Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

Size: px

Start display at page:

Download "Test Venue: Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh"

Clifton Bradley
5 years ago
Views:

1 Test Venue: Lajpat hawan, Madhya Marg, Sector 15-, handigarh Dr. Sangeeta Khanna Ph.D 1 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

2 RED THE INSTRUTINS REFULLY 1. The test duration is of hour.. The maximum marks are There are total 51 questions. 4. Fill your particulars correctly and completely on the MR NSWERSHEET. SETIN (Single orrect hoice Type) Negative Marking [-1 This Section contains 3 multiple choice questions. Each question has four choices ), ), ) and D) out of which NLY NE is correct. 3 4 = 1 Marks 1. The cell reaction for the given cell is spontaneous if Pt, (1 M) (1 M) Pt, l (P 1 ) (P ) a. P 1 > P b. P 1 < P c. P 1 = P d. P = 1 atm. P Sol..59 P 1 P Ecell E oxi E red log Pt, l l l l, Pt P1 =.59 P E cell log P1 Now E cell will be + ve if P > P 1. The hydrogen electrode is dipped in a solution of ph = 3 at 5. The potential of the cell would be (the value of.33 RT/F is.59 V) a..177 V b..87 V c V d..59 V Sol. H + + e 1 H.59 1 E = E - log n [H.59 = - ph = = V 1 3. If the pressure of hydrogen gas is increased from 1 atm. to 1 atm., keeping the hydrogen ion concentration constant at 1 M, the voltage of the hydrogen half cell at 5 will be a..59 V b V c..95 V d..118 V Sol. H + + e 1 H, 1/.59 p H E = E - log 1.6 (1) 1/ = - log = -.59 V n [H E for the cell Zn Zn + (aq) u + (aq) u is 1.1 V at 5. The equilibrium constant for the cell reaction Zn + u + (aq) u + Zn + (aq) is of the order of a b c d Sol Ecell log Keq or log K eq = n.59 = 37.8 K eq = antilog 37.8 = Dr. Sangeeta Khanna Ph.D HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

3 5. n aqueous solution of conc. H S 4 is electrolysed using platinum electrodes. The products at the anode and cathode are a., H b. S 8, Na c., Na d. S 8, H D 6. hoose the incorrect statement: a. reaction is spontaneous from left to right if K eq > q in which case E cell > b. reaction occurs from right to left if K eq < q, in which case E cell < c. If the system is at equilibrium no net reaction occurs d. E cell is temperature independent D 7. Given E values for some reactions as (i) I + e I ; E =.54 V (ii) Mn 4 + 8H+ + 5e Mn + + 4H ; E = 1.5 V (iii) Fe 3+ + e Fe + ; E =.77 V (iv) Sn 4+ + e Sn + ; E =.1 V The strongest reductant and oxidant respectively are a. Sn +, Mn 4 b. Mn 4, Sn4+ c. I, Fe 3+ d. I, Sn + Sol. Larger the value of E red larger is tendency for reduction and consequently stronger will be the oxidant. Similarly, smaller the value of E red larger the tendency for oxidation and consequently stronger will be the reductant. 8. In the electrolysis of copper chloride solution using copper electrodes, the mass of cathode increases by 3.5 g. In the anode, a..5 mol of u will go into solution as u + ions b. 56 ml of at S.T.P. will be liberated c. 11 ml of l will liberate d. 3. mol of copper metal will dissolve Sol. ul u + + l opper deposited = =.5 mol The u anode will give same amt. of u + ions in solution. 9. Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume of the two gases (dry and at STP) produced will be approximately (in litres) a..4 b c. 67. d Sol. 1 H H + or H + + e H i.e., moles of e produce 1 mole of H i.e.,.4 L. Hence 4 moles of e will give H = 44.8 L. produced = 1 of H =.4 L Total volume = = 67. L 1. Red hot carbon will remove oxygen from the oxide and but not from M, while will remove oxygen from. The activity of metals, and M in decreasing order is: a. > > M b. > > M c. M > > d. M > > Dr. Sangeeta Khanna Ph.D 3 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

