Electrochemistry Mock Test

MOCK TEST SECTION - I : STRAIGHT OBJECTIVE TYPE 1. 0.45 V ClO- ClO3- 1 Cl 2 2 1.07 V Cl- 0.71 V E0 The E0 in the given diagram is, (A) 0.5 2. (B) 0.6 (C) 0.7 (D) 0.8 What is cell entropy change of the following cell ? Pt(s) | H2(g) | CH3COOH, HCl | |KCl (aq) | Hg2Cl2 (s) | Hg P = 1 atm 0.1 M 0.1 M Emf of the cell ius found to be 0.045 V at 298 K and temperature coefficient is 3.4 × 10 –4 V K–1 5 Given K a (CH3COOH)  10 M (A) 60 3. (B) 65.2 (C) 69.2 (D) 63.5 Following cell has EMF 0.7995 V. Pt | H2 (1 atm) | HNO3 (1M) | |AgNO3 (1M) | Ag If we add enough KCl to the Ag cell so that the final Cl– is 1M. Now the measured emf of the cell is 0.222 V. The Ksp of AgCl would be : (A) 1 × 10–9.8 4. (B) 1 × 10–19.6 The solubility of [Co(NH3)4Cl2] ClO4 if the  CO(NH (C) 2 × 10–10 3 ) 4 Cl2  (D) 2.64 × 10–14  50 ,  ClO   70 and the measured resistance was 33.5  in 4 a cell with cell constant of 0.20 is __________ (A) 59.7 mmol/L 5. (B) 49.7 mmol/L (C) 39.7 mmol/L (D) 29.7 mmol/L At 298K the standard free energy of formation of H2O() is –237.20 kJ/mole while that of its ionisation into H+ ion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be (Take F = 96500 C] H2 (g, 1 bar) | H+ (1M)| |OH– (1M) | O2 (g, 1 bar) (A) 0.40 V 6. (B) 0.81 V (C) 1.23 V (D) –0.40 V A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio a : b and another hydrogen electrode Y was placed in a buffer solution of sodium acetate and acetic acid in the ratio b : a. If reduction potential values for two cells are found to be E1 and E2 respectively w.r.t. standard hydrogen electrode, the pKa value of the acid can be given as (A) E1  E 2 0.118 (B)  E1  E 2 0.118 (C ) E1  0.118 E2 (D) E 2  E1 0.118 ELECTROCHEMISTRY 7. 8. Which of the following cell can produce more electrical work. (A) pt, H2 |NH4Cl| | 0.1 M CH3COOH | H2, pt (B) pt, H2 |0.1 M HCl| | 0.1 M NaOH | H2, pt (C) pt, H2 |0.1 M HCl| | 0.1 M CH3COOK | H2, pt (D) pt, H2 | 0.1 M CH3COOK | |0.1 M HCl| H2, pt A hydrogen electrode is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution. (Take p H2  1 atm ) T = 298 K. 9. (A) increase by 0.41 V (B) increase by 59 mV (C) decrease by 0.41 V (D) decrease by 59 mV Calculate the EMF of the cell at 298 K pt, H2 |(1 atm)| NaOH (xM), NaCl (xM)| AgCl (s)| Ag 0 If E Cl / AgCl/Ag  0.222V (A) 1.048 V (B) –0.04 V (C) –0.604 V (D) emf depends on x and cannot be determined unless value of x is given 10. A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition. (Cu – 63.5) (A) 79% 11. (B) 39.5% (C) 63.25% (D) 63.5% A flashlight cell has the cathodic reaction 2MnO2 (s) + Zn+2 + 2e– Zn Mn2 O4(s) If the flashlight cell is to give out 4.825 mA, how long could it run if initially 8.7 g of the limiting reagent MnO 2 is present ? [Mn = 55, O = 16] (A) 2 × 106 sec 12. (C) 6 × 106 sec (D) 8 × 106 sec A resistance of 50 is registered when two electrodes are suspended into a beaker containing a dilute solution of a strong electrolyte such that exactly half of the them are submerged into solution. If the solution is diluted by adding pure water (negligible conductivity) so as to just completely submerge the electrodes, the new resistance offered by the solution would be (A) 50  13. (B) 4 × 106 sec (B) 100  (C) 25  (D) 25  The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is (A) 0.5 mmol (B) 1.0 mmol (C) 2.0 mmol (D) 2.5 mmol 14. It is observed that the voltage of a galvanic cell using the reaction M(s)  xH    Mx  x H 2 varies linearly with 2 the log of the square root of the hydrogen pressure and the cube root of the Mx+ concentration. The value of x is 15. Acetic acid has Ka = 1.8 × 10–5 while formic acid has Ka = 2.1 × 10–4 . What would be the magnitude of the emf of the cell Pt(H 2 ) 0.1 M acetic acid + (A) 0.0315 volt 16. (B) 0.0629 volt 18. Pt(H 2 ) at 25° C ? (C) 0.0455 volt (D) 0.545 volt Consider the cell Ag(s)|AgBr(s)|Br– (aq)| | AgCl(s) | Cl– (aq) | Ag (s) at 25° C. The solubility product constants of AgBr & AgCl are respectively 5 × 10–13 & 1 × 10–10 . For what ratio of the concentrations of Br– & Cl– ions would the emf of the cell be zero ? (A) 1 : 200 17. 