CLASS 12th
Electrochemistry
Electrochemistry
01. Electrochemical Cells An electrochemical cell consists of two electrodes (metallic conductors)in contact with an electrolyte (an ionic conductor). An electrode and its electrolyte comprise an Electrode Compartment. Electrochemical Cells can be classified as: (a) Electrolytic cells in which a non-spontaneous reaction is driven by an external source of current. (b) Galvanic cells which produce electricity as a result of a spontaneous cell reaction. An electrochemical cell is a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses of generates an electric current. A voltaic of galvanic cell is an electrochemical cell in which a spontaneous reaction generates an electric current. A voltaic cell consists of two half-cells that are electrically connected. Each half cell is the portion of an electrochemical cell in which a half cell reaction take place. In a voltaic cell, two half-cells are connected in such a way that electrons flow from one metal electrode to another through an external circuit, and ions flow from one half-cell to another through an internal cell connection. Figure given below illustrates an atomic view of a voltaic cell consisting of a zinc electrode and a copper electrode. As long as there is an external circuit, electrons can flow through it from one electrode to another. Because zinc tends to lose electrons more readily than copper, zinc atoms in the zinc electrode lose electrons to produce zinc ions. These electrons flow through the external circuit to the copper electrode, where they react with the copper ions to produce copper metal, and an electric current flows through the external circuit.
The two half-cells of a voltaic cell are connected by a salt bridge. A salt bridge is a tube of an electrolyte in a gel that is connected to the two half-cells of a voltaic cell; the salt bridge allows the flow of ions but prevents the mixing of the different solutions that would allow direct reaction of the cell reactants. The half-cells are connected externally so that an electric current flows. The two half-cell reactions, as noted earlier, are Zn(s) → Zn2+(aq) + 2e− (oxidation half cell-reaction) Cu2+(aq) + 2e− → Cu(s) (reduction half cell-reaction) Note that the sum of the two half cell-reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) The net reaction that occurs in the voltaic cell, it is called the cell reaction.
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Electrochemistry NOTE ☞
(i)
The salt-bridge contains solution of strong ionic salts like NaCl, NaNO3, KNO3, KCl etc., which is soaked in a colloidal solution of agar-agar gel which permits the movement of ions of salts only. (ii) Salt bridge carries whole of the current across the boundary ; more over the K+ and NO3− ions have same speed. Hence, salt bridge with uniform and same mobility of cations and anions completes the electrical circuit & permits the ions to migrate. (iii) It maintains the electrical neutrality of the solutions in the two half-cells. In the absence of salt bridge, a reverse potential difference is set up in the two half-cells which results in breaking the continuous supply of voltage.
02. Representation of a cell (IUPAC Conventions) Let us illustrate the convention taking the example of Daniel cell. (i) Anodic half cell is written on left and cathodic half cell on right hand side. Zn(s) | ZnSO4(sol) || CuSO4(sol) | Cu(s) (ii) Two half cells are separated by double vertical lines: Double vertical lines indicate salt bridge of any type of porous partition. (iii) EMF (electromotive force) may be written on the right hand side of the cell. (iv) Single vertical lines indicate the phase separation between electrode and electrolyte solution. Zn | Zn2+ || Cu2+ | Cu (v) Inert electrodes are represented in the bracket Zn | ZnSO4 || H+ | H2, (Pt) Write cell reaction of the following cells: (i) Ag | Ag+(aq) || Cu2+(aq) | Cu (ii) Pt | H2 | H+(aq) || Cd2+(aq) | Cd
Example
03. Electrode Potential When a metal is placed in a solution of its ions, the metal acquires either a positive or negative charge with respect to the solution. On account of this, a definite potential difference is developed between the metal and the solution. This potential difference is called electrode potential. The potential difference is established due to the formation of electrical double layer at the interface os metal and the solution. The development of negative charge (as on zinc plate) of positive charge (as on copper plate) can be explained in the following manner. Depending on the nature of the metal electrode to lose or gain electrons, the electrode potential may be of two types :
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Electrochemistry Oxidation potential : When electrode is negatively charged with respect to solution, i.e., it acts as anode. Oxidation occurs. M Mn+ + ne− Reduction potential : When electrode is positively charged with respect to solution i.e., it acts as cathode. Reduction occurs. Mn+ + ne− M It is not possible to measure the absolute value of the single electrode potential directly. Only the difference in potential between two electrodes can be measured experimentally.
