ELECTROCHEMISTRY

1. A direct current of 1.25A was passed through 200 ml of 0.25M Fe2(SO4)3 solution for a period of 1.1 hour. The resulting solution in cathod chamber was analysed by titrating against acidic solution of 25 ml KMnO4 was required to reach the end point. What is the molarity of KMnO4 solution

02. Conductivity of a saturated solution of If the

value of solubility product of is the value of x is ( use

the following data in

ANS:5

08. 4.6 gms of a non radioactive nuclide  present in a sample of salt, is mixed with a very small quantity of radioactive nuclide  and an aqueous solution is prepared. The whole element is removed as precipitate and we get 25 ml of a sample of aqueous solution with an activity  times of its initial activity then its molar conductivity  is

 then y= ? (specific conductivity ) (ANS-5)

09. How long (in sec) a current of 9.65 ampere be passed through 80ml of a 0.1M solution in order to make its , assuming no volume change? (Ans-8)11. The equilibrium constant for the following reaction is . The value of is

_______.

(ANS-5)12. The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver

electrode is 0.79 V. The number of millimoles of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is

13. The emf of the cell M/Metal nitrate (0.01 M)//Metal nitrate (0.1 M)/M is found to be 0.0591 V at 298 K. The valency of metal is (Ans-1)

25. The standard emf of the cell, is 1.53 V and at

is 1.55 V. The value of for the overall reaction is

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* A) B) C) D)

26. During electrolysis under suitable conditions, 0.015 mol of chromium is deposited on the

cathode when 0.090 mol of electrons is passed through a chromium containing electrolyte. Which of the following substances could have been the electrolyte?

A) *B) C) D)

Q.No 3-5 Chiufen, the old mining town located within the hills in the north-east Taiwan, is a place where you can really experience Taiwan’s historical legacy. It was the site of one of the largest gold mines in Asia. Accordingly, Chiufen is often referred to as gold capital of Asia. The compound KCN is traditionally used to extract gold from ore. Gold dissolves in cyanide solutions in the presence of air to form , which is stable in aqueous solution. 3. How many grams, approximately, of KCN are needed to extract 20g of gold from ore? (Au = 197) A. 20g B. 6.5g C. 13g D. 8g Aquaregia, a 3:1 mixture (by volume) of conc. HCl and HNO3 was developed by the alchemists as a means to dissolve gold. The process is actually a redox reaction. Gold is too noble to react with nitric acid. However, gold does react with aquaregia because the complex AuCl4

- forms.

4. Calculate the formation constant, approximately, of AuCl4- at 250C.

A. 105 B. 1025 C. 1012 D. 1042

5. The function of HCl is to provide Cl-. What is the purpose of the Cl- in the above reaction. Select your choice from the following A. it is an oxidising agent B. it is a reducing agent C, it is a complexing agent D. it is a catalyst

Q.No 6-8

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The specific conductivity (κ) of a 0.1 M NaOH solution is 0.0221 S cm. On addition of an equal volume (V) of 0.1 M HCl solution, the value of κ falls to 0.0056 S cm. On further addition of the same volume V of 0.1M HCl, the value of κ rises to 0.0170 S cm-1.

6. Calculate the equivalent conductivity ( ) for NaOH A. 112 S cm2 mol-1 B. 221 S cm2 mol-1

C. 432 S cm2 mol-1 D. 295 S cm2 mol-1

7. Calculate the equivalent conductivity ( ) for HCl A. 450 S cm2 mol-1 B. 398 S cm2 mol-1

C. 550 S cm2 mol-1 D. 112 S cm2 mol-1

8. . Calculate the equivalent conductivity ( ) for H+ and OH- together. A. 450 S cm2 mol-1 B. 307 S cm2 mol-1