0.1 M formic acid + 0.1M sodium acetate 0.1M sodium formate (B) 1 : 100 (C) 1 : 500 (D) 200 : 1 Value of  m  for SrCl2 in water at 25°C from the following data : Conc. (mol/lt) 0.25 1  m ( 1cm 2 mol1 ) 260 250 (A) 270 (B) 260 (C) 250 (D) 255 Which of these ions Cu+, Co3+, Fe2+ is stable in aqueous medium. 0 0 0 0 Given : E Cu 2 /Cu   0.15 volt ; E Cu  /Cu  0.53 V ; E Co3 /Co2  1.82 V ; E Fe3 /Fe2  0.77 V 0 E 0Fe2 /Fe  0.44 V ; E O2 ,H  /H 2O  1.23 V (A) Cu+ , Co3+ 19. (B) Co3+ (C) Cu+ (D) Co3+, Cu+ , Fe2+ What is the value of pKb ( CH3COO–) if  m  390 & m = 7.8 for 0.04 of a CH3COOH at 25°C (A) 9.3 (B) 9.2 (C) 4.7 (D) 4.8 SECTION - II : MULTIPLE CORRECT ANSWER TYPE 20. 0 In which the following cell (s) : E cell  E cell ? 2  (A) Cu(s) Cu (0.01M Ag (0.1M) Ag(s) 2 (B) Pt(H 2 ) pH  1 Zn (0.01M) Zn(s) 2 (C) Pt(H 2 ) pH  1 Zn (1M) Zn(s)  2 (D) Pt(H 2 ) H  00.1M Zn (0.01M Zn(s) ELECTROCHEMISTRY 21. Which one is/are correct among the followings ? 0 0 Given, the half cell emf’s E Cu 2 |Cu  0.337, E Cu 1|Cu  0.521 (A) Cu+1 disproportionates (B) Cu and Cu2+ comproportionates (reverse of disproportionation into Cu+). 0 0 (C) E Cu|Cu 2  E Cu 1|Cu is positive (D) All of these 22. For the cell (at 298 K) Ag(s) |AgCl(s) | Cl– (aq)| |AgNO3 (aq) | Ag(s) Which of the following is correct (A) The cell emf will be zero when [Ag+]a = [Ag+]c ([Ag+] in anodic compartment = [Ag+] in cathodic compartment) (B) The amount of AgCl(s) precipitate in anodic compartment will decrease with the working of the cell. (C) The concentration of [Ag+] = constant, in anodic compartment during working of cell. 0 0 (D) E cell  E Ag |Ag  E Cl |AgCl|Ag  23. 0.059 1 log  1 [Cl ]a The standard redox potentials E0 of the following systems are System E°(volts) (i) MnO 4  8H   5e    Mn 2  4H 2 O 1.51 (ii) Sn 2   Sn 4  2e –0.15 (iii) Cr2 O 72  14H   6e    2Cr 3  7H 2 O 1.33 (iv) Ce3   Ce4  e –1.61 The oxidising power of the various species are related as 2  (A) Cr2 O 7  MnO 4 24. (B) Ce 4  Sn 4 (C) Ce 4  MnO 4 (D) MnO 4  Sn 4 Acetic acid is titrated with NaOH solution. Which of the following statement is correct for this titration ? (A) conductance increase upto equivalence point, then it decreases (B) conductance decreases upto equivalence point, then it increases (C) first conductance increases slowly upto equivalence point and then increases rapidly (D) first conductance increases slowly upto equivalence point and then drops rapidly 25. During an electrolysis of conc. H2SO4, perdisulphuric acid (H2S2O8) and O2 from in equimolar amount. The amount of H2 that will form simultaneously will be (2H2SO4  H2S2O8 + 2H+ + 2e–) (A) thrice that of O2 in moles (B) twice that of O2 in moles (C) equal to that of O2 in moles (D) Half of that of O2 in moles SECTION - III : ASSERTION AND REASON TYPE 26. Statement - 1 : Zinc protect the iron better than tin even after it cracks. Statement - 2 : E 0OP Zn  E 0OP Fe But E 0OP Sn  E 0OP Fe (A) Statement-1 is True, Statement-2 is True and Statement-2 is correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True and Statement-2 is not correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 27. Statement - 1 : For the reaction : Ni   2e   Ni Fe   2e   Fe  