04. Concept of Electromotive Force (EMF) of A Cell Electron flows from anode to cathode in external circuit due to a pushing effect called or electromotive force (e.m.f.). EMF is called as cell potential. Unit of e.m.f. of cell is volt. EMF of cell may be calculated as: Ecell = Reduction potential of cathode − Reduction potential of anode Similarly, standard e.m.f. of the cell (E°) may be calculated as E°cell = Standard reduction potential of cathode − Standard reduction potential of anode Ecell = R.P.(Cathode) − R.P.(Anode) = R.P.(Cathode) − O.P.(Anode) For the cell reaction 2Ce4+ + Co 2Ce3+ + Co2+ o o ECo E°cell is 1.89V. If ECo Co is−0.28V, what is the value of Ce
Example
o o E°cell = ECo Ce −ECoCo
Solution
o 1.89 = ECo Ce −(−.28) o ECo Ce = 1.61V
05. Relationship Between ∆G and Electrode Potential Let n, Faraday charge is taken out from a cell of e.m.f.(E), than electrical work done by the cell my be calculated as, Work done = Charge × Potential = nFE From thermodynamics, we know that decrease in Gibbs free energy of a system is a measure of reversible or maximum obtainable work by the system ∴ ∆G = −nFE Under standard state ∆Go = −nFEo (i) (i) From thermodynamics we know, ∆G = negative for spontaneous process. Thus from eq.(i) it is clear that the EMF should be+be for a cell process to be feasible or spontaneous. (ii) When ∆G = positive, E = negative and the cell process will be non spontaneous.
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Electrochemistry
Reactions Spontaneous Non-spontaneous Equilibrium
∆G (−) (+) 0
E (+) (−) 0
Standard free energy change of a cell may be calculated by electrode potential data. Substituting the value of Eo (i.e., standard reduction potential of cathode-standard reduction potential of anode) in eq. (i) we may get ∆Go.
06. Nernst Equation Walter Nernst derived a relation between cell potential and concentration or Reaction quotient. ∆G = ∆Go + RT lnQ ...(1) where ∆G and ∆Go are free energy and standard free energy change; ‘Q’ is reaction quotient. ∴ −∆G = nFE and −∆Go = nFEo Thus from Eq.(i), we get−nFE = nFEo + RT lnQ At 25°C, above equation may be written as E = Eo −
0.0591 log Q n
Where ‘n’ represents number of moles of electron involved in process. In general, for a redox cell reaction involving the transference of n electrons aA + bB → cC + dD, the EMF can be calculated as: Ecell = Eocell−
0.0591 [C]c[D]d log n [A]a[B]b
07. Thermodynamic Treatment of Nernst Equation Prediction and feasibility of spontaneity of a cell reaction. Let us see whether the cell (Daniel) is feasible or not: i.e. whether Zinc will displace copper or not. Zn | (s) | ZnSo4(1M) || CuSO4(1M) | Cu(s) o o volt ECu volt EZn Zn Cu o o EZn Ecell ECu Cu Zn
= 0.34−(−0.76) = +1.10 volt Since E = +ve, hence the cell will be feasible and zinc will displace copper from its salt solution. In the other words zinc will reduce copper. o
Determination of equilibrium constant : E = Eo −
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0.0591 log Q n
...(i)
Electrochemistry At equilibrium, the cell potential is zero because cell reactions are balanced, i.e. E = 0 ∴ From Eq. (i), we have 0 = Eo −
0.0591 log Keq n
or
Keq = anti log
08. Different types of half-cells and their reduction potential (i)
Gas-Ion Half Cell: In such a half cell, an inert collector of electrons, platinum or graphite is in contact with gas and a solution containing a specified ion. One of the most important gas-ion half cell is the hydrogen-gas-hydrogen ion half cell. In this half cell, purified H2 gas at a constant pressure is passed over a platinum electrode which is in contact with an acid solution. H+(aq) + e− 1/2 H2 pH EH H EH H log H (ii) Metal-Metal Ion Half Cell: This type of cell consist of a metal M in contact with a solution containing Mn+ ions. Mn+(aq) + ne− M(s) EM n M EM n M log n M n (iii) Metal-Insoluble Salt-Anion Half Cell: In this half cell, a metal coated with its insoluble salt is in contact with a solution containing the anion of the insoluble salt eg. Silver Chloride Half Cell: This half cell is represented as Cl−/AgCl/Ag. The equilibrium reaction that occurs at the electrode is AgCl(s) + e− Ag(s) + Cl−(aq) log Cl EClAgClAg ECl AgClAg Potential of such cells depends upon the concentration of anions. Such cells can be used as Reference Electrode. (iv) Oxidation-reduction Half Cell: This type of half cell is made by using an inert metal collector, usually platinum, immersed in a solution which contains two ions of the same element in different states of oxidation. eg. Fe2+-Fe3+ half cell. Fe3+(aq) + e− Fe2+(aq) Fe EFeFe EFe log Fe Fe
