PRACTISE QUESTIONSELECTROCHEMISTRY

1 For the Galvanic cell,Ag | AgCl(s) | KCl (0.2 M) || KBr (0.001 M) | AgBr(s) | AgCalculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell at 25°CGiven Ksp(AgCl) = 2.8 ´ 10–10, Ksp (AgBr) = 3.3 ´ 10–13

1. AgCl Ag+ + Cl–

Ksp(AgCl) [Ag+] [Cl–]

\[Ag+]LHS =

AgBr Ag++ Br–

Ksp(AgBr) =[Ag+] [Br–]

[Ag+]RHS =

Cell Cell Reaction E0

LHS Ag ¾® Ag++ e– = xVLHS

RHS Ag+ + e– ¾® AgRHS–––––––––––––––––––––––

Net Ag+(RHS) ¾® Ag+(LHS) E0cell= 0.00 V

K =

Ecell = E0cell – = = – 0.0371 V

To make Ecell positive, LHS cell should be cathode (+ve half cell)

2. Assume that impure copper contains only iron, silver, and gold as impurities. After passage of 140 A, for 482.5 s, the mass of the anode decreased by 22.260 g and the cathode increased in mass by 22.011 g. Estimate the % iron and % copper originally present.

2. The increase in mass of the cathode is solely due to copper. Hence, there is, 22.011 g of copper (equivalent to 0.3464 mol) and therefore a total of 0.249 g of iron, gold and silver. Only the iron and copper are oxidized. The gold and silver fall to the bottom in the anode mud. Since each of the active metals requires 2 mol electrons per mol metal,

there must be(482.5s) = 0.3500 moles of M2+

The no. of moles of iron is therefore 0.0036, and the mass of iron is 0.20g. The metal is 98.88% copper and 0.90% iron.

3. When a rod of metallic lead was added to a 0.01 M solution of [Co(en)3]3+, it was found that 68% of the cobalt complex was reduced to [Co(en)3] 2+ by lead.i) Find the value of K for Pb + 2[Co(en)3]3+ Pb2+ + 2[Co(en)3] 2+

ii) What is the value of Eo [Co(en)3]3+|[Co(en)3] 2+

Given: Eo (Pb2+|Pb) = - 0.126 V

3. i) [Co(en)3]3+ = 0.0032, [Co(en)3]2+ = 0.0068 , [Pb2+] = 0..0034

K =

On putting the various known values , we get K = 0.0154

ii) DG10 = -nFE°cell = -2.303 RT log K.

From here we get, E°cell = -0.0536 V From which we can calculate E [Co(en)3

3+/[Co(en)32+] = -0.18V

4. a) Write down the half cell reaction involved in the following electrode.

b) What is the maximum value of ratio for which the following will act as

electrochemical cell?

[log 6 = 0.778]

4. a) The half cell reaction is

b) The cell reaction is

to act as electrochemical cell,

Here

or,

or,

should be less than

5. A conductance cell was calibrated by filling it with a 0.02 M solution of potassium chloride (specific conductance = 0.2768 ) and measuring the resistance at 298K which was found to be 457.3 ohms. The cell was then filled with calcium chloride containing 0.555g

per litre. The measured resistance was 1050 ohm. Calculate the molar conductance of solution.

5. For KCl solution

Specific conductance =

= 126.6 …(1) For solutionSpecific conductance = Conductance Cell constant …(2) Putting the value of cell constant from equation (1) in equation (2)

specific conductance of solution.

=

conc. of solution

=

= = 5 mole

= 6. Iron is corroded by atmospheric oxygen under acidic condition to product Fe2+ (aq) ions

initially. The standard reduction potential

E0Fe2+/Fe = -0.44 V and for the reaction H2O ( ) ¾® 2H+ (g) +

E0 = -1.23 V Find whether the formation of Fe2+ (aq) is thermodynamically favorable or not.

6. The reactions are (i) Fe (s) ¾® Fe+2 (aq) + 2e–,

(ii) 2H+ + (g) + 2e- ¾® H2O (l), E0 = +1.23V

Adding equation (i) and (ii), we get

Fe (s) + 2H+ + (g) ¾® Fe+ (aq) + H2O (l), = 1.67V

\ = positive = 1.67V\ DG0 is negative. So, the reaction is thermodynamically favourable or spontaneous.