E 0Fe |Fe  E 0Ni  |Ni and E 0red  0  So Fe electrode is cathode and Ni electrode is anode Statement - 2 : Because DG° < 0 and E°Cell > 0, so cell is possible. (A) Statement-1 is True, Statement-2 is True and Statement-2 is correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True and Statement-2 is not correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 28. KMnO4 liberates O2 from water in presence of an acid. Given, E 0MnO 4,Mn 2 ,H  |Pt  1.51V E 0H  |O |Pt  1.223V 2 29. The standard potentials of some metal-insoluble salt-anion electrode are as given below Ag – AgCl E 0red  0.222 Ag – AgBr = 0.03 Ag – AgI = – 0.151 Ag – Ag2S = – 0.69 Then the KSP values must follow : KSP (AgCl) > KSP (AgBr) > KSP (AgIP) > KSP (Ag2S) 30. At 300 K specific conductivity of ethanol is 4 × 10–10 mho cm–1 . The ionic conductances of H+, C2H5O– at this temperature is 300 and 100 mho cm2 equivalent–1 respectively. Then the negative logarithm of ionic product of alcohol will be 18. 31. The process of AgCn + KCN   K[Ag(CN) 2 ] involves the oxidation of Ag. ELECTROCHEMISTRY SECTION - IV : COMPREHENSION TYPE Read the following comprehensions carefully and answer the questions. Comprehension # 1 Strong Acid Versus Strong Base The principle of conductometric titrations is based on the fact that during the titration, one of the ions is replaced by the other and invariably these two ions differ in the ionic conductivity with the result that the conductivity of the solution varies during the course of the titration. Take, for example, the titration between as strong acid, say HCl, and a strong base 2, say NaOH. Before NaOH is added, the conductance added, H+ ions are replaced by relatively slower moving Na+ ions. Consequently, the conductance of the solution decreases and this continues right upto the equivalence point where the solution contains only NaCl. Beyond the equivalence point, if more of NaOH is added, then the solution contains in excess of the fast moving OH– ions with the result that its conductance is increased and it continues to increase as more and more of NaOH is added. If we plot the conductance value versus the amount of NaOH added, we get a curve of the type shown in Fig. conductance A D B E C volume of NaOH The descending portion AB represents the conductances before the equivalence point (solution contains as mixture of acid HCl and the salt NaCl) and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCl and the excess of NaOH). The point E which represents the minimum conductance is due to the solution containing only NaCl with not free acid or alkali and thus represents the equivalence point. This point can, however, be obtained by the extrapolation of the lines AB and DC, and therefore, one is not very particular in locating this point experimentally as it is in the case of ordinary acid-base titrations involving the acid-base indicators. Weak Acid versus Strong Base Let us take the specific example of acetic acid being titrated against NaOH. Before the addition of alkali, the solution shows poor conductance due to feeble ionization of acetic and. Initially the addition of alkali causes not only the replacement of H+ by Na+ but also suppresses the dissociation of acetic acid due to the common ion Ac– and thus the conductance of the solution decreases in the beginning. But very soon the conductance starts increasing as addition of NaOH neutralized the undissociated HAc to Na+Ac+ thus causing the replacement of non-conduction HAc with strong-conducting electrolyte Na+Ac–. The increase in conductance continues right up to the equivalence point. Beyond this point conductance increases more