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Electrochemistry
09. Concentration Cell The cells in which electrical current is produced due to transport of a substance from higher to lower concentration. Concentration gradient may arise either in electrode material or in electrolyte. Thus there are two types of concentration cell. For such cell, E°Cell = 0. (i) Electrode Gas concentration cell: Pt, H2(P1) | H+(C) | H2(P2), Pt Here, hydrogen gas is bubbled at two different partial pressures at electrode dopped in the solution of same electrolyte. Cell process: 1/2H2(p1) → H+ (c) + e− (Anode process) H+ (c) + e− → 1/2H2(p2) (cathode process) 1/2H2(p1)
1/2H2(p2) (Net reactions) p RT E log F p
∴
or
(ii)
At 25°C,
p E log p
For spontanity of such cell reaction, p1>p2 Electrolyte concentration cells: Zn(s) | ZnSO4(C1) || ZnSO4(C2) | Zn(s) In such cells, concentration gradient arise in electrolyte solutions. Cell process may be given as, Zn(s) → Zn2+(C1) + 2e (Anodic process) 2+ Zn (C2) + 2e → Zn(s) (Cathodic process) Zn2+(C2) Zn2+(C1) (Over all process) ∴ From Nernst equation, we have C RT E log F C For spontanity of such cell reaction, C2>C1
10. Commercial Voltaic Cells Batteries can be classified as primary and secondary. Primary batteries can not be returned to their original state by recharging, so when the reactants are consumed, the battery is “dead” and must be discarded. Secondary batteries are often called storage batteries or rechargeable batteries. The reactions in these batteries can be reversed; thus, the batteries can be recharged.
11. Primary Batteries Dry cells and alkaline batteries Zinc serves as the anode, and the cathode is a graphite rod placed down the center of the device. These cells are often called “dry cells“ because there is no visible liquid phase. However, water is present, so the cell contains a moist paste of NH4Cl, ZnCl2 and MnO2.
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Electrochemistry The moisture is necessary because the ions present must be in a medium in which they can migrate from one electrode to the other. The cell generates a potential of 1.5 V using the following half-reactions: Cathode, reductions: 2NH4+(aq) + 2e− → 2NH3(g) + H2(g) Anode, Oxidation: Zn(s) → Zn2+(aq) + 2e− The two gases formed at the cathode will build up pressure and could cause the cell to rupture. This problem is avoided, however, by two other reactions that take place in the cell. Ammonia molecules bind to Zn2+ ions, and hydrogen gas is oxidized by MnO2 to water. Zn2+(aq) + 2NH3(g) + 2Cl−(aq) → Zn(NH3)2 Cl2(s) 2MnO2(s) + H2(g) → Mn2O3(s) + H2O(l)
12. Secondary Or Rechargeable Batteries An automobile battery − the lead storage battery − is probably the best − known rechargeable battery figure. The 12-V version of this battery contains six voltaic cells, each generating about 2V. The lead storage battery can produce a large initial current, an essential feature when starting an automobile engine. When the cell supplies electrical energy, the lead anode is oxidized to lead (II) sulfate, an insoluble substance that adheres to the electrode surface. The two electrons produced per lead atom move through the external circuit to the cathode, where PbO2 is reduced to Pb2+ ions that, in the presence of H2SO4, also form lead (II) sulfate. Cathode, reduction: PbO2(s) + 4H+(aq) + SO42−(aq) + 2e− → PbSO4(s) + 2H2O(l) Anode, oxidation: Pb(s) + SO42−(aq) → PbSO2(s) + 2e− Net cell reaction Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) The discharging process of the storage cell is based on the principles of electrochemical cell, whereas the charging process is based upon the principles of electrolytic cells.
