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# Numerical Problems

Chemistry : Electro Chemistry : Book Back, Exercise, Example Numerical Question with Answers, Solution

Example

A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area of each electrode is 4.5 sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15 Ōä” . Find the specific conductance of the solution.

Solution

l = 1 5. cm = 1.5├Ś10-2 m

A = 4.5 cm2 = 4.5├Ś(10-4)m2

R = 15Ōä”

╬║ = 1/R (l/A)

╬║ = [ 1/15Ōä” ] ├Ś [1.5├Ś10-2 m  / 4.5├Ś(10-4)m]

= 2.22 Sm-1

### Example

Calculate the molar conductance of 0.025M aqueous solution of calcium chloride at 25┬║ C. The specific conductance of calcium chloride is 12.04 ├Ś 10-2 Sm-1.

Molar conductance = ╬ø m = = 481.6 ├Ś10-5 Sm2mol-1

### Example

The resistance of a conductivity cell is measured as 190 Ōä” using 0.1M KCl solution (specific conductance of 0.1M KCl is 1.3 Sm-1 ) . When the same cell is filled with 0.003M sodium chloride solution, the measured resistance is 6.3KŌä”. Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of NaCl solution.

Given that

╬║ = 1.3 Sm-1 (for 0.1M KCl solution)

R = 190 Ōä” ╬øm = 13.04 ├Ś 10-3 Sm2 mol-1

### Example

A solution of silver nitrate is electrolysed for 20 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode.

Electrochemical reaction at cathode is Ag+ +eŌåÆ Ag (reduction) 8. The conductivity of a 0.01M solution of a 1 :1 weak electrolyte at 298K is 1.5 ├Ś10-4 S cmŌłÆ1.

i) molar conductivity of the solution

ii) degree of dissociation and the dissociation constant of the weak electrolyte

Given that

╬╗┬║cation = 248.2 S cm2 molŌłÆ1

╬╗┬║cation = 51.8 S cm2 molŌłÆ1

Given

C = 0.01M

╬╗┬║cation = 248.2 S cm2 molŌłÆ1

K= 1.5 ├Ś 10ŌłÆ4 S cmŌłÆ1

╬╗┬║anion = 51.8 S cm2 molŌłÆ1.

1. Molar conductivity

K = 1.5 ├Ś10ŌłÆ4 S cmŌłÆ1

╬ø┬║m =  K(smŌłÆ1 ) ├Ś10ŌłÆ3 / C (in M) mol-1 m3

= (1.5 ├Ś102 ├Ś10ŌłÆ3 ) / 0.01 S mol ŌłÆ1 m2      1 cmŌłÆ1 = 102mŌłÆ1

= 1.5 ├Ś10ŌłÆ3 S m2 molŌłÆ1 = 1.5 ├Ś102

2. Degree of dissociation ╬▒ = ╬ø┬║/╬ø┬║Ōł×

╬øoŌł× =╬╗ocation +╬╗aniono

= (248.2 + 51.8) S cm2 molŌłÆ1

= 300 S cm2 molŌłÆ1

= 300 ├Ś10ŌłÆ14 s m2 molŌłÆ1

╬▒ = 1.5 ├Ś10ŌłÆ3 S m2 molŌłÆ1  / 300├Ś10ŌłÆ4 S m2 molŌłÆ1

╬▒ = 0.05

Ka = ╬▒2c / 1-╬▒

= [ (0.05)2 (0.01) ] / 1-0.05

= 25 ├Ś10ŌłÆ4 ├Ś10ŌłÆ2 /  95 ├Ś10ŌłÆ2

= 0.26 ├Ś10ŌłÆ4

= 2.6 ├Ś10ŌłÆ5.

13. A current of 1.608A is passed through 250 mL of 0.5M solution of copper sulphate for 50 minutes. Calculate the strength of Cu2+ after electrolysis assuming volume to be constant and the current efficiency is 100%.

Given

I = 1.608A; t = 50 min = 50 ├Ś 60

= 3000S

╬Ę = 100%

V=250 mL

C=0.5M

Calculate the number of faradays of electricity passed through the CuSO4 solution

ŌćÆ Q=It

Q = 1.608 ├Ś3000

Q = 4824C

Ōł┤ number of Faradays of electicity = 4824 C  / 96500C = 0.05F

Electrolysis of CuSO4

Cu2+ (aq)+2eŌłÆ ŌåÆ Cu(s).

The above equation shows that 2F electricity will deposit 1 mole of Cu2+ to Cu.