rapidly with the addition of NaOH due to the highly conducting OH– ions. The graph near the equivalence point is curved due the hydrolysis of the salt NaAc. The actual equivalence point can, as usual, be obtained by the extrapolation method. In all these graphs it has been assumed that the volume change due addition of solution from burette is negligible, hence volume change of the solution in beaker the conductance of which is measured is almost constant throughout the measurement. conductance The nature of curve obtained for the titration between weak acid versus strong base as described in the above passage will be conductance 32. (A) (B) conductance volume of NaOH conductance volume of NaOH (C) (D) volume of NaOH conductance conductance The most appropriate titration curve obtained when a mixture of a strong acid (say HCl) and weak acid (say CH3COOH) is titrated with a strong base (say NaOH) will be (A) (B) V1 V2 volume of NaOH V1 + V2 volume of NaOH conductance conductance 33. volume of NaOH (C) (D) V1 V2 volume of NaOH V1 V2 volume of NaOH ELECTROCHEMISTRY 34. If a 100 mL solution of 0.1 M HBr is titrated using a very concentrated solution of NaOH, then the conductivity (specific conductance) of this solution at the equivalence point will be (assume volume change is negligible due to addition of NaOH). Calculate your answer in terms of 10–1 Sm–1. 0 0 [Given  (Na  ) = 8 × 10–3 Sm2 mol–1  (Br  ) = 4 × 10–3 S m–1 mol–1] (A) 6 (B) 12 (C) 15 (D) 24 Comprehension # 2 Redox titrations Titrations are one of the methods we can use to discover the precise concentrations of solution. A typical titration involves adding a solution from a buretic to another solution in a flask. The endpoint of the titration is found by watching a colour change taking place. However, a problem arises when a suitable indicator cannot be found or when the colour changes involved are unclear. In these cases redox potentials may sometimes come to the rescue. A particularly well known example (Fig.) is a method of discovering the concentration of iron (II) ions in a solution 0 by titrating them with a solution of cerium (IV). The redox potentials that are of interest here are E Fe3 /Fe2 = + 0.77 0 V and E Ce4 /Ce3 = + 1.61 V. These tell us that cerium (IV) ions are the oxidising agents, and iron (II) ions are the reducing agents. They should react according to the equation. Fe2+(aq) + Ce4+ (aq)  Fe3+ (aq) + Ce3+ (aq) Now imagine that we know the concentration of the cerium (IV) ions solution in the burette. We want to measure the concentration of the iron (II) solution. If we add just one drop of the cerium (IV) solution from the burette, some of the iron (II) ions will be oxidised. As a consequence the beak would now contain a large number of unreacted iron (II) ions, but also some iron (III) ions as well. All of the cerium (III). The solution in the beaker now represents an iron (III) / iron (II) half cell, although not at standard conditions. Thus the e.m.f. of the cell will be near, but not equal, to E Fe3 /Fe2 . Ions Present Fe After the first few drops of Ce4+ (aq) Near the endpoint Just after 2+ Fe 3+ Ce 3+ Ce 4+ Cell E.m.f/V Many Some Some None Fe /Fe 3+ 2+ Some Many Many None Fe /Fe 3+ 2+ None Many Many Some Fe /Fe 4+ 3+ A little less than 0.77 A little more than 0.77 A little more than 1.61 If we continue to add cerium (IV) solution, the number of iron (II) ions is gradually reduced and eventually only a very few are left (Table). At this stage the next few drops of cerium (IV) solution convert all the remaining iron (II) ions into iron (III) and some of the cerium (IV) ions are left unreacted. Once this