7. The emf of the cell reaction,

Calculate the entropy change. Given that enthalpy of the reaction is - 216.7 KJ mol-1 And and = + 0.34V.

7. For the cell reaction

,

= 0.76 + 0.34= 1.1V

\ DG0 = - nF = - 2 ´ 96,500 ´ 1.1 = - 212.3 KJ

= - 14.76 J K-1 mol-1.

8. At 25°C, the emf of the cellPb | PbCl2.HCl (0.5 M) || HCl (0.5 M) | AgCl(s) | Ag is 0.49 volts and its temperature

coefficient . Calculate

a) the entropy change when 1 gm mol of silver is deposited and b) the heat of formation of AgCl, if the heat of formation of lead chloride is – 86000 cal.

8. The net cell reaction

= – 11260 cal

= = – 4.14 cal/degree

DH of the reactionDH = DG + TDS

= – 11200 + 298 ´ (–4.14) = – 12494 cal

This heat of reaction is the algebraic sum of the heats of formation of the components.

= – 30506 cal/mole

9. 25 mL of a solution of HCl (0.1M) is being titrated potentiometrically against 0.1 M NaOH solution using a hydrogen electrode as the indicator electrode and saturated calomel electrode (SCE) as the reference electrode. What would be the EMF of the cell initially and after the addition of 20 mL of alkali at 25°C? Given Reduction potential of SCE = 0.2422V. [log 9 = 0.95].

9. The galvanic cell formed in this case is as follows:

= = 0.2422 - 0.0591 log [= 0.2422 + 0.0591 pH at

Initial pH of the 0.1 HCl: = 0.2422 + 0.0591 = 0.3013V

pH after addition of 20 mL alkali:Amount of HCl initially present = Amount of NaOH added = Amount of HCl left unreached = = 0.5 millimole

= 0.3574V

10. At , the values for and ions are and respectively and the specific conductance of a saturated solution of AgCl after substracting the specific conductance of water is . Calculate of AgCl at .

10.

=

=

11. At 25°C the specific conductance of a saturated solution of AgCl after substracting the specific conductance of water is 1.82 ´ 10–4 S m–1. The molar conductance at infinite dilution of AgNO3, HNO3 and HCl are respectively,133.0´10–4, 421.0´10–4 and 426.0´10–4S m2mol–

1. Write down half cell reaction and calculate standard reduction potential for the half cell: Pt | 50 mL 0.5 M KCl solution, in which few drops of 0.1 M solution of AgNO3 is added to precipitate out some AgCl.

0Ag,AgE = 0.80V

11. From Kohlrausch’s law= = (133 + 426 – 421)10–4 = 138 ´ 10–4 S m2 mol–1

=

\ C =

= = 1.32 ´ 10–2 mol m–3 = 1.32 ´ 10–5 mol dm–3

C is the concentration of saturated solution of AgCl and hence its solubilityKsp = [Ag+] [Cl–] = C2 = (1.32 ´ 10–5)2 = 1.74 ´ 10–10 M2

The given half cell is:Pt | 0.5 M KCl saturated with AgCl and also in contact with AgCl(s)

The half-cell reaction reduction:AgCl(s) + e Ag(s) + Cl–

Calculation of E0

AgCl(s)  + e Ag(s) + Cl– (reduction)

Ag(s) Ag+ + e (oxidation)––––––––––––––––––––––––––––––––AgCl(s) Ag+ + Cl– (cell reaction which is solubility equilibrium controlled by

Ksp)= (RHE where reduction occur) – (LHE where oxidation occurs)=

0.059 log Ksp =

= 0.224 V

12. Calculate the equilibrium constant for the reaction . The standard reduction potentials in acid conditions are 0.77 and 0.54 V respectively for and

couples. (antilog 7.7966 = ).

12. For the chemical change

At equilibrium

As,

Thus,

13. 0.5 N solution of a salt placed between two platinum electrode 2.0 cm apart and of area of cross-section 4.0 sq. cm has a resistance of 25 ohms. Calculate the equivalent conductance of solution.

