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Chapter: 12th Chemistry : UNIT 9 : Electro Chemistry

Numerical Problems

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

Short Answer Questions


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 Ω . Find the specific conductance of the solution.

Solution

l = 1 5. cm = 1.5×10-2 m

A = 4.5 cm2 = 4.5×(10-4)m2

R = 15Ω

κ = 1/R (l/A)

κ = [ 1/15Ω ] × [1.5×10-2 m  / 4.5×(10-4)m]

= 2.22 Sm-1

Example

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

Molar conductance = Λ m =


 = 481.6 ×10-5 Sm2mol-1

Example

The resistance of a conductivity cell is measured as 190 Ω 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 6.3KΩ. Both these measurements are made at a particular temperature. Calculate the specific and molar conductance of NaCl solution.

Given that

κ = 1.3 Sm-1 (for 0.1M KCl solution)

R = 190 Ω


Λm = 13.04 × 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+ +e→ Ag (reduction)



 

Book Back Questions and Answers


8. The conductivity of a 0.01M solution of a 1 :1 weak electrolyte at 298K is 1.5 ×10-4 S cm1.

i) molar conductivity of the solution

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

Given that

λºcation = 248.2 S cm2 mol1

λºcation = 51.8 S cm2 mol1

Answer:

Given

C = 0.01M 

λºcation = 248.2 S cm2 mol−1

K= 1.5 × 10−4 S cm−1

λºanion = 51.8 S cm2 mol−1.

1. Molar conductivity

K = 1.5 ×10−4 S cm−1

Λºm =  K(sm−1 ) ×10−3 / C (in M) mol-1 m3

= (1.5 ×102 ×10−3 ) / 0.01 S mol −1 m2      1 cm−1 = 102m−1

= 1.5 ×10−3 S m2 mol−1 = 1.5 ×102

2. Degree of dissociation α = Λº/Λº

Λo =λocation +λaniono

= (248.2 + 51.8) S cm2 mol1

= 300 S cm2 mol1

= 300 ×1014 s m2 mol1

α = 1.5 ×103 S m2 mol1  / 300×104 S m2 mol1

α = 0.05

Ka = α2c / 1-α

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

= 25 ×10−4 ×10−2 /  95 ×10−2

= 0.26 ×10−4

= 2.6 ×10−5.

 

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%.

Answer:

Given

I = 1.608A; t = 50 min = 50 × 60

= 3000S

 η = 100%

V=250 mL

C=0.5M

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

Q=It

 Q = 1.608 ×3000

Q = 4824C

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

Electrolysis of CuSO4

Cu2+ (aq)+2e Cu(s).

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

0.05F electricity will

deposit  1 mol /2F × 0 .05F = 0.025 mol

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

= 0.125 mol

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

= 0.1 mol

Concentration of Cu2+ = [ 0.1 mol/ 250 mL ] ×1000 mL

=0.4M

 

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

Given: EFe 3+|Fe2+ = 0.771

EBr2|Br = 1.09V.

Answer:

Required half cell reaction

2 Br- Br2 + 2e

( Eoox )= −1.09V

2 Fe3+ + 2e 2Fe2+

( Eored )= +0.771V

2Fe3+ + 2Br 2Fe2+ +Br2

( Eocell )= ?

Eocell = ( Eoox )+ ( Eored )

 = −1.09 + 0.771

0.319V

Ecell is – ve; ∆G 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 = −0.44V .

Answer:

 ( 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.

Answer:

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 EºM12+|M1 = − 2.3V and E EºM22+|M2 = 0.2V  Predict which one is better for coating the surface of iron. Given : E Fe2+ |Fe = −0.44V

Answer:

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.

Answer:

Cell reactions:

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

( Eoox) Cu|Cd2+ = 0.4V

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

( Eored )Cu2+|Cu = 0.34V

Cd ( s ) + 2e- Cd2+ (aq) + Cu(s)

Ecell = ( Eoox )+ ( Eredo )cathode

= 0.4 + 0.34

= 0.74V.

emf is +ve, so ∆G 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?

Answer:

Oxidation at anode:

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

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

no. of moles of hydrogen gas produced = [1 mole / (22.4 litres)] × 44.8 litres

 = 2 moles of hydrogen

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

= 4×96500 C / 10×60 s

I=643.33 A

Electro deposition of copper

Cu2+ (aq)+2e- → 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× 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.

Answer:

Ni2+ (aq) + 2e- Ni (s)

Cr2+ (aq)+3e 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

The amount of charge passed through the cell = (2F/58.7 g) × 2.935g

= 0.1F

if 0.1F charge is passed through chromium nitrate the amount of chromium deposited = 52g/3F × 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 ].

Answer:

Given that

[ Cu2+] = 0.1M

EºCu2+ |Cu = 0.34

Ecell = ?

Cell reaction is

Cu2+ (aq) + 2e → Cu (s)

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

= 0.34 – [0.0591/2] log (1/0.1)

 = 0.34 0.0296

 = 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 : Eº Mg 2+ | Mg = −2.37V and EºAg + | Ag = 0.80V.

Answer:

oxidation at anode

Mg → Mg2+ +2e-    .............. (1)   ( Eºxo) = 2.37V

Reduction at cathode

Ag+ + e- → Ag ........... (2) ( Eºred) = 0.80V

Eºcell = ( Eºox )anode + ( Eºred )cathode

= 2.37+0.80

= 3.17V

Overall reaction

Equation (1) + 2 × equation ( 2 )

Mg+ 2Ag2+ Mg2+ + 2Ag

G = -nfE

=−2 × 96500 × 3.17

=−611810 J

Go = −6.12 ×105 J

W = 6.12 ×105 J

Go =−2.303 RT logKC

Go =−2.303 RT logKC

log Kc = (6.12 ×105) /  (2.303 × 8.314 × 298)

Kc = Antilog of (107.2)

 

23. 8.2 ×1012 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 ×106 Cs1 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.

Answer:

Electrolysis of water

At anode:

2H2O 4H+ + O2 + 4e−      ......... (1)

At cathode:

2H2 O+2e H2 + 2OH-

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

(or)

Equation (1) +(2) ×2 2H2 O 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 × 96500 C.

When the whole water which is available on the lake is completely electrolysed the amount of charge generated is equal to (4 × 96500 C / 36 mL)  × 9 ×1012 L

= ( 4 × 96500 × 9 ×1012 ) / ( 36 ×103 ) C

= 96500 ×1015 C

Given that in 1 second, 2 ×106 C is generated therefore, the time required to generate

 96500 ×1015 C is = [ 1 S/  2 ×106C]  × 96500 ×1015 C

= 48250 ×109 S

1 year = 365 days

= 365 × 24 hours

= 365×24 × 60 min

1 year = 365×24 × 60 × 60 sec.

Number of years = 48250 ×109 / 365×24×60×60

 = 1.5299 ×106 years 

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