happens we no longer have an iron (III) / Iron (II) half-cell. Instead we have a solution in which there is a large number of cerium (III) ions and a smaller number of cerium (IV) ions. The solution in the beaker now behaves as a cerium (IV)/ cerium (III) half-cell (although no a standard one). Just before all the iron (II) ions are converted into iron (III) we have a cell with an e.m.f. of around + 0.77V. After all the iron (II) ions are oxidised, we have a cell with an e.m.f. of about +1.61 V. This rapid rise in e.m.f. occurs with the addition of just one drop of cerium (IV) solution. You should be able to understand why a graph of cell e.m.f. against volume of cerium (IV) solution added looks like that of Fig. The end point of the titration can be read from the graph and the concentration of the iron (II) solution calculated in the usual way 35. When an ion is converted into a complex ion, the redox potential changes. You can see this in the case of the e.m.f. of the iron (III) / iron (II) system (+0.77V) and the hexacyanoferrate (III) /hexacyanoferrate(II) system (+0.36V). The cyanide ion is said to stabilise the oxidation stale of the iron. If you were to make up a cell 3– 4– Pt(s) Fe(CN) 6 (aq.),Fe(CN) 6 (aq), 1mol dm 3 3+ 2+ Fe (aq), Fe (aq), Pt(s) 1mol dm 3 What would be the e.m.f. and what would be the cell reaction? (A) 0.41 V, Fe(CN)64–(aq.) + Fe3+(aq.)   Fe(CN)63– (aq.) + Fe2+(aq.) (B) 0.13 V, Fe(CN)64–(aq.) + Fe3+(aq.)   Fe(CN)63– (aq.) + Fe2+(aq.) (C) 0.41 V, Fe(CN)63–(aq.) + Fe2+(aq.)   Fe(CN)64– (aq.) + Fe3+(aq.) (D) 0.13 V, Fe(CN)63–(aq.) + Fe2+(aq.)   Fe(CN)64– (aq.) + Fe3+(aq.) ELECTROCHEMISTRY 36. The cell shown below was set up What would be the cell e.m.f. ? If potassium cyanide solution were added to the left hand half cell (with due care!), what would you expect to happen to the e.m.f. of the cell? E 0Br2 Br = 1.07 V and use data of previous question, if required. 37. (A) 0.30 V, emf will increase from 0.30 V to 0.41 V (B) 1.84 V, emf will decrease from 1.84 V to 1.43 V (C) 0.30 V, emf will increase from 0.30 V to 0.71 V (D) 0.30 V, emf will increase from 0.30 V to 0.43 V Imagine you were given a solution of potassium dichromate (VI) in a beaker and a solution of iron (II) sulphate in a burette. You do not know the concentration of dichromate (VI) ions but the concentration of the iron (II) solution is known. You take is to carry out a redox titration using the two solutions in order to determine the above titration. E 0Cr O2 /Cr3 = 1.33 V, E 0Fe3 /Fe2 = 0.77 V. 2 7 (A) (C) (B) (D) Comprehension # 3 A fuel cell is a cell that is continuously supplied with an oxidant and a reductant so that it can deliver a current indefinitely. Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work. Internal combustion engines at best convert only the fraction (T 2 – T1)/T2 of the heat G , H where G is the Gibbs energy change for the cell reaction and H is the enthalpy change of the cell reaction. This efficiency can be upto 80% also in contrast to normal heat engine efficiencies which are generally about 40%. of combustion into mechanical work. While the thermodynamic efficiency of the fuel cell is given by,   Fuel cells may be classified according to the temperature range in which they operate : low temperature (25 – 100°C), medium temperature (100 to 500°C), high temperature (500 to 1000°C) and very high temperature (above 1000°C). The advantage of using high temperature is that catalysts for the various steps in the process are not so necessary. Polarization of a fuel cell reduces the current. Polarization is the result of slow reaction or processes such as diffusion in the cell. The