C. 507 S cm2 mol-1 D. 112 S cm2 mol-1

9. Standard potential of the following reaction can be expressed as

Q.No: 11-13Redox potential and Gibb’s free energyThe proton, neutron, and electron are the three sub-atomic particles important in chemistry. The particles occupy two regions. Proton and neutron occupy the central place of the nucleus and electron the vast space outside the nucleus. Neutron transfer does not take place in ordinary chemical reactions. Proton transfer constitutes acid-base reactions. Electron transfer constitutes redox reactions. Redox reactions are essential for life. Photosynthesis and respiration are two prime examples. Redox reactions also allow key thermodynamic quantities to be measured as demonstrated in this problem. Given the following information

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11. What is ‘X’ in the above Latimer diagram ? A. 1.634 V B. 1.098 V C. 3.075 V D. 1.441 V

12. A galvanic cell using the standard hydrogen electrode as an anode is constructed in which the overall reaction is

Silver ions are added until AgBr precipitates at the cathode and reaches 0.06 M. The cell voltage is measured to be 1.721 V. Calculate E0 for the galvanic cell approximately. ( log 8 = 0.9030 ) A. 1.734 V B. 2.19 V C. 1.065 V D. 0.32 V

13. Using the data given in the comprehension and/or in the above two questions which of the following can be calculated ?I. Solubility of AgBr in a 0.1M aqueous solution of ammonia at 250C.II. III. Solubility of Br2 in water at 250C.IV. KSP of AgBr(s) at 250C. A. III & IV B. I & IV C. I, III, & IV D. I, II, III, & IV14. Consider the following standard reduction potentials,

half reaction Eo, VNi2+(aq) + 2e- <==> Ni(s) -0.23Co2+(aq) + 2e- <==> Co(s) -0.28Fe2+(aq) + 2e- <==> Fe(s) -0.41Cr3+(aq) + 3e- <==> Cr(s) -0.74Mn2+(aq) + 2e- <==> Mn(s) -1.03

which of the following metals could be used successfully to galvanize steel? A. Ni only B. Ni and Co d

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C. Fe only D. Mn and Cr

15. Consider the following reduction-oxidation reaction, Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq), Eo = +1.10 VWhen the components of the two half-reactions involved in this reaction are correctly separated, this reaction can be used to do electrical work and the change in internal energy of the system decreases. Which of the following statements best describes what will happen if Zn(s) is added to a solution containing Cu2+(aq) in an insulated container at constant volume at 25oC? A. The reaction will occur and the temperature of the solution will increase. B. The reaction will occur and the temperature of the solution will decrease. C. The reaction will occur and the temperature of the solution will remain the same. D The reaction will not occur and the temperature of the solution will remain the same.

Problem 8: Acidified water is electrolysed using an inert electrode. The volume of gases liberated at STP is 0.168L. The quantity of charge passed through the acidified water would be(a) 96,500C (b) 9,650C(c) 965C (d) 168C

Solution: (c) 2H2O ¾® 2H2(g)  + O2(g)

2x x 3x = 0.168 x = 0.056L

= 2x = 0.112L, = x = 0.056L11.2L of H2 at STP 1F0.112L of H2 at STP 0.01F0.056L of O2 at STP = 0.01F The amount of electricity passed = 0.01F = 965C

Problem 15: Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below :

+ 8H+ (aq) + 5e Mn2+(aq) + 4H2O (l); E0 = 1.51 VCr2 (aq) + 14H+ (aq) + 6e 2Cr3+(aq) + 7H2O (l); E0 = 1.33VFe3+(aq) + e Fe2+(aq); E0 = 0.77 VCl2 (g) + 2e 2Cl(aq); E0 = 1.40 VIdentify the only correct statement regarding the quantitative estimation of aqueous Fe(NO3)2

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(a) Mn can be used in aqeous HCl (b) Cr2 can be used in aqueous HCl(c) can be used in aqueous H2SO4 (d) Cr2 can be used in aqueous H2SO4

Solution: (b, c, d)