4 onductance onductance onductance onductance Dr. Sangeeta Khanna Ph.D Sol. removes oxygen from and, is better reducing agent than and ; remove oxygen from, is better reducing agent than. etter reducing means more active. Hence the correct order is M > > >. 11. Point out the correct statement: 1. Electrolysis of aqueous solution of Lil shows ph > 7. xidation number of in carbon suboxide is +/3 3. xidation number of P in sodium hypophosphite is The oxidation number and covalency of oxygen in molecule are and a. 1, 3 b. 1,, 4 c. 1,, 3, 4 d. 1, 3, 4 D x Sol. arbon suboxide is 3 3x 4 = 3x = +4 or x = +4/3 1. onductometric titration curve of a equimolar mixture of a Hl and HN with NaH (aq) is: a. b. Vol of NaH Vol of NaH c. d. Vol of NaH Vol of NaH D Sol. Molar conductivity of H + and H are very high as compare to other ions. Initially conductance of solution sharply decreases due to consumption of free H + then increases due to formation of salt (NaN) and fter complete neutralization further sharply increases due to presence of H. 13. onsider the cell given below: u/u + g + /g g + + e g; E = x u + + e u; E = y E of the cell is: a. x y b. x y c. y x d. y x 14. For a cell reaction involving two-electron change, the standard emf of the cell is found to be.95 V at 5. The equilibrium constant of the reaction at 5 will be: a. 1. x 1 1 b. 9.5 x 1 c. 1. d. 1. x 1 1 D.59 Sol. E log K.95 log K 1.59 Dr. Sangeeta Khanna Ph.D 4 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

5 15. Equal quantities of electricity are passed through three voltameters containing FeS 4, Fe (S 4 ) 3 and Fe(N 3 ) 3. onsider the following statements in this regard: 1. The amount of iron deposited in FeS 4 and Fe (S 4 ) 3 is equal. The amount of iron deposited in Fe(N 3 ) 3 is two-third of the amount of iron deposited in FeS 4 3. The amount of iron deposited in Fe (S 4 ) 3 and Fe(N 3 ) 3 is equal a. 1 alone is correct b. 1 and are correct c. and 3 are correct d. 3 alone is correct 16. onsider following electrolytes: 1. us 4. Fe (S 4 ) 3 3. ll 3 4. gn 3 The quantity of electricity needed to electrolyse completely 1 M, 1 lit. solution of these electrolytes will be: a. F, 6F, 3F and 1 F respectively b. 6F, F, 3F and 1F respectively c. F, 6F, 1F and 3F respectively d. 6F, F, 1F and 3F respectively 17. Zn Zn + (a =.1 M)II Fe + (a=.1 M) Fe The emf of the above cell is.95 V. The equilibrium constant for the cell reaction is: a. 1. x 1.3/.591 b. 1. x 1.3/.95 c. 1. x 1.6/.95 d. e.3/ Sol..95 = E - log = E.3 = E.59 E = log K E.3 log K For which of these electrodes will the reduction potential vary with ph (I) Sol. e m / m, (II) m / m 4, ( III ) m 4 / m a. (I) only b. (II) only c. (I) and (II) only d. (I), (II) and (III) m m H e m m H 6 4 [No H + is used e + m +4 m In the diagram given below the value of X is: u V u V u [E will depend on H + a..35 V b..65 V c V d V Sol. G 3 G 1 G = -.15 F.5 F = -.65 F G nfe F = 3 E 3 =.35 V FE 3 E = X volt Dr. Sangeeta Khanna Ph.D 5 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

6 . The ionization constant of a weak electrolyte is 5 x 1 6 while the equivalent conductance of its.1 M solution is 19.6 s cm eq -1. The equivalent conductance of the electrolyte at infinite dilution (in s cm eq 1 ) will be: a. 5 b. 196 c. 39 d. 384 Sol. K = 5 x 1 6 ; eq = 19.6 S cm eq 1 ; = K The ionic conductance of l 3+ and S 4 ions at infinite dilution are x and y ohm -1 cm mol -1 respectively. If Kohlrausch s Law is valid, the molar conductance of aluminium sulphate at infinite dilution will be: a. 3x + y b. 3y + x c. x + y d. 3x + 3y 3 x Sol. l(s 4 ) 3 l 3S. m 3y x 4 y Electrolyte Kl KN 3 Hl Nac Nal (S cm mol 1 ) alculate Hc using appropriate molar conductance of the electrolytes listed above at infinite dilution in H at 5. a b c d Sol. ch cna Hl Nal = = 39.7 Scm mol 1 3. The molar conductivity of acetic acid at infinite dilution is 39.7 and for.1 M acetic acid solution is 5. mho cm mol 1.[H + in solution will be a..133 M b..133 M c. 133 M d M Sol. The degree of dissociation is given by v or 1.33% H 3 H H 3 + H + efore dissociation 1 fter dissociation (1 - ) [H + = =.1 x.133 =.133 M 4..5 N solution of a salt occupying a volume between two platinum electrodes separated by a distance of 1.7 cm and having an area of 4.5 cm has a resistance of 5 ohm. alculate the equivalent conductance of the solution. a ohm -1 cm eq -1 b ohm -1 cm eq -1 c. 356 ohm -1 cm eq -1 d ohm -1 cm eq -1 Dr. Sangeeta Khanna Ph.D 6 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