figure, indicates the construction of a hydrogen-oxygen fuel cell with a solid electrolyte, which is an ion exchange membrane. The membranes is impermeable to the reactant gases, but is permeable to hydrogen ions, which carry the current between the electrodes. To facilitate the operation of the cell at 40 to 60°C, the electrodes are covered with finely divided platinum that function as a catalyst. Water is drained out of the cell during operation. Fuel cells of this general type have been used successfully in the space program and are quite efficient. Their disadvantages for large-scale commercial application are that hydrogen presents storage problems and platinum is an expensive catalyst. Cheaper catalysts have been found for higher temperature operation of hydrogen-oxygen fuel cells. Fuel cells that use hydrocarbons and air have been developed, but their power per unit weight is too low to make them practical in ordinary automobiles. Batter catalysts are needed. A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte. The half-cell reactions are 1 O (g) + 2H+ + 2e– = H2O() 2 2 2H+ + 2e– = H2(g) H2(g) + or 1 O (g) = H2O() 2 2 E° = 1.2288 V E° = 0 E° = 1.2288 V ELECTROCHEMISTRY 1 O (g) + H2O() + 2e– = 2OH– 2 2 E° = 0.4009 V 2H2O() + 2e– = H2(g) + 2OH– H2(g) + E° = – 0.8279 V 1 O (g) = H2O () 2 2 E° = 1.2288 V To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized. Since fused salts have lower electrolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications. High temperature also allow the use of liquid metal electrode, which make possible higher current densities than solid electrodes. 38. If 560 mL of H2 gas at STP is fed into and is consumed by the fuel cell in 10 minutes, then what is the current output of the fuel cell? (A) 4 A 39. (C) 16 A (D) 12 A For a Hydrogen – Oxygen fuel cell if H (H2O,) = 285 kJ/mol, then what will be thermodynamic efficiency under standard conditions (use data given in the passage if required) 0 f (A) 0.91 .40. (B) 8 A (B) 0.41 (C) 0.63 (D) 0.83 Why are fuel cells not being used in daily life despite their very high efficiency. (A) The fuels needed for their operation are rarely found in nature (B) Their operation always needs very high temperature (C) The catalyst used in fuel cells at normal temperature are very expensive (D) The design of fuel cells is so complicated that it is not feasible to construct these for daily applications. 41. What is the substance X taken out of the bottom of the fuel cell shown in the figure. (A) H2O 42. (B) H2O2 (C) Both (A) and (B) (D) neither The advantage of using fuel cell in a motorcar could be (A) emissionless operation (B) light weight (C) freedom from refueling (D) low initial cost Comprehension # 4 Metallurgy of copper Electrolytic purification of copper is required since the product of the Bessemer converter is impure. Sheets of pure copper are made the cathodes and blister copper is used as the anodes. The bath contains CuSO4, H2SO4 and NaCl Metallic copper goes into solution as Cu+2 ions at the anode and deposit at the cathode. Zinc and iron go into solution and do not deposit at the cathode along with copper. Silver goes into solution as Ag+ ions, but forms a sludge in the electrolysis cell. The noble metals such as gold and platinum do not dissolve, they fall to the bottom when released from the copper anode. They help to pay for the copper refining process. 