@ Write-up I Oxidation-reduction titrations are important in analytical chemistry and a discussion of them provides considerable insight into the operation of galvanic cells as well as the use of the Nernst equation.For Nernst equation calculation we will use ceric ion, Ce4+, because it is a one electron oxidising agent, and consider the titration of 50 ml of 0.2M Fe2+ by a volume V(ml) of 0.2M Ce4+. The reaction isFe2+ + Ce4+ ¾® Fe3+ + Ce3+

The ceric ion is added from burret to a right-hand beaker containing the ferrous ion. Into this beaker are placed a salt bridge and a platinum wire, so that the right breaker may be operated as a half-cell against the standard hydrogen electrode. In the right beaker either of the two half-reactions,Ce4+  + e–  Ce3+

Fe3+ + e– Fe2+ can take place.

1. After addition of some amount of Ce4+ such that an equilibrium is established between Fe3+, Fe2+, Ce4+ and Ce3+, which among the following can be regarded as half cell:(a) Fe3+ – Fe2+ half cell (b) Ce4+ – Ce3+ half cell(c) Fe3+ – Ce3+ half cell (d) either of (a) or (b) can be taken

2. When chemical equilibrium exists between the ions then which is true for equilibrium half cell potential (Ecell) of this solution:(a) (b)(c) (d) none of these

3. Which among the following expression indicates equivalence point of above titration. (Eep ® Emf at equivalence point of right half cell).

(a) (b)

(c) (d) none of these

Solution :1. (d)2. (d) At equilibrium

3. (b) At equivalence point

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, where

or

@ Write-up IIThe E0 values for the changes given below are measured against NHE at 270C.Cu2+ + e Cu+ ; E0 = +0.15 VCu+ + e Cu ; E0 = +0.50 VZn2+ + 2e Zn ; E0 = -0.76 VThe temperature coefficient of emf a cell designed as Zn| Zn2+| | Cu2+ | Cu is –1.4 10-4V

1M 0.1Mper degree. For a cell reaction in equilibrium G = 0 and G0 = -2.303 RT log10 KC. The heat of reaction and entropy change during the reaction are related by G = H - TS.

1. The heat of reaction for the change Zn + Cu2+ Zn2+ + Cu at 270C is: 0.1M 1 M (a) –2.118 105 J (b) –2.197 105 J(c) –2.237 105 J (d) –1.818 105 J

2. Change in entropy for the reaction: Zn + Cu2+ Zn2+ + Cu at 270C is: 0.1M 1 M (a) 14 J (b) 27 J(c) –14J (d) –27 J

3. The equilibrium constant for the disproportionation of Cu+ is:(a) 8.5 10-5 (b) 8.5 105

(c) 8.5 10-4 (d) 8.5 106

Solution :

1. (a)

= 2 96500 [300 (-1.4 10-4) – 1.0555= -2.118 105 J

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2. (d) G = H - T S-2.037 105 = -2.118 105 – 300 SS = -27 J

3. (b)

or 0.35 =

Kc = 8.5 105

Subjective

Problem 1: Calculate the solubility product and solubility of AgCl in the following cell which has an emf of 0.455 volts at 25°C.Ag/AgCl in 0.1 M KCl | 0.1 M AgNO3 | Ag a1 = ? a2

Solution: Ag ¾® + e–

+ e– ¾® Ag–––––––––––––––

¾®

E =

= 0.0591 log = 0.059 log = 0.455

= 7.71

7.71 = 5.129 107

a1 = = 1.95 10–9

[Cl–] = 10–1

Solubility product of AgCl = [Ag+] [Cl–]= 1.95 10–9 10–1 = 1.95 10–10

Solubility of AgCl = [Ag+] = [Cl–] = = 1.396 10–5 moles/litre

Problem 2: For the galvanic cellAg | AgCl(s) KCl (0.2M) || KBr (0.001 M) AgBr(s) | Ag

