liberate 10 g of bromine from KBr solution in half an hour? Solution: Bromine is liberated by the reaction at anode as follows

14 i = ampere Example 8: How many atoms of calcium will be deposited from a solution of CaCl 2 by a current of 25 milliamperes flowing for 60 s? Solution: Number of Faraday of electricity passed moles of Ca atoms = atoms of calcium. atoms of Ca Example 9: A 100 W, 110 V incandescent lamp is connected in series with an electrolyte cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hr? (Gram atomic mass of Cd = g)

Solution: Watt = Volt Current 100 = 110 Current But we know, Q = i t 10 10 60 60 Number of equivalent = = 0.339 11 96500 Mass of cadmium deposited 0.339 112.4 = 19.

15 Solution: Watt = Volt Current 100 = 110 Current But we know, Q = i t Number of equivalent = = Mass of cadmium deposited = g 2 Example 10: The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. What will be the ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells? Solution: Ratio of iron deposited at cathode will be in their ratio of equivalents

Electrolysis The process of decomposition of an electrolyte by the

In electrolysis electrical energy is used to cause a chemical

For example, Electrolysis of molten sodium chloride Electrolysis of

16 Electrolysis The process of decomposition of an electrolyte by the passage of electricity is called electrolysis. In electrolysis electrical energy is used to cause a chemical reaction. For example, Electrolysis of molten sodium chloride Electrolysis of aqueous sodium chloride NaCl (aq) Na + (aq) + Cl - (aq) H 2 O (aq) H + (aq) + OH - (aq) REMEMBER: anode is where OXIDATION occurs and at cathode REDUCTION occurs

17 H + ions are discharged at cathode because discharge potential of H + ion is much lower than Na + ion 1 H + e At anode, Cl is discharge as Cl 2 (gas) because discharge potential of Cl is much lower than that of OH ion. 1 Cl Cl2 g e Electrolysis of aqueous copper sulphate solution using inert electrodes 2+ 2 CuSO (aq.) Cu (aq.) + SO (aq.) 4 4 H 2 O (aq) H + (aq) + OH - (aq) H 2 (gas) 2 2 At cathode, Cu 2+ ions are discharged in preference to H + ions because discharge potential of Cu 2+ is much lower than H + ions. 2 Cu aq. 2e Cu aq. At anode, OH ions are discharged in preference to SO 2 4 ions because discharge potential of OH is much lower than SO 2 4 ions. 4OH 2H O ( ) O (g) 4e 2 2

19 Higher the standard reduction potential, the lesser will be its reducing strength. Li is strongest reducing agent in aqueous solution. The lesser the standard reduction potential of an element, greater will be its activity. A more active metal displaces a less active one from its salt solution. Those metals which have positive oxidation potential will displace hydrogen from acids. The metals above hydrogen are easily rusted and those situated below are not rusted. Example 11: E 0 values of Mg +2 /Mg is 2.37V, Zn +2 /Zn is 0.76V, Fe +2 /Fe is 0.44V.Using this information predict whether Zn will reduce iron or not? Solution: Zinc has lower reduction potential than iron. Therefore, it can reduce iron.

20 Galvanic cell Daniel cell Cell reaction: Zn Cu Zn Cu 2 2 (s) (aq) (aq) (s) 3.gif

21 Electromotive force(emf) of a cell As per IUPAC convention if we consider standard reduction potential of both the electrodes then Emf = E R.H.Electrode - E L.H.Electrode Emf Ecathode Eanode Salt Bridge Salt bridge Agar agar mixed with KCl, KNO 3, NH 4 NO 3) etc. Eliminates liquid - liquid junction potential Maintains the electrical neutrality of solutions. Completes the circuit. NOTE: Emf is ALWAYS positive

22 he galvanic cell is represented in a short IUPAC cell notation as: t is important to note that: First of all the anode (electrode of the anode half cell) is written. In the above case, it s Zn. After the anode, the electrolyte of the anode should be written with concentration. In his case it is ZnSO 4 with concentration in moles/litre. A slash ( ) is put in between the Zn rod and the electrolyte. This slash denotes a urface barrier between the two as they exist in different phases. Then we indicate the presence of a salt bridge by a double slash ( ). Now, we write the electrolyte of the cathode half-cell which is CuSO 4 with its oncentration in moles/l. Finally we write the cathode electrode of the cathode half cell. A slash (/) between the electrolyte and the electrode in the cathode half cell. In case of a gas, the gas to be indicated after the electrode in case of anode and efore the electrode in case of cathode. Example: Pt, H 2 /H + or H + H 2, Pt.

23 Example 12: The standard oxidation potential E o for the half reaction are given below. o Zn 2e Zn E 0.76 V 2 o Fe 2e Fe E 0.41 V Calculate E 0 for the following cell reaction Zn + Fe ++ Zn ++ + Fe Solution: o o o E E E cell cathode anode = 0.41 ( 0.76) = V NERNST EQUATION We know, ΔG o = nfe Also, ΔG=ΔG +RTlnQ Hence, nfe= nfe +RTlnQ E=E RTlnQ E=E 2.303RTlogQ E=E logQ (at 25 C) nf nf n

24 Example 13: Find the cell potential of a galvanic cell based on the following reduction half-reactions at 25 C Cd e- Cd E = V Pb e- Pb E = V where [Cd2+] = M and [Pb2+] = M. Solution: The total cell reaction is: Pb 2+ (aq) + Cd(s) Cd 2+ (aq) + Pb(s) and E 0 cell = V V = V Q = [Cd 2+ ]/[Pb 2+ ] = M / M = E cell = E 0 cell - (RT/nF) * lnq E cell = V V * ln(0.100) E cell = V V * E cell = V V E cell = V Example 14: For Zn Zn e- E = V (assume) and Cu + + e- Cu E = V (assume) Find E of the cell reaction