43. In what form will silver be recovered from the plating bath? (A) Ag+ (B) AgCl (C) Ag (D) Ag – Cu alloy 44. The different chemical behaviour of various metal ions (for example only Cu is deposited on the cathode) involved are explainable best using 45. (A) their amounts in impure copper (B) the electrochemical series (C) their ionization potentials (D) their hydration energies The voltage is kept low enough during the operation. This is to ensure that (A) Fe+2 and Zn+2 ions are kept in solution only (B) there is no risk of electric shock to the plant operators (C) gold and platinum do not dissolve in solution and are recovered as solids. (D) temperature control is maintained to effect reaction rate control 46. The most important role of various electrolytes added to the electrorefining bath in this case is (A) to reduce the temperature of bath (B) to result in good electrical connectivity so as to have high production rates (C) so that the anode acts as an attackable electrode (D) to enhance the viscosity of electrolyte thus preventing leakage from bath. SECTION - V : MATRIX - MATCH TYPE 47. Match the column Column I Column II (A) Zn / Zn+2 | |Mg2+ / Mg (p) E0Cell = 0 c1 c2 (c1 = c2) (B)Zn / Zn+2 | | Ag+ / Ag (q) E0Cell  0 at. equilibrium (C) Ag / Ag+ | |Ag+ / Ag c1 c2 (c1 = c2) (D) Fe / Fe+2 | |Ag / Ag+ c1 48. (r) E0Cell = +ve c2 (s) E0Cell = 0 (c1 = c2) Match the following (1 to 1 matching) : Column I Column II O (A) Concentration cell (p) OH + +2H + 2e O (B) Electrode reversible with respect to anion (q) Pt |Ce+4 , Ce+3 (C) Quin hydrone electrode (r) Hg | Hg2Cl2, KCl (D) Redox-electrode (s) Na(Hg) | NaCl | Na (Hg) OH ELECTROCHEMISTRY SECTION - VI : SUBJECTIVE TYPE 49. The E°cell = 1.18 V for Zn(s) | Zn+2 (1M) || Cu+2 (1M) |Cu(s)|. Determine the value of x if when excess granulated zinc is added to 1 M Cu+2 solution the [Cu+2]eq. becomes 10–x M. (T = 298 K, 50. 2.303RT = 0.059) F For the cell in opposition, Zn(s) | ZnCl2(sol.) | AgCl(s) | Ag | AgCl(s) | ZnCl2 (sol.) | Zn(s) | C1 = 0.02 M C2 = 0.5 M Find out the emf (in millivolt) of the resultant cell? (Take log 2 = 0.3, 51. RT at = 298 K = 0.060) F The following two cells with initial concentration as given are connected in parallel with each other. (1) Fe(s) | Fe(NO3)2 (aq.) (1 M ) || SnCl2(1M) | Sn(s) (2) Zn(s) |ZnSO4(aq.) (1M) || Fe(NO3)2(aq.) (1M) | Fe(s) After sufficient time equilibrium is established in the circuit. What will be the concentrations (in mmoles/L) of Fe 2+ ions in first and second cells respectively. 0 0 0 [Take E sn 2 /Sn = – 0.14 V, E Zn 2 /Zn = – 0.76V,, E Fe 2 /Fe = –0.44 V, 2.3 × RT = 6433, log2 = 0.3] 52. For the cell (at 1 bar H2 pressure) Pt/H2(g) H X (m1), NaX(m2), NaCl(m3) / AgCl / Ag / Pt it is found that the value of  m HX .m cl   approaches 0.2814 in the limit of zero concentration. E – E° + RTF–1 n  m  x  Calculate Ka for the acid HX at 25°C expressing your answer as 107 Ka. 53. The standard EMF of the cell Pt | H2(g) | H+ (A) || OH– (aq) | O2 (g) | Pt is 0.3900 V at 27°C and 0.4000 V at 7°C. If entropy change during formation of H2O() from H2(g) and O2(g) is – 193 J/K, then find the S° (magnitude only, in mJ/K) for  H2O (). H+ (aq) + OH– (aq)  54. Calculate the emf of the cell in mV (atleast in fist two digit must match with correct answer) (giving your answer in magnitude only) Ag (s), AgIO3 (s) | Ag+ (xM), HIO3 (1M)| | Zn+2 (1M)| Zn(s) If Kap = 3 × 10–8 for AgIO3 and Ka = (log 3 = 0.48) (Take RT = 0.059) F 1  2Ag+ + Zn is – 1.56 V. for HIO3 and E 0cell for 2Ag + Zn+2  6 ELECTROCHEMISTRY MOCK - TEST 1. B 2. B 3. A 4. 14. B 15. B 16. A 26. C 27. D 28. True 38. B 39. D 40. C B 17. A 5. A 18. B 29. True 41. A 48. As ; B r ; C p ; D q 42. A 6. B 19. B 7. 8. C 9. A 20. A, B, D 21. A, C 30. True 43. B D 31. False 44. B 45. A 10. A 11 A 12. A 13. B 22. A 23. B, C, D 24. C 35. A 25. A 32. A 33. C 34. B 36. C 37. B 46. B 47. Aq,s ; B p ; C p ; D q,r