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Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at 25°C[Ksp(AgCl) = 2.8 10–10; Ksp (AgBr) = 3.3 10–13]

Solution: The given cell may be written as underAg|Ag+(C1)Cl– (0.2M) || Ag+ (C2), Br– (0.001 M) | AgThe above cell in a concentration cell with the cell reaction[Ag+]RHS ¾® [Ag+]LHS

and cell potential = – 0.059

Ecell = – 0.0591 log

= – 0.0591 log = – 0.0591log

= – 0.0591 log 4.242 = – 0.0591 0.6276 = – 0.03709 voltsSince cell potential is negative, the cell reaction will be spontaneous in the reverse direction i.e. AgCl + Br– ¾® AgBr + Cl–

with Cl– | AgCl | Ag serving as cathode (+ve terminal) and Br– | AgBr | Ag as anode ( –ve terminal). The spontaneity is due to the fact that AgBr is less soluble than AgCl.

Problem 4: Calculate EMF of the cell

Pt, H2 (0.1 atm) | solution (pH = 4) || KCl solution (saturated with AgCl) | Ag

and equilibrium constant of the cell reaction at 25°C. Given = 0.80 V and Ksp (AgCl) = 4.36 10–11 at 25°C.

Solution: Cell reaction: + Ag+ ¾® H+ + Ag(s)

Q = = = 2.29 104.5

(Q pH = 4 [H+] = 10–4 M. [H2] = = 0.1 atm, and

Ksp = [Ag+] [Cl–], [Ag+] = = = 4.36 10–9 M)

Ecell = – 0.059 Q

= – 0.059 log Q

= (0.80 – 0) – 0.059 log (2.29 104.5 )= 0.80 – 0.059 (log 2.29 + 4.5)= 0.80 – 0.059 (0.3598 + 4.5) =0.514V

Alternate method:

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= + 0.059 log [Ag+]= 0.80 + 0.059 log (4.36 10–9)= 0.80 + 0.059 (0.6493 – 9)= 0.80 + 0.0377 – 0.531 = 0.3067

=

= 0 – 0.059 log

= – 0.059 3.5 = – 0.2065

Ecell =

= 0.3067 + 0.2065 = 0.5132 VE = Eo – 0.059 log QAt equilibrium: E = 0 and Q = K = equilibrium constant So, Eo = 0.059 log K0.80 = 0.059 log KK = 3.622 1013

Problem 5: Construct the cell corresponding to the reaction:3Cr2+ (1M) ¾¾® 2Cr3+ (1M) + Cr(s)and predict if the reaction is spontaneous. Also calculate the following i) H and S of the reaction at 25°C.ii) G of the reaction at 25°C when Cr2+ = 0.5 M and Cr3+ = 0.25 MGiven = 0.5 V, = – 0.41 VG° of the reaction at 35°C = – 270.50 kJ

Solution: The cell corresponding to the given reaction is as follows.Pt |Cr2+(1M), Cr3+(1M) || Cr2+ (1M) | Cr(s)I Cr3+ + 3e ¾¾® Cr(s),

II Cr3+ + e ¾¾® Cr2+ , Eq. I – Eq. II: Cr2+ + 2e ¾¾® Cr(s), G° = – 3 0.5 F – 0.41 F = – 1.91 FG° = – 2 = – 1.91 F

= = 0.955 V

– = 0.955 + 0.41 = 1.365 V

is +ve so G° is – Ve and hence the given reaction is spontaneous

i) G° = – nF = – 2 96500 1.365 J= – 263.44 kJFrom Gibb's – Helmholtz equation

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G = H + T

– 263.44 = H + 298

= H – 298 0.706 H = - 53.05 kJ

S = =

= 0.706 kJ K–1 = 706 kJ K–1

ii) G = G° + RT ln Q, Q = = 0.5

= – 263.44 + 8.314 10–3 298 2.303 log 0.5= – 263.44 – 5.705 .30103 = – 263.44 – 1.717 = – 264.717 kJ

Problem 6: Consider the cell: Pb | PbSO4 | Na2SO410H2O The temperature coefficient of the emf of above cell is 0.000174V deg–1 and heat of reaction is – 176.146 kJ. Calculate emf of the cell.