Solution: Specific conductance (k) = Conductance ´ Cell constant= = = 0.02 ohm–1 cm–1

Ùc = = = 40 ohm–1 cm2 eq–1

14. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298K if the emf of the cell:Ag | Ag+ (sat Ag2CrO4 solution) || Ag+ (0.1 M) | Ag is 0.164 at 298K

14. For cell Ag | Ag+ (saturated solution of Ag2CrO4) || Ag+ (0.1 M) | AgEcell = 0.164V

Ecell = E0cell –

Þ 0.164 =

\ [Ag+]LHS = 1.66 ´ 10–4MKsp for Ag2CrO4 2Ag+ + CrO4

2–

Ksp = [Ag+]2 [CrO42–] = [1.66 ´ 10–4]2

\ Ksp = 2.287 ´ 10–12 mol3 liltre–3

15. A constant current was flown for 2 hour through a solution of KI. At the end of experiment liberated iodine consumed 21.75 mL of 0.0831 M solution of sodium thiosulphate following the reaction: I2 + 2S2O3

2– ¾® 2I– + S4O62–. What was the average rate of

current flow in ampere?

15: 2S2O32– ¾® S4O6

2– + 2en-factor of sodium thiosulphate = no. of electrons lost per molecule = 1\ Molarity of Na2S2O3 solution = Normalitymeq. of I2 liberated = m equiv. of Na2S2O3 = 21.75 ´ 0.0831 = 1.807Thus,

= 1.807

= 1.87 ´ 10–3 =

\ i = 0.0242A

15. The half cell potential of a half cell Ax+, A(x+n)+/Pt were found to be as follows Percent of reduced form 24.4 48.8Cell potential /V 0.101 0.115Determine the value of n

15. The half cell reaction is A(x+n)+ + ne– ¾¾® Ax+

Its Nernst equation

E = E° –

Substituting the given values, we get

0.101 = E° – log --------- (i)

0.115 = E° – ----------- (ii)

Substracting equation (i) from (ii) and solving for n n» 1.98 » 2

16. Two weak acid solutions and each with same concentration and having pKa values 3 and 5 are placed in contact with hydrogenelectrode (1 atm, 25°) and are interconnected through a salt-bridge. Find e.m.f. of cell.

16. The cell is represented asPt, H2(1 atm) | HA2 || HA1 | H2 (1 atm), Pt

At L.H.S.

=

At R.H.S.

= For acid H+ +

[H+] = \ (pH)1 = log C

Similarly (pH)2 =

[Since concentration of both acids HA1 and HA2 are same i.e. c]

\ Ecell = = 0.0592 ´

= Ecell = 0.0592

17. The standard reduction potential for the half-cell+ 2H+(aq) + e– ¾® NO2(g) + H2O is 0.78 volt

i) Calculate the reduction potential in 8 M H+ ii) What will be the reduction potential of the half cell in a neutral solution? Assume

all other species to be at unit concentration.

17. The half cell is (aq) = 2H+ (aq) + e– ¾® NO2(g) + H2

=O

Given – log

\ - log

= 0.78 –

= 0.78 –

= 0.78 + 0.0592 log 64= 0.78 + 0.0592 log26= 0.88969 voltIn neutral solution[H+] = 10–7 M

\ Eel = – log

= = 0.78 –

= 0.78 – = 0.78 – 0.8288

= – 0.0488 volt

18. Calculate the potential of an indicator electrode versus the standard hydrogen electrode which originally 0.1M MnO4

– and 0.8M H+ and which has been treated with 90% of the Fe2+ necessary to reduce all the MnO4

– to Mn2+.MnO4

– + 8H+ + 5e– ¾® Mn2+ + 4H2O E° = 1.51 V18. Let us consider Galvanic cell is

H+ (1M) | H2(1atm), Pt || MnO4– (H+) | Mn+2, Pt

Anode half cell : 2H+ (1M) ¾® H2 (1atm) + 2e–

Cathode half cell: MnO4– + 8H+ + 5e– ¾® Mn+2 + 4H2O

Initial Conc.: 0.1 0.8 0 0

Alter Complete

reaction with Fe+2

(0.01) (0.08) (0.09) So, electrode potential of indicator electrode

=

= 1.51 –

= 1.51 –

= 1.51 – (5.36 ´ 109)

= 1.51 – 0.1149= 1.395 V

Thus, potential of an indicator electrode versus the SHE is 1.395 V because ESHE = 0

19. The e.m.f of cell Zns|ZnSO4|| CuSO4|Cus at 25°C is 0.03 V and temperature coefficient of e.m.f is (–1.4´10-4 V) per degree. Calculate the heat of reaction for the change taking place inside the cell.