13. Fuel Cells and Hybrid Cars An advantage of voltaic cells is that they are small and portable, but their size is also a limitation. The amount of electric current produced is limited by the quantity of reagents contained in the cell. When one of the reactants is completely consumed, the cell will no linger generate a current. Fuel cells avoid this limitation because the reactants (fuel and oxidant) can be supplied continuously toe the cell from an external reservoir. In a Hydrogen – Oxygen fuel cell figure, hydrogen is pumped into the anode of the cell, and O2 (or air) is directed to the cathode where the following reactions occur : Cathode, reduction : O2(g) + 2H2O(l) + 4 e− → 4OH−(aq) Anode, Oxidation : H2(g) → 2H+(aq) + 2 e−
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Electrochemistry
14. Corrosion Cells and Reactions The special characteristic of most corrosion processes is that the oxidation and reduction steps occur at separate locations on the metal. This is possible because metals are conductive, so the electrons can flow through the metal from the anodic to the cathodic regions. The presence of water is necessary in order to transport ions to and from the metal, but a thin film of adsorbed moisture can be sufficient. A corrosion system can be regarded as a short-circuited electrochemical cell in which the anodic process is something like Fe(s) → Fe2+(aq) + 2 e− and the cathodic steps can be any of O2 + 2 H2O + 4e− → 4 OH− H+ + e− → 1/2 H2(g) M2+ + 2 e− → M(s) where M is a metal. Which parts of the metal serve as anodes and cathodes can depend on many factors, as can be seen from the irregular corrosion patterns that are commonly observed. Atoms in regions that have undergone stress, as might be produced by forming or machining, often tend to have higher free energies, and thus tend to become anodic.
15. Electrolysis The decomposition of electrolyte solution by passage of electric current, resulting into deposition of metals or liberation of gases at electrodes is known as electrolysis.
16. Electrolytic Cell This cell converts electrical energy into chemical energy. The entire assembly except that of the external battery is known as the electrolytic cell.
17. Electrolysis of Molten Sodium Chloride Reaction at
NaCl (molten) → Na+ + Cl− anode (oxidation) : − − 2Cl → Cl2(g) + 2e :
cathode (reduction) 2Na+ + 2e− → 2Na(l)
There are two types of electrodes used in the electrolytic cell, namely attackable and non – attackable. The attackable electrodes participitate in the electrode reaction. They are made up of reactive metals like Zn, Cu, Ag etc. In such electrodes, atom of the metals gets oxidised into the corresponding cation, which is passed into the solution. Thus, such anodes get dissolved and their mass decreases. On the other hand, non-attackable electrodes do not participate in the electrode reaction as they made up of unreactive elements like Pt, graphite etc. Such electrodes do not dissolve and their mass remain same.
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Electrochemistry
18. Electrolysis of Sodium Chloride Solutions When you electrolyze an aqueous solution of sodium chloride, NaCl, the possible species involved in half-reactions are Na+, Cl− and H2O. The possible cathode half-reactions are Na+(aq) + e− → Na(s) Eo = −2.71V − − 2H2O(l) + 2e → H2(g) + 2OH (aq) Eo = −0.83V Under standard conditions, you expect H2O to be reduced in preference to Na+, which agrees with what you observe. Hydrogen is evolved at the cathode. 2Cl−(aq) → Cl2(g) + 2e− Eo = −1.36V 2H2O(l) → O2(g) + 4H+(aq) + 4e Eo = −1.23V Under standard-state conditions, you might expect H2O to be oxidized in preference to Cl−. However, the potentials are close and over voltages at the electrodes could alter this conclusion. It is possible nevertheless to give a general statement about the product expected at the anode. Electrode potentials, as you have seen, depend on concentrations. It turns out that when the solution is concentrated enough in Cl−, Cl2 is the product; but in dilute solution, O2 is the product. To see this, you would simply apply the Nernst equation of the Cl− |Cl2 half reaction. 2Cl−(aq) → Cl2(g) + 2e− Starting with very dilute very negative, so H2O concentration, you would is oxidized in preference
NaCl solutions, you would find that the oxidation potential of Cl− is is reduced in preference to Cl−. But as you increase the NaCl find that the oxidation potential of Cl− increases until eventually Cl− to H2O. The product charges from O2 to Cl2.