Solution: The cell reaction isAt anode Pb(s) + SO4

2– ¾® PbSO4(s) + 2e–

At cathode Hg2SO4(s) + 2e– ¾® 2Hg(l) + SO42–

––––––––––––––––––––––––––––––Pb(s) + Hg2SO4(s) ¾® PbSO4(s) + 2Hg(l)

i.e. the number of electrons involved are two, substituting the values of known terms in equation (3)– 176146 J = – (2) (96500 J V–1) E0 + (298 K ) (96500 J V–1) (2) (0.00174 VK–1)E0= 0.9645V

Problem 7: The solubility product of Fe(OH)3 at 25°C is 10–36.4 mol4 dm–12 and E0 (Fe3+/Fe) = – 0.036 V. Calculate the standard emf of the reaction Fe(OH)3(s)  ® Fe3+ + 3OH–.

Solution: The cell can be formed as Fe|Fe3+; OH– | Fe(OH)3(s) | FeThe electrode reactions can be written asFe(s) ¾® Fe3+ + 3e–

Fe(OH)3(s) + 3e– ¾® Fe(s) + 3OH–

Overall reaction is i.e., by addingi.e.

can be calculated if we know E0 and E0 can be calculated as follows

=

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Problem 10: It is desired to constructed the following voltaic cell to have Ecel = 0.0860V.

What [Cl–] must be present in the cathode half cell to achieve this result?Ag(s) | Ag+ (satd. AgI(aq)) || Ag+ (satd. AgCl, xMCl–) Ag+(s)Ksp[AgCl = 1.8 × 10–10, AgI = 8.5 × 10–17];

Solution: At L.H.S. half cell Ag(s) ¾® AgA+ (aq) + e–

At R.HS. half cell Agc+(aq) ¾® Ag(s)

R.H.S.––––––––––––––––––––––Agc

+ (aq) ¾® AgA+ (aq) E0

cell = 0.00VAg+ (L.H.S.) is from Ag( (satd. Aq)AgI(s) Ag+

(aq) + I–(aq)

Ksp = [Ag+] [I–] = [Ag+]2

[Ag+]AgI = [Ag+]A =

= Ag+ (R.H.S.) is from AgCl in presence of [Cl–] = xMAgCl(s) Ag+

(aq) + Cl–(aq)

Ksp= [Ag+] [Cl–]

[Cl–]

0.0860 = – 0.0591

– 1.4562 = log 51.22 + logx = 1.7095 + logxlogx = – 3.1647x = 7 × 10–4M[Cl–] = 7 × 10–4M

Assignments (New Pattern)SECTION – I Single Choice Questions

LEVEL – I

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13. The titration curve shown in the parenthesis represents that for the:(a) Titration of HCl vs NaOH(b) Titration of CH3COOH vs NaOH(c) Titration of HCOOH vs KOH(d) Titration of HCl vs NH4OH

15. The following facts are available: 2A– + B2 ¾® 2B– + A2; 2C– + B2 ¾® No reaction;

2D– + A2 ¾® 2A– + D2. Which of the following statement is correct.