7 Sol. Specific conductance = conductance cell constant K = 1 = R = = ohm -1 cm e = K N = = 3.56 ohm -1 cm eq The normal oxidation potential of zinc referred to the standard hydrogen electrode is.76 volt and that of copper is.34 volt at 5. 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 Zn + to u + ions at equilibrium? a : b : 1 c : 1 d : 1 Sol. Zn + us 4 u + ZnS 4 Zn + u + u + Zn [Zn E cell = E cell - log [u t equilibrium E cell =.591 [Zn [Zn E cell E cell = log or log [u [u.591 [ E cell = = 1.1 volt Zn 1.1 log = u.591 = 37.5 [Zn [u = : 1 6. NaH is added in R.H.S. of the given cell to consume all the H + present in R.H.S. of the cell of emf +.71 volt at 5 before its use. What is the emf of the cell after addition of NaH. Zn Zn + (.1 M) Hl (1 litre) Pt(H g).76 V 1 atm E Zn/ Zn a volt b volt c volt d volt Sol. Zn + H + Zn + + H.591 [Zn pplying Nernst eq. E cell = E - log [H fter addition of NaH, the solution becomes neutral i e, the concentration of H + ions in cathodic solution becomes 1-7, pplying again Nernst equation E cell = E cell [Zn log [H =.76 - log (1 7 ) =.3759 volt 7. onsider the following reaction: 5H + xl +H xl + y + 6H a. x= 5, y = b. x=, y = 5 c. x= 4, y = 1 d. x= 5, y = 5 Dr. Sangeeta Khanna Ph.D 7 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

8 8. alance the following redox reaction by using following correct coefficients: 1 H 11 + HN 3 H 4 + H +N a. 1 : 18 : 6 : 18 : 19 b. 1 : 36 : 6 : 36 : 3 c. 1 : 36 : 6 : 18 : 36 d. 1 : 36 : 6 : 3 : 36 D 9. The order of the oxidation state of the phosphorus atom in H 3 P, H 3 P 4, H 3 P 3 and H 4 P 6 is a. H 3 P 4 > H 4 P 6 > H 3 P 3 > H 3 P b. H 3 P 3 > H 3 P > H 3 P 4 > H 4 P 6 c. H 3 P > H 3 P 3 > H 4 P 6 > H 3 P 4 d. H 3 P 4 > H 3 P > H 3 P 3 > H 4 P 6 Sol. H 3 P 4 H P H H ; oxidation state of P = +5 H P 6 H P H P H H ; oxidation state of P = +4 H 3 P 4 H P H H ; oxidation state of P = +3 H 3 P H 3. For the following cell: Zn(s) ZnS 4 (aq.) us 4 (aq.) u(s) when the concentration of Zn + is 1 times the concentration of u +, the expression for G (in J mol -1 ) is [F is Faraday constant, R is gas constant, T is temperature, E (cell) = 1.1V a..33 RT + 1.1F b..33 RT.F c. 1.1 F d.. F Sol. G = G +.33 RT log Q G = -nfe +.33RT log Q Given: E = 1.1 V and n = G = (- 1.1 F) +.33RT log P H G = -. F +.33 RT H ; oxidation state of P = +1 [Zn [u Dr. Sangeeta Khanna Ph.D 8 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

9 SETIN (ssertion and Reason) Negative Marking [-1 This Section contains 5 multiple choice questions. Each question has four choices ), ), ) and D) out of which NLY NE is correct. 5 4 = Marks (a) If both assertion and reason are RRET, and reason is the RRET explanation of the assertion. (b) If both assertion and reason are RRET, but reason is NT the RRET explanation of the assertion. (c) If assertion is RRET, but reason is INRRET. (d) If both assertion & reason are INRRET. 1. ssertion:- t the end of electrolysis using platinum electrodes, an aqueous solution of copper sulphate turns colourless. Reason: opper in copper sulphate is converted to copper hydroxide during the electrolysis. a. (a) b. (b) c. (c) (d). d. ssertion: The conductivity of an electrolytic solution decreases with dilution whereas equivalent conductance increases with dilution. Reason: onductivity is for 1 cc of the solution whereas equivalent conductance is for any volume of the solution containing one gram equivalent of the electrolyte. a. (a) b. (b) c. (c) (d). d 3. ssertion: In electrolysis, the quantity of electricity needed for depositing 1 mole of silver is different from that required for 1 mole of copper. Reason: The molecular weights of silver and copper are different. a. (a) b. (b) c. (c) (d). d 4. ssertion: When aqueous solution of NaN 3 is electolysed, sodium is liberated at cathode: (using Hg cathode) Reason: Na + is neutralized at cathode. a. (a) b. (b) c. (c) (d). d 5. ssertion: Electrolysis of an aqueous solution of a bromide or iodide gives r or I at the anode whereas that of NaF gives and not F. Reason: F is highly reactive whereas r and I are less reactive gases. a. (a) b. (b) c. (c) (d). d Dr. Sangeeta Khanna Ph.D 9 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