19. F= 96500 CE = 0.03 VT = 273+25 = 298 Kn = 2

= - 1.4´10-4 V per degree

As DH = nF

= 2´96500 [298´(-1.4´10-4)-0.03]= -13842 J/mol = –13.842 KJ/mol

20. A test for complete removal of Cu2+ ions from a solution of Cu2+(aq) is to add NH3(aq). A blue

colour signifies the formation of complex [Cu(NH3)4]2+ having Kf = 1.1 ´ 1013 and thus

confirms the presence of Cu2+ in solution. 250 ml of 0.1M CuSO4(aq) is electrolyzed by passing a current of 3.5 ampere for 1350 seconds. Then sufficient quantity of NH3(aq) is added to the electrolyzed solution maintaining [NH3] = 0.1 M. If [Cu(NH3)4]2+ is detectable upto its concentration as low as 1 ´ 10–5, would a blue colour be shown by the electrolyzed solution on addition of NH3?

20. Cu2+ + 4NH3 [Cu(NH3)4]2+

Kf  =

Blue colour will be noticed upto [Cu(NH3)4]2+= 1 ´ 10–5

Thus at this stage

[Cu2+] = = 9.1 ´ 10–15M

Cu deposited (w) = = = 1.5546 gms

Millimoles of Cu2+ present (initial) = 250 ´ 0.1

Weight of Cu2+ = = 1.5875 gms

Weight of Cu2+ left in solution = 1.5875 – 1.5546 = 0.0329 gms

[Cu2+] left = = 2.07 ´ 10–3

Thus solution will show blue colour, as it will provide appreciable Cu2+ to form complex.

21. A sample of lead weighing 1.05 g was dissolved in a small quantity of nitric acid to produce aqueous solution of Pb2+ and Ag+ (which is present as impurity). The volume of the solution was increased to 300 ml by adding water, a pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen electrode was found to be 0.503 V at 25°C. What was the % of Ag in the lead metal? Given =0.799 V. Neglect amount of Ag+ converted to Ag.

21. E = E° – log

0.503 = 0.799 – [Ag+] = 9.62 ´ 10–6 M

Moles of Ag+ = 9.62 ´ 10–6 ´ = 2.89 ´ 10–6

Mass of Ag = 2.89 ´ 10–6 ´ 108 = 3.11 ´ 10–4 g \ Percentage of Ag = 0.0296%

22. The EMF of the cell Pt |H2 (1 atm), HA (0.1 M, 30 ml) || Ag+ (0.8 M) | Ag is 0.9V. Calculate the EMF when 0.05 M NaOH (40 ml) is added to the cathode compartment. HA is a weak acid. [ , log 6.3=0.8, log 2=0.301]

22. Since E = E0C – E0

A –

0.9 = 0.77 – 0 – \ Ka = 2.5 ´ 10–4

When 40 ml and NaOH is added, the [H+] is given by

pH = pKa + logpH = 4 – log 2.5 + 0.30120 = 3.9[H+] = 1.25 ´ 10–4

E = 0.77 - = 0.994 V

23. Specific conductance of a decinormal solution of KCl is 0.0224 . The resistance of a cell containing the solution was found to be 64. What is the cell constant?

23. We known that sp. conductance = cell constant ´ conductance

Cell constant = = sp conductance ´ resistance = 0.0224´64= 1.4336 cm-1

24. Calculate the electrode potential of a copper electrode dipped in a 0.1 M solution of copper sulphate at the standard electrode potential of system is 0.34 volts at 298 K.

24. We known that

Putting the values of = 0.34V, n = 2 and [Cu2+] = 0.1 M

Ered =0.38 + = 0. 34+0.02955´(-1)= 0.34 – 0.02955 = 0.31045 volt

25. The standard reduction potential of electrodes are 0.34 and 0.80 volt respectively. Construct a galvanic cell using these electrodes so that its standard emf is positive. For what concentration of will the emf of the cell at be zero. If concentration of is 0.02 M?

25. Given volt and volt the standard emf will be positive if

Cu/Cu2+ is anode and Ag+/Ag is cathode. The cell can be represented as Cu/Cu+2||Ag+|AgThe cell reaction isCu+2Ag+ ¾¾® Cu2+ + 2Ag

= oxi potential of anode + red potential of cathode = -0.34 + 0.80 = 0.46 volt applying next equation

When Ecell = 0

log

[Ag+] = 2.154´10-9 M