Ōł┤0.05F electricity will

deposit  1 mol /2F ├Ś 0 .05F = 0.025 mol

Initial number of molar of Cu2+ in 250 ml of solution = (0.5 / 1000 mL)  ├Ś 250mL

= 0.125 mol

Ōł┤ number of moles of Cu2+ after electrolysis = 0.125 - 0.025

= 0.1 mol

Ōł┤ Concentration of Cu2+ = [ 0.1 mol/ 250 mL ] ├Ś1000 mL

=0.4M

14. Can Fe3+ oxidises bromide to bromine under standard conditions?

Given: EFe 3+|Fe2+ = 0.771

EBr2|BrŌłÆ = 1.09V.

Required half cell reaction

2 Br- ŌåÆ Br2 + 2eŌłÆ

( Eoox )= ŌłÆ1.09V

2 Fe3+ + 2eŌłÆ ŌåÆ 2Fe2+

( Eored )= +0.771V

2Fe3+ + 2BrŌłÆ ŌåÆ 2Fe2+ +Br2

( Eocell )= ?

Eocell = ( Eoox )+ ( Eored )

= ŌłÆ1.09 + 0.771

ŌłÆ0.319V

Ecell is ŌĆō ve; ŌłåG is +ve and the cell reaction is non spontaneous. Hence Fe3+ cannot oxidises Br- to Br2

15. Is it possible to store copper sulphate in an iron vessel for a long time?

Given : ECu 2+|Cu = 0.34 V and EFe 2+|Fe = ŌłÆ0.44V .

( Eoxo )Fe|Fe2+ = 0.44V and ( Eredo )Cu 2+|Cu = 0.34V .

These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve.

Hence it is not possible to store copper sulphate in an iron vessel.

16. Two metals M1 and M2 have reduction potential values of -xV and +yV respectively. Which will liberate H2 and H2SO4.

Metals having higher oxidation potential will liberate H2 from H2 SO4 . Hence, the metal M1 having + xV, oxidation potential will liberate H2 from H2SO4 .

17. Reduction potential of two metals M and M2 are E┬║M12+|M1 = ŌłÆ 2.3V and E E┬║M22+|M2 = 0.2V  Predict which one is better for coating the surface of iron. Given : E Fe2+ |Fe = ŌłÆ0.44V

oxidation potential of M1 is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting

18. Calculate the standard emf of the cell: Cd | Cd 2+ || Cu 2+ | Cu and determine the cell reaction. The standard reduction potentials of Cu2+ | Cu and Cd2+ | Cd are 0.34V and -0.40 volts respectively. Predict the feasibility of the cell reaction.

Cell reactions:

Oxidation at anode: Cd (s) ŌåÆ Cd2+ (aq) + 2eŌłÆ

( Eoox) Cu|Cd2+ = 0.4V

Reduction at cathode: Cu2+ (aq) + 2e- ŌåÆ Cu (s)

( Eored )Cu2+|Cu = 0.34V

Cd ( s ) + 2e- ŌåÆ Cd2+ (aq) + Cu(s)

Ecell = ( Eoox )+ ( Eredo )cathode

= 0.4 + 0.34

= 0.74V.

emf is +ve, so ŌłåG is (-)ve, the reaction is feasible.

19. In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidised at the anode and O2 at cathode. If 44.8 litre of H2 at 25oC and 1atm pressure reacts in 10 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu2+ , how many grams of Cu deposited?

Oxidation at anode:

2H2 (g) + 4OH- (aq) ŌåÆ 4H2O (l) + 4eŌłÆ

1 mole of hydrogen gas produces 2 moles of electrons at 25┬║ C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres

Ōł┤ no. of moles of hydrogen gas produced = [1 mole / (22.4 litres)] ├Ś 44.8 litres

= 2 moles of hydrogen

Ōł┤ 2 of moles of hydrogen produces 4 moles of electron i.e., 4F charge.

We know that Q= It

I= Q/t

= 4F / 10 mins

= 4├Ś96500 C / 10├Ś60 s

I=643.33 A

Electro deposition of copper

Cu2+ (aq)+2e- ŌåÆ Cu(s)

2F charge is required to deposit

1 mole of copper i.e., 63.5 g

If the entire current produced in the fuel cell ie., 4 F is utilised for electrolysis, then 2├Ś 63.5 i.e., 127.0 g copper will be deposited at cathode.

20. The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited in the first cell. The amount of Cr deposited in the another cell? Give : molar mass of Nickel and chromium are 58.74 and 52gm-1 respectively.