(a) (b)

(c) (d)

16. Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below:

E° = 1.51 VE° = 1.38 VE° = 0.77 VE° = 1.40 V

Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2

(a) can be used in aqueous HCl (b) can be used in aqueous HCl(c) can be used in aqueous H2SO4 (d) can be used in aqueous H2SO4

4. Cu2+ + 2e– ¾® Cu; log[Cu2+] vs Ered graph is of the type shown in figure where OA = 0.34V, then electrode potential of the half cell of Cu/Cu2+ (0.1 M) will be

(a) – 0.34 + V (b) 0.34 + 0.059V

(c) 0.34 V (d) None

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5. Which of the following statements is (are) not correct?(a) Metallic conduction is due to the movement of electrons whereas the electrolytic

conduction is due to the movement of ions.(b) Both metallic and electrolytic conductions involve transfer of matter(c) Electrolytic conduction decreases with rise in temperature(d) Metallic conduction involves no chemical change

6. Which among the following statement(s) is/are correct?(a) If is negative, H+ will be reduced to H2 by the element ‘A’(b) Compounds of (Zn, Na, Mg) are reduced by hydrogen (H2) whereas those of noble

metals (Cu, Ag, Au) are not reducible.(c) If is positive, An+ will be reduced to A by H2

(d) M | Mn+ || H+ | H2 (Pt) will be spontaneous if is negative

7. Electrote potential data are given below

Based on the given data which statement/s is/are true ?(a) Fe2+ is stronger reducing agent than Br (b) Fe2+ is stronger reducing agent than Al(c) Al is stronger reducing agent than Fe2+ (d) Br is stronger reducing agent than Al

8. The standard emf of the cell, in which the cell

reaction is 0.6195 V at 0°C and 0.6753 V at 25° C. The value of H of the reaction at 25°C is(a) 167.26 kJ/mol (b) 167.26 kJ/mol(c) 40 K/mol (d) 40 Kcal/mol

9. The standard electrode potentials, and are respectively + 0.54

V, 1.09 V and 0.44V as the basis of given data which of following is/are spontaneous (a) (b) (c) (d)

10. We observe blue colour if : (a) Cu electrode is placed in the AgNO3 solution (b) Cu electrode is placed in the ZnSO4 (c) Cu electrode is placed in the dil HNO3

(d) Cu electrode is placed in the NiSO4

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1. Match column I with column II select the correct answer using the codes given below the lists.

Column – I Column – II(a) Electrode reversible with respect to cation

(p) Pt/Fe2+, Fe3+

(b) Electrode reversible with respect to anion

(q) Pt, H2 (1 atm)/H+ (a = 1)

(c) Redox electrode (r) Ag/AgCl(s), HCl (aq)

(d) Reference electrode (s) Ag/AgNO3

@ Write-up I

A acidic solution of Cu+2 salt containing 0.4gm of Cu+2 is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volumes of solution kept at 100 ml and the current at 1.2 amp.

1. In the beginning of the electrolysis the substances evolved at cathode and anode are respectively(a) Cu and O2 (b) H2 and O2(c) Cu and H2 (d) H2 and Cu

2. During another seven minutes of electrolysis the substances evolved at cathode and anode are respectively(a) Cu and O2 (b) H2 and O2(c) Cu and H2 (d) Only O2 is evolved at cathode

3. The total volume of gases evolved at STP at cathode during the entire electrolysis(a) 58.4 ml (b) 48.23 ml(c) 60.35 ml d) 35.58 ml

4. The total volume of gases evolved at STP at anode during the entire electrolysis (a) 58.94 ml (b) 70.44 ml(c) 29.28 ml (d) 99.68ml

1. In refining of silver by electrolytic method, what will be the weight of 100 gm of Ag anode if 5A current is passed for 2 hour. The purity of silver anode is 95% by weight.

2. A current of 1.70Å is passed through 300 ml of 0.160 M solution of ZnSO4 for 230- sec with current efficiency of 90%. Find the molarity of Zn2+ after the deposition of Zn. Assume the volume of solution remains constant during electrolysis.

3. Determine the standard potential of a cell in which the reaction Co3+(aq) + 3Cl–(aq) + 3Ag(s) ¾¾® 3AgCl(s) + Co(s) from the standard potentials of the couples Ag/AgCl, Cl (+0.22V), Co3+/Co2+ (+1.81V) and Co2+/Co(–0.28V).