10 SETIN (Statement ased) Negative Marking [-1 This Section contains 5 multiple choice questions. Each question has four choices ), ), ) and D) out of which NLY NE is correct. 5 4 = Marks In each of the following questions (Q.1 Q.5), three statements I, II, III are given. Mark (a) if all the statements are correct (b) if II and III are correct (c) if I and III are correct (d) if only II is correct 1. I: onductance of electrolytic solution increases with temperature. II: Resistivity is reciprocal of molar conductivity of electrolyte. III: ell constant has a unit cm 1. a. (a) b. (b) c. (c) d. (d) Sol. onductance increases with temp. due to increase in mobility of ions ell constant = a = cm 1. I: xidation number is same as the formal charge. II: The numerical value of.n. and valency may differ. III: bsolute value of electrode potential cannot be determined a. (a) b. (b) c. (c) d. (d) Sol..N is residual charge and is different for same element in different compounds. The valency of an element remains the same in all the compounds 3. In r 5, I: xidation number of r is +6 II: Four atoms of molecule are involved in peroxide linkage III: nly one atom has.n. = - a. (a) b. (b) c. (c) d. (d) Sol. The structure of r 5 is r 1 4. I: The conductivity of molten Nal is due to movement of Na + and l ions II: Solid Nal is also good conductor of electricity III: Sodium is a good conductor because of mobile electrons a. (a) b. (b) c. (c) d. (d) Sol. Molten Nal and molten Na are good conductors. 5. I: athode is ve terminal both electrochemical and electrolytic cell. II: Reduction occurs at cathode both in galvanic as well as electrolytic cell III: hemical change in electrolytic cell is non-spontaneous. a. (a) b. (b) c. (c) d. (d) Dr. Sangeeta Khanna Ph.D 1 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

11 SETIN D (Multiple orrect hoice Type) No Negative Marking This Section contains 5 multiple choice questions. Each question has four choices ), ), ) and D) out of which NE R MRE may be correct. 5 5 = 5 Marks 1. Which of the following facts is true? a. If E (M n+ /M) is negative, H + will be reduced to H by the metal M b. If E (M n+ /M) is positive, M n+ will be reduced to M by H c. In a cell M n+ /M assembly is attached to hydrogen half cell. To produce spontaneous cell reaction, metal M will act as negative electrode, if the potential M n+ /M is negative. It will serve as positive electrode, If M n+ /M has a positive potential d. ompounds of active metals (Zn, Na, Mg) are reducible by H whereas those of noble metals (u, g, u) are not reducible,,,. For a strong electrolyte, increased slowly with dilution and can be represented by the equation: = 1/ Which electrolyte (s) have the same value of the constant? a. Nal b. al c. Znl d. ll 3, Sol. oth are 1 type electrolyte 3. Using the standard potential values given below, decide which of the statements I, II, III, IV are correct. hoose the right answer from (a), (b), (c) and (d) Fe + + e = Fe, E = -.44 V u + + e = u, E = +.34 V g + + e = g, E = +.8 V I. opper can displace iron from FeS 4 solution II. Iron can displace copper from us 4 solution III. u can displace g from gn 3 solution IV. Iron can displace silver from gn 3 solution a. I and II b. II and III c. II and IV d. I and IV, g / g Mg /Mg 4. Given E.8 V, E.37 V, E.34 V, E u / u Hg /Hg.79 V. Which of the following statement is/are correct? a. gn 3 can be stored in copper vessel b. u(n 3 ) can be stored in silver vessel c. ul can be stored in silver vessel d. Hgl can be stored in copper vessel, 5. The values of E of some of the reactions are given below: I +e I ; E = +.54 volt l + e l ; E = volt Fe 3+ + e Fe + ; E = +.76 volt e 4+ + e e 3+ ; E = +1.6 volt Sn 4+ + e Sn + ; E = +.15 volt Which of the following statement is correct? a. Fe +3 will oxidise e +3 b. Fe + will be oxidised by e +4 c. I does not displace l from Kl d. Fel 3 can oxidise Snl,, D Dr. Sangeeta Khanna Ph.D 11 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