Ni2+ (aq) + 2e- ŌåÆ Ni (s)

Cr2+ (aq)+3eŌłÆ ŌåÆ Cr (s)

The above reaction indicates that 2F charge is required to deposit 58.7g of Nickel form nickel nitrate and 3F charge is required to deposit 52g of chromium.

Given that 2.935 gram of Nickel is deposited

Ōł┤The amount of charge passed through the cell = (2F/58.7 g) ├Ś 2.935g

= 0.1F

Ōł┤ if 0.1F charge is passed through chromium nitrate the amount of chromium deposited = 52g/3F ├Ś 0.1F

= 1.733g

21. A copper electrode is dipped in 0.1M copper sulphate solution at 25oC . Calculate the electrode potential of copper. [Given: ECu 2+|Cu = 0.34 V ].

Given that

[ Cu2+] = 0.1M

E┬║Cu2+ |Cu = 0.34

Ecell = ?

Cell reaction is

Cu2+ (aq) + 2eŌłÆ ŌåÆ Cu (s)

Ecell = Eo ŌĆō [0.0591/n] log {[Cu] [Cu2+]}

= 0.34 ŌĆō [0.0591/2] log (1/0.1)

= 0.34 ŌłÆ0.0296

= 0.31V

22. For the cell Mg (s) | Mg2+ (aq) || Ag+ (aq) | Ag (s), calculate the equilibrium constant at  25oC and maximum work that can be obtained during operation of cell. Given : E┬║ Mg 2+ | Mg = ŌłÆ2.37V and E┬║Ag + | Ag = 0.80V.

oxidation at anode

Mg ŌåÆ Mg2+ +2e-    .............. (1)   ( E┬║xo) = 2.37V

Reduction at cathode

Ag+ + e- ŌåÆ Ag ........... (2) ( E┬║red) = 0.80V

Ōł┤ E┬║cell = ( E┬║ox )anode + ( E┬║red )cathode

= 2.37+0.80

= 3.17V

Overall reaction

Equation (1) + 2 ├Ś equation ( 2 ) ŌćÆ

Mg+ 2Ag2+ ŌåÆ Mg2+ + 2Ag

ŌłåG = -nfE

=ŌłÆ2 ├Ś 96500 ├Ś 3.17

=ŌłÆ611810 J

ŌłåGo = ŌłÆ6.12 ├Ś105 J

W = 6.12 ├Ś105 J

ŌłåGo =ŌłÆ2.303 RT logKC

ŌłåGo =ŌłÆ2.303 RT logKC

ŌćÆ log Kc = (6.12 ├Ś105) /  (2.303 ├Ś 8.314 ├Ś 298)

Kc = Antilog of (107.2)

23. 8.2 ├Ś1012 litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of 2 ├Ś106 CsŌłÆ1 at an appropriate voltage. How many years would it like to completely electrolyse the water in the lake. Assume that there is no loss of water except due to electrolysis.

Electrolysis of water

At anode:

2H2O ŌåÆ 4H+ + O2 + 4eŌłÆ      ......... (1)

At cathode:

2H2 O+2eŌłÆ ŌåÆ H2 + 2OH-

Overall reaction 6H2O ŌåÆ 4H+ + 4OH- + 2H2 + O2

(or)

Equation (1) +(2) ├Ś2 ŌćÆ 2H2 O ŌåÆ 2H2 + O2

According to faradays Law of electrolysis, to electrolyse two mole of Water (36g = 36 mL of H2O), 4F charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = 4 ├Ś 96500 C.

When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to (4 ├Ś 96500 C / 36 mL)  ├Ś 9 ├Ś1012 L

= ( 4 ├Ś 96500 ├Ś 9 ├Ś1012 ) / ( 36 ├Ś10ŌłÆ3 ) C

= 96500 ├Ś1015 C

Ōł┤Given that in 1 second, 2 ├Ś106 C is generated therefore, the time required to generate

96500 ├Ś1015 C is = [ 1 S/  2 ├Ś106C]  ├Ś 96500 ├Ś1015 C

= 48250 ├Ś109 S

1 year = 365 days

= 365 ├Ś 24 hours

= 365├Ś24 ├Ś 60 min

1 year = 365├Ś24 ├Ś 60 ├Ś 60 sec.

Ōł┤ Number of years = 48250 ├Ś109 / 365├Ś24├Ś60├Ś60

= 1.5299 ├Ś106 years

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12th Chemistry : UNIT 9 : Electro Chemistry : Numerical Problems | Question with Answers, Solution | Electro Chemistry