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4. A current of 2 amperes is passed for 11 hours through 500 mL of 2M solution of Ni(NO3)2 using platinum electrodes. What will be the molarity of the solution at the end of the electrolysis ?

5. A current of 0.193 amp is passed through 100 ml of 0.2M NaCl for an hour. Calculate pH of solution after electrolysis if current efficiency is 90%. Assume no volume change.

6. Calculate the EMF of the cellMg(s) | Mg2+ (0.1 M) || Ag+ (1 × 10–4M) | Ag

What will be the effect on EMF if concentration of Ag+ is increased to 1 × 10–3 M?

7. If and Ksp of AgCl is given, find the half cell potential of Cl–(x)/AgCl/Ag electrode.

8. Silver is electrodeposited on a metallic vessel of surface area 800 cm2 by passing a current of 0.20 ampere for 3.0 hours. Calculate the thickness of silver deposited, given its density is 10.47g/cc [Atomic weight of Ag = 107.92]

9. How many coulombs are required to reduce 9.48 g to ions?(At. Masses of K = 39, Mn = 55, O = 16 amu)

10. When a current of 1.5 A was passed for 30 minutes through a solution of a salt of a trivalent metal, 1.071 g of the metal were deposited at the cathode. Calculate the atomic mass of the metal.

LEVEL – II1. The Edison storage cell is represented as

Fe(s) | FeO(s) | KOH(aq) | Ni2O3(s) | Ni(s)The half cell reactions areNi2O3(s) + H2O(l) + 2e ¾® 2NiO(s) + 2OH–

(aq) E0 = 0.40 VFeO(s) + H2O(l) + 2e ¾® Fe(s) ¾® Fe(s) + 2OH– E0= –0.87Vi) What is the cell reaction?ii) What is emf of the cell? How does it depend on the concentration of KOH?iii) What is the maximum amount of electrical energy that can be obtained from one mol

of Ni2O3.

2. Calculate the equilibrium constant for the reaction 2Fe2++ 3I– 2Fe2+ + I3–. The

standard reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for Fe3+/Fe2+ and I3

–/I–.

3. In a fuel cell H2 and O2 react to produce electricity,. In the process H2 gas is oxidised at the anode and O2 at cathode. If 67.2 litre of H2 at STP reacts in 15 minutes. What is the average current produced? If the entire current is used for electro deposition of Cu from Cu+2, how many grams of Cu are deposited.

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4. 0.11 Faraday is required to reduce an aqueous solution containing 0.02 mol of a mixture of and to and respectively. Calculate the amount of each in the solution. (At. Masses of K = 39, Mn = 55, Cr = 52) = 16)

5. Given the cellCd | Cd(OH)2(s) | NaOH (0.01 mol kg-1) | H2(1 atm) | PtWith Ecell

= 0.0 V at 298 K. If = – 0.40 V, Calculate Ksp for Cd(OH)2

6. The density of copper is 6.94 g mL–1. Find the number of coulomb needed to plate an area of 10 10 cm2 to a thickness of 10–2 cm using CuSO4 solution as an electrolyte. Atomic weight of Cu is 63.5.

7. 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 the measured e.m.f. be zero at 25°C? Assume [Ni2+] = 1M and .

8. If we want the galvanic cell Cu(s) | 1 atm. to have an EMF of 1.136 V at 25°C, what value should be used for , the concentration of ion ? (

9. a) Calculate the equilibrium constant for the following reaction at 25°C?

b) If an excess of tin metal is added to 1.0 M what is the concentration of at equilibrium?

10. How many grams of silver could be plated out on a serving tray by electrolysis of a solution containing Ag(l) 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 platting is 0.00254 cm? Density of silver is 10.5 g/ .

LEVEL – III (Judge yourself at JEE level)

1. Copper sulphate solution (250 ml) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 16 minute. It was found that after electrolysis, the absorbance of the solution was reduced to 50% of its original value. Calculate the concentration of copper sulphate in the solution to begin with.