The half-reactions and cell reaction for the electrolysis of aqueous sodium chloride to chlorine and hydroxide ion are as follows: 2H2O(l) + 2e− → H2(g) + 2OH−(aq) (cathode) 2Cl−(aq) → Cl2(g) + 2e− 2H2O(l) + 2Cl−(aq) → H2(g) + Cl2(g) + 2OH−(aq) Because the electrolysis started with sodium chloride, the cation in the electrolyte solution is Na+. When you evaporate the electrolyte solution at the cathode, you obtain sodium hydroxide NaOH.
19. Faraday’s Laws of Electrolysis (i)
First law of electrolysis: Amount of substance deposited or liberated at an electrode is directly proportional to amount of charge passed (utilized) through the solution. w∝Q W = weight liberated, Q = charge in coulomb w = ZQ Z = electrochemical equivalent when Q = 1 coulomb, then w = Z Thus, weight deposited by 1 coulomb charge is called electrochemical equivalent. Let 1 ampere current is passed till ’t’ seconds. Then, Q = It ∴w = ZIt
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Electrochemistry
(ii)
1 Faraday = 96500 coulomb = charge of one mole electrons One faraday is the charge required to liberate or deposit one gm equivalent of a substance at corresponding electrode. Second law of electrolysis: When same amount of charge is passed through different electrolyte solutions connected in series then weight of substance deposited or dissolved at anode or cathode are in ratio of their equivalent weights. i.e. w1/w2 = E1/E2
20. Conductance Both metallic and electrolytic conductors obey Ohm’s law i.e. V = IR where V = Potential difference in volt; I = Current in ampere; R = resistance in Ohm We know, resistance is directly proportional to length of conductor and inversely proportional to cross sectional area of the conductor. R∝
l or A
R = ρ
l A
(ρ = Specific resistance)
Specific resistance is the resistance of a conductor having lengths of 1 cm and cross sectional area of 1 cm2. Reciprocal of resistance is called as conductance and reciprocal of specific resistance is called as specific conductance. 1 1 A = R ρ l
or
C = K
A l
where C = conductance ohm−1; K = specific conductance ohm−1 cm−1. Mho and siemens are other units of conductance l C A
K =
Specific conductance = Cell constant × Conductance k = G* × G
21. Equivalent Conductance Equivalent conductance is the conductance of an electrolyte solution containing 1 gm equivalent of electrolyte. it is denoted by ∧. ∧ = Kx V (∧ = ohm−1 cm−1x cm3 = ohm−1 cm2) Usually concern ration of electrolyte solution is expressed as C gm equivalent per litre. Thus,
V =
1000 C
{Volume having 1 gm equivalent electrolyte in the solution} Thus, ∧ = K×
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1000 C
Electrochemistry
22. Molar Conductance Molar conductance may be defined as conductance of an electrolyte solution having 1 gm mole electrolyte in a litre. It is denoted by ∧m. ∧m = K × V Usually concentration of electrolyte solution is expressed as ‘M’ gm mole electrolyte per litre. Thus,
V =
Hence,
1000 M ∧m = Kx
1000 M
Relation between ∧ and ∧m :
∧m = n × ∧
23. Determination of ∧m0 or ∧0 C as found experimentally is as shown below graphically. A plot of ∧m vs
C plot of strong electrolyte being linear it can be extrapolated to zero The ∧m vs concentration. Thus, ∧m values of the solution of the test electrolyte are determined at various concentrations the concentrations should be as low as good. C when a straight line is obtained. This is the ∧m values are then plotted against extrapolated to zero concentration. The point where the straight line intersects ∧m axis is ∧m0 of the strong electrolyte. However, the plot in the case weak electrolyte being non linear, shooting up suddenly at some low concentration and assuming the shape of a straight line parallel to ∧m axis. Hence extrapolation in this case is not possible. Thus, ∧0 of a weak electrolyte cannot be determined experimentally.