12 SETIN E (Matrix Type) No Negative Marking This Section contains questions. Each question has four choices (,, and D) given in olumn I and five statements (p, q, r, and s) in olumn II. ny given statement in olumn I can have correct matching with one or more statement(s) given in olumn II. 8 Marks Each 1. Match the olumn I with olumn II. [Single Match olumn I olumn II () No. of Faradays of electricity produced by combustion of 1 mol benzene (p) 1 () No. of electrons gained by mol of KMn 4 in acidic medium (q) () Eq. Wt. of NH 4 N 3 in given reaction (r) 3 NH4 N 3 N H (D) No. of electron exchanged in the given reaction: (s) 5 3l + 6NaH 5Nal + Nal 3 + 3H Sol. R; P; Q; D S H H x = 3. KMn H + + 1e Mn + + 8H + K + x = 1. eq. NH 4 = eq. mass NH 4 N 3 = = D. e + l l ) 5 l l 3 + 1e 6l 1l + l 3 3l 5l + l 3 (5e ). The standard reduction potential data at 5 is given below. E (Fe 3+, Fe + ) = +.77 V; E (Fe +, Fe) = -.44 V E (u +, u) = +.34 V; E (u +, u) = +.5 V E [ (g) + 4H + + 4e H = +1.3 V; E [ (g) + H + 4e 4H = +.4 V E (r 3+, r) = -.74 V; E (r +, r) = -.91 V Match E of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists. (Single Match nly ) olumn I olumn II () E (Fe 3+, Fe) (p) -.36 V () E (4H 4H + + 4H ) (q) -.4 V () E (u + + u u + ) (r) -.4 V (D) E (r 3+, r + ) (s) -.83 V Sol. r; s; p; D q () Fe e Fe...(1) () 4H 4H + + 4H...(1) Fe +3 + e Fe +...() + 4H + + 4e H...() Fe + + e Fe...(3) + H + 4e 4H...(3) G 1 = G + G 3 G 1 = G 3 - G 3FE1 = 1 FE FE E E E E E E E 1 E3 E = = -.83 Dr. Sangeeta Khanna Ph.D 1 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

13 () u + + u u +...(1) (D) r +3 + e r + u + + e u...() r e r (u + + e u)...(3) r + + e r G 1 = G G 3 G 1 = G - G 3 E SETIN E (Integer Type) No Negative Marking This Section contains 4 questions. The answer to each question is a single digit integer ranging from to 9. The correct digit below the question number in the RS is to be bubbled. 4 5 = Marks 1. How many of the following have eq lower than m (molar conductance) (a) Kl (b) a(h) (c) Nal (d) Na 3 l 3 (e) K (HgI 4 ) (f) NaH 4 (g) FeS 4 (h) K r 7 Sol. (5) eq. M x ; x = charge eq M if x > 1 b, d, e, g, h. How many of the following Half cell have Reduction electrode potential more than standard Reduction potential (E ). (a) (d) Pt H H atm (1M) 3 Zn Zn Fe, Fe Pt M 1M M Sol. () b, f (a) H + + e H.59 p log [H H E ; PH (b) Pt (e) 1atm E E H H 1atm ( M) H Pt H 1atm ph (b) H + > 1 M E > E (c) u + + e u ;.59 E E log(u ) u + < 1 M ; E < E Zn + Fe +3 Zn + + Fe + (d) (e).59 [Zn [Fe E E log 3 [Fe [Zn + & [Fe + > 1 M E < E H e H 1 1atm.59 [1 E log ; E E 1 (f) l + e l.59 E E log E > E 1 Dr. Sangeeta Khanna Ph.D 13 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc (c) (f) u (.1M ) u Pt l l atm (1M)

14 3. How many Faraday s will be required to convert nitrobenzene to nitrosobenzene Sol.() +3 N +1 N = e + 4. The conductance of a.15 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 1 cm with an area of cross section of 1 cm. The conductance of this solution was found to be S. The ph of the solution is 4. The value of limiting molar conductivity of this weak monobasic acid in aqueous solution is Z 1 S cm -1 mol -1. The value of Z is. Sol. (6) M 1 G 1 m 1 a For w eakacid [H c c c c m m m m Z 6 Z 1 Dr. Sangeeta Khanna Ph.D 14 HEMISTRY HING IRLE D:\Important Data\17\+\Physical\Grand Test\+ Grand Test \+ Grand Test-3-Redox & Electrochemistry.doc

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