25 Solution: Zn Zn e- E = V (Cu + + e- Cu) * 2 E = V Zn + 2Cu + Zn Cu E = V NOTE that we did not multiply 2 with E of reduction half cell (why?? Because voltage is the quotient of two extensive quantities, it is itself intensive. When we multiply the anodic and cathodic half-reactions by the stoichiometric factors required to ensure that each involves the same quantity of charge, the free energy change and the number of coulombs both increase by the same factor, leaving the potential (voltage) unchanged.) Example 15: Calculate E for the electrode Fe 3+ /Fe(s) from the standard potential of the couples Fe 3+ /Fe 2+ and Fe 2+ /Fe(s) Fe 3+ + e Fe 2+ E 1 = v Fe e Fe(s) E 2 = v Solution: DONOT add the emfs!!! Instead you have to add their ΔG values and then find the corresponding E value

For Fe 3+ + e Fe 2+, ΔG = -nfe = 1*(-0.771) F = -0.771F And for Fe 2+ + 2 e Fe, ΔG = -nfe = 2*(0.44)F = 0.88F Hence, for Fe 3+ + 3 e Fe ΔG = (-0.771 + 0.88)F = +0.109 F Therefore, E= -ΔG /nf = 0.

26 For Fe 3+ + e Fe 2+, ΔG = -nfe = 1*(-0.771) F = F And for Fe e Fe, ΔG = -nfe = 2*(0.44)F = 0.88F Hence, for Fe e Fe ΔG = ( )F = F Therefore, E= -ΔG /nf = 0.109/3 = v (-ve sign indicates the cell is nonspontaneous) DRY CELL (Primary cell - cannot be recharged) (emf= 1.5V) cell.jpg

STANDARD HYDROGEN ELECTRODE CALOMEL ELECTRODE http://glencoe.

jpg In calomel, if it cathode, reaction: If it is anode,

27 STANDARD HYDROGEN ELECTRODE CALOMEL ELECTRODE In calomel, if it cathode, reaction: If it is anode, opposite of this reaction takes place

28 LEAD STORAGE BATTERIES (Secondary voltaic cell ie can be recharged) (Emf = 2V)

FUEL CELLS (emf= 1.23V) http://www.hybridcars.

29 FUEL CELLS (emf= 1.23V)

30 Reactions: At anode: 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e- At cathode: O 2 (g) + 2H 2 O (l) + 4e- Net: 2H 2 (g) + O 2 (g) Advantages: 2H 2 O (l) 4OH - (aq) On combustion produces H 2 O. Hence pollution free. Reacting substances continuously supplied to the electrodes and fuel needn t be discharged when chemicals are consumed High efficiency Disadvantages: Hydrogen gas is hazardous to use High cost

At anode : Mechanism of rusting At anode : Fe (s) Fe 2+ (aq) + 2e- Thus, the metal atoms in the lattice pass into the solution as ions, leaving electrons on the metal itself.

31 At anode : Mechanism of rusting At anode : Fe (s) Fe 2+ (aq) + 2e- Thus, the metal atoms in the lattice pass into the solution as ions, leaving electrons on the metal itself. These electrons move towards the cathode region through the metal. At cathode : O 2 (g) + 4 H + (aq) + 4e- 2H 2 O(l) Overall: 2Fe (s) + O 2 (g) + 4 H + (aq) 2Fe 2+ (aq) + 2H 2 O(l) Fe 2+ is further oxidised to Fe 3+ by oxygen 4Fe 2+ (aq) + O 2 (g) + 4 H + (aq) 4Fe 3+ (aq) + 2H 2 O(l) Fe 3+ ions form an insoluble hydrated oxide (ie rust) 3Fe 3+ (aq) + 4H 2 O(l) Fe 2 O 3.xH 2 O (s)

Prevention of rusting Use of protective layer Painting Used in cars, ships, bridges Greasing Tools & machine parts Zinc plating(galvanising) Zinc roofs Tin plating Food cans

Magnesium or zinc is attached to iron or steel Protects by sacrificing itself, corrodes first since it is more reactive Iron will not rust in the presence of a more reactive metal

32 Prevention of rusting Use of protective layer Painting Used in cars, ships, bridges Greasing Tools & machine parts Zinc plating(galvanising) Zinc roofs Tin plating Food cans (Creates barrier around the metal preventing contact with oxygen and water) Sacrificial protection: More reactive metal, eg. Magnesium or zinc is attached to iron or steel Protects by sacrificing itself, corrodes first since it is more reactive Iron will not rust in the presence of a more reactive metal Used in underground pipes, ships, steel piers ontentblockphotos/display /53/anti%20rust%20packagi ng%20vci-paper-volatile- Corrosion-Inhibitor.png

Unit - 3 ELECTROCHEMISTRY VSA QUESTIONS (1 - MARK QUESTIONS) 3. Mention the purpose of salt-bridge placed between two half-cells of a galvanic cell?

Unit - 3 ELECTROCHEMISTRY VSA QUESTIONS (1 - MARK QUESTIONS) 3. Mention the purpose of salt-bridge placed between two half-cells of a galvanic cell? Unit - 3 ELECTROCHEMISTRY 1. What is a galvanic cell? VSA QUESTIONS (1 - MARK QUESTIONS) 2. Give the cell representation for Daniell Cell. 3. Mention the purpose of salt-bridge placed between two half-cells