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Electrochemistry
24. Kohlrausch’s Law of Independent Migration of Ions Kohlrausch determined ∧0 values of pairs of some strong electrolytes containing same cation say KF and KCl, NaF and NaCl etc. and found that the difference in ∧0 values in each case remains the same: ∧m0 (KCl) − ∧m0 (KF) = ∧m0 (NaCl) − ∧m0 (NaF) He also determined ∧0 values of pairs of strong electrolytes containing same anion say KF and NaF, KCl and NaCl etc. and found that the difference in ∧0 values in each case remains the same. ∧m0 (KF) − ∧m0 (NaF) = ∧m0 (KCl) − ∧m0 (NaCl) This experimental data led him to formulate the following law called Kohlrausch’s law of independent migration of ions. At infinite dilution when dissociation is complete, every ion makes some definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion which with it is associated and that the molar conductance at infinite dilution for any electrolyte is given by the sum of the contribution of the two ions. Thus, ∧m0 =
25. Application of Kohlrausch’s Law (i)
Determination of ∧m0 of a weak electrolyte: In order to calculate ∧m0 of a weak electrolyte say CH3COOH, we determine experimentally ∧m0 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. ∧m0 of CH3COOH is then given as: ∧m0 (CH3COOH) = ∧m0 (HCl) + ∧m0 (CH3COONa) − ∧m0 (NaCl) Proof: ∧m0 (HCl) = H Cl ..........(i) ∧m0 (CH3COONa) = CH COO Na
∧m0 (NaCl) = Na Cl
..........(ii) ..........(iii)
Adding equation (i) and equation (ii) and subtracting (iii) from them: ∧m HCl ∧CH COONa ∧NaCl H CH COO CH COOH
(ii)
Determination of degree of dissociation (α) : α =
No. of molecules ionised total number of molecules dissolved
=
∧m ∧m0
(iii) Determination of solubility of sparingly soluble salt The specific conductivity of a saturated solution of the test electrolyte (sparingly soluble) made in conductivity water is determined by the method as described above. Form this the specific conductivity of conductivity water is deducted.
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Electrochemistry The molar conductance of the saturated solution is taken to be equal to ∧m0 as the saturated solution of a sparingly soluble salt is extremely dilute. Hence from equation (4). ∧m0 =
1000k C
where C is the molarity of solution and hence the solubility.
26. Electrochemical Series
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Electrochemistry
NEET Pattern Exercise (1) 1. A standard hydrogen electrode has zero electrode potential because (a) hydrogen is easier to oxidise (b) this electrode potential is assumed to be zero (c) hydrogen atom has only one electron (d) hydrogen is the lighest element 2. Faraday’s laws of electrolysis are related to the (a) atomic number of the cation (b) atomic number of the anion (c) equivalent weight of the electrolyte (d) speed of the cation 3. If x is specific resistance of the electrolyte solution and y is the molarity of the solution, then ∧ is given by (a) (b) (c) (d) 4. During electrolysis of a concentrated aqueous solution of NaCl, the product at the cathode is (a) Na (b) Cl2 (c) O2 (d) H2 5. Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 u) (a) 0 g (b) 63.5 g (c) 2 g (d) 127 g 6. Two Pt electrodes fitted in a conductance cell are 1.5 cm apart and the cross-sectional area of each electrode is 0.75 cm2. The cell constant is (a) 1.25 (b) 0.5 cm (c) 2.0 cm−1 (d) 0.2 cm−1
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Electrochemistry 7. The charge required for the oxidation of one mole Mn3O4 into MnO42− in presence of alkaline medium is (a) 5×96500 C (b) 95600 C (c) 10×96500 C (d) 2×96500 C 8. The standard reduction potential values of three metallic cations, X, Y, Z are 0.52, −3.03 and −1.18 V respectively. The order of reducing power of the corresponding metals is (a) Y > Z > X (b) X > Y > Z (c) Z > Y > X (d) Z > X > Y 9. The standard reduction potentials 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 10. The (a) (b) (c) (d)
electric charge for electrode deposition of one gram equivalent of a substance is one ampere per second 96,500 coulombs per second one ampere for one hour charge on one mole of electrons
ANSWER Q1 (b)
Q2 (c)
Q3 (c)
Q4 (d)
Q5 (b)
Q6
Q7
Q8
Q9
Q10
(c)
(c)
(a)
(c)
(d)
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