2. Calculate the quantity of electricity that would be required to reduce 12.3g of nitrobenzene to aniline, if current efficiency is 50%. If the potential drops across the cell is 3.0 volt, how much energy will be consumed?

3. The following electrochemical cell has been set up.

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If an ammeter is connected between the two platinum electrode, predict the direction of flow of current. Will the current increase or decrease with time?

5. The standard reduction potential for Cu2+/Cu is +0.34V. Calculate the reduction potential at pH = 14 for the above couple. Ksp of Cu(OH)2 is 1.0 × 10–19.

6. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298K if the e.m.f. of the cell Ag+|Ag+ (saturated Ag2CrO4 solution) || Ag+ (0.1M) | Ag is 0.164V at 298K.

7. For the galvanic cellAg | AgCl(s), KCl || KBr, AgBr(s) | Ag

0.2M 0.001MCalculate the e.m.f. generated and assign correct polarity to each electrode for a spontaneous process after taking on account of cell reaction at 25°C. Given:

8. The standard reduction potential of Ag+/Ag electrode at 298 K is 0.799V. Given that for AgI, Ksp = 8.7 × 10–17, evaluate the potential of Ag+/Ag electrode in a saturated solution of AgI. Also calculate the standard reduction potential of the I–/AgI/Ag electrode.

9. A cell Ag | Ag+ || Cu2+ | Cu initially contains 1M Ag+ and 1M Cu2+ ions. Calculate the change in the cell potential after the passage of 9.65 ampere of current for 1 hour.

10. Cu2+  + In2+ ¾→ Cu+ + In3+

Calculate the equilibrium constant. Given that

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Answers to Assignments SECTION - I

LEVEL - I

1. (b) 2. (d) 3. (d) 4. (c)5. (c) 6. (a) 7. (b) 8. (b)9. (a) 10. (a) 11. (c) 12. (b)13. (d) 14. (c) 15. (b) 16. (a)17. (b) 18. (d) 19. (b) 20. (d)

LEVEL - II

1. (b) 2. (c) 3. (b) 4. (a)5. (d) 6. (a) 7. (d) 8. (a)9. (c) 10. (c)

SECTION - II

1. (a), (d) 2. (a), (c) 3. (b), (c), (d)4. (a), (d) 5. (b), (c) 6. (a), (c), (d)7. (a), (c) 8. (b), (d) 9. (a), (b), (c)10. (a), (c)

SECTION - III

1. (d) 2. (c) 3. (a) 4. (b)5. (a) 6. (d) 7. (b) 8. (b)9. (d) 10. (d) 11. (d) 12. (c)13. (a) 14. (a) 15. (a)

SECTION - IV

1. (a - s); (b - r); (c – p); (d – q)

2. (a - s); (b - r); (c – q); (d – p)

SECTION - V

1. (a) 2. (b) 3. (a) 4. (d)5. (b) 6. (b) 7. (c) 8. (a)9. (b) 10. (b) 11. (a) 12. (c)

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SECTION - VI

LEVEL – I1. 57.59 gm 2. 0.154M3. 0.20V 4. 1.1795. 12.82 6. 2.96V, increases to 3.02V

7. 8. 2.88 × 10–4 cm

9. 2.895 × 104C 10. 114.83

LEVEL – III

1. 2.45 102 kJ/mol 2. 6.26 107

3. 190.5 g 4. 1.58gm, 2.94gm5. 3.36 10–15 M2 6. 27171.96 coloumb7. 4 8. 0.01M9. (a) 2.18, (b) 0.315M 10. 272.2g, 1.02 104 cm2

LEVEL – III

1. 7.95 × 10–5M2. 115800 coulomb, 347.4 kJ3. decrease with time5. –0.2205V6. 2.287 × 10–12 mol3litre–3

7. Polarity of the cell should be reversed – 0.03%8. 0.32V, –0.149V9. 0.010V10. .

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