Choose the correct answer

Chemistry : Electro Chemistry

EVALUATION

Choose the correct answer:

1. The number of electrons that have a total charge of 9650 coulombs is

a) 6.22 ×1023

b) 6.022 ×1024

c) 6.022 ×1022

d) 6.022 ×10−34

Solution:

1F = 96500 C = 1 mole of e− = 6.023 ×1023 e−

∴ 9650 C = [ 6.22 ×1023 / 96500 ] × 9650 = 6.022×1022

= 6.022×1022

Option (C)

2. Consider the following half cell reactions:

Mn2+ + 2e− → Mn   Eº = -1.18V

Mn2+ → Mn3+ + e-   Eº = -1.51V

The E for the reaction 3Mn2+ → Mn+2Mn3+ , and the possibility of the forward reaction are respectively.

a) 2.69V and spontaneous

b) -2.69 and non spontaneous

c) 0.33V and Spontaneous

d) 4.18V and non spontaneous

Solution:

Mn2+ + 2e− → Mn (Eºred) = -1.18V

2[Mn2+ → Mn3+ + e- ](Eºox) = −1.51V

3Mn2+ → Mn + 2Mn3+        Eºcell?

Eºcell = ( Eºox )+ ( Eºred )

= −1.51 −1.18 and non spontaneous

= −2.69V

Since Eo is –ve ∆G is +ve and the given forward cell reaction is non – spontaneous.

(Option (b))

3. The button cell used in watches function as follows

Zn (s) + Ag2 O (s) + H2 O (l) ↔ 2 Ag (s) + Zn2+ (aq) + 2OH− (aq) the half cell potentials are Ag2 O (s) + H2O (l) + 2e− → 2Ag (s) + 2 OH− (aq) E = 0.34V The cell potential will be

a) 0.84V

b) 1.34V

c) 1.10V

d) 0.42V

Solution:

Anodic oxidation: (Reverse the given reaction)

( Eoox ) = 0.76V cathodic reduction

∴ Ecell = ( Eoox )+ ( Eredo )

 = 0.76 + 0.34 = 1.1V

= 1.1V

(Option (c))

4. The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity of 5.76 ×10−3 S cm−1 at 298 K is

a) 2.88 S cm2 mol-1

b) 11.52 S cm2 mol-1

c) 0.086 S cm2 mol-1

d) 28.8 S cm2 mol-1

Solution:

Λ = κ/M ×10−3 mol−1 m3

= [( 5.76 × 10−3 S cm−1 ×10−3 ) / (0. 5)]  mol−1m3

= [(5.76 × 10−3 × 10−3 ×106 ) / (0.5)] S cm−1mol−1 cm3 .

= 11.52 S cm2 mol−1

(Option (b))

5.

Calculate ΛºHOAC using appropriate molar conductances of the electrolytes listed above at infinite dilution in water at 25oC .

a) 517.2

b) 552.7

c) 390.7

d) 217.5

Solution:

( Λº )HoAC = ( Λº)HCl + ( Λº)NaOAC - ( Λº)NaCl

= (426.2 + 91) −(126.5)

= 390.7

(Option (c))

6. Faradays constant is defined as

a) charge carried by 1 electron

b) charge carried by one mole of electrons

c) charge required to deposit one mole of substance

d) charge carried by 6.22 ×1010 electrons.

Solution:

1F = 96500 C = charge of 1 mole of e- =

charge of 6.022 ×1023 e−

(Option (b))

7. How many faradays of electricity are required for the following reaction to occur MnO-4 → Mn2+

a) 5F

b) 3F

c) 1F

d) 7F

Solution:

 7MnO-4 + 5e− → Mn2+ + 4H2O

5 moles of electrons i.e., 5F charge is required.

(Option (a))

8. A current strength of 3.86 A was passed through molten Calcium oxide for 41minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g / mol and 1F = 96500C).

a) 4

b) 2

c) 8

d) 6

Solution:

 m=ZIt

(41min 40sec = 2500 seconds)

Z = m/ (n × 96500) = 40 / (2 × 96500)

m = Zit

 = ( 40 × 3.86 × 2500 ) / (2 × 96500)

= 2g

 (Option (b))

9. During electrolysis of molten sodium chloride, the time required to produce 0.1mole of chlorine gas using a current of 3A is

a) 55 minutes

b) 107.2 minutes

c) 220 minutes

d) 330 minutes

Solution:

m=ZIt

(mass of 1 mole of Cl2 gas = 71)

t = m/ ZI  (∴mass of 0.1mole of Cl2 gas = 7.1 g mol−1 )

 = 7.1 / [71/(2 ×96500) × 3 ]

 (2 Cl- → Cl2 +2e- )

= (2 × 96500 × 7.1) / (71× 3)

 = 6433.33sec

 = 107.2 min

(Option (b))

10. The number of electrons delivered at the cathode during electrolysis by a current of 1A in 60 seconds is (charge of electron = 1.6 ×10−19 C )

a) 6.22 × 1023

b) 6.022 ×1020

c) 3.75 ×1020

d) 7.48 ×1023

Solution:

 Q =It

 = 1A×60S

96500 C charge ≡ 6.022 ×1023 electrons

60 C charge ≡ [ 6.022 ×1023 /  96500  ] × 60

 = 3.744 ×1020 electrons

(Option (C))

11. Which of the following electrolytic solution has the least specific conductance

a) 2N

b) 0.002N

c) 0.02N

d) 0.2N

Solution:

In general, specific conductance of an electrolyte decreases with dilution. So, 0.002N solution has least specific conductance.

(Option (b))

12. While charging lead storage battery

a) PbSO4 on cathode is reduced to Pb

b) PbSO4 on anode is oxidised to PbO2

c) PbSO4 on anode is reduced to Pb

d) PbSO4 on cathode is oxidised to Pb

Solution:

Charging : anode : PbSO4 (s)+ 2e− → Pb (s) + SO4-2 (aq)

Cathode : PbSO4 (s)+ 2H2 O (l) → PbO2 (s) + SO4-2 (aq)+2e−

(Option (C))

13. Among the following cells

I) Leclanche cell

II) Nickel – Cadmium cell

III) Lead storage battery

IV) Mercury cell

Primary cells are

a) I and IV

b) I and III

c) III and IV

d) II and III

Solution: Option (a)

14. Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because

a) Zinc is lighter than iron

b) Zinc has lower melting point than iron

c) Zinc has lower negative electrode potential than iron

d) Zinc has higher negative electrode potential than iron

Solution:

EoZn2+|Zn = − 0.76V and EoFe2+|Fe = −0.44V Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

Option (d)

15. Assertion : pure iron when heated in dry air is converted with a layer of rust.

Reason : Rust has the composition Fe3 O4

a) if both assertion and reason are true and reason is the correct explanation of assertion.

b) if both assertion and reason are true but reason is not the correct explanation of assertion.

c) assertion is true but reason is false

d) both assertion and reason are false.

Solution:

Both are false

i) Dry air has no reaction with iron

ii) Rust has the composition Fe2 O3 . x H2O

(Option (d))

16. In H2 -O2 fuel cell the reaction occurs at cathode is

a) O2 (g) + 2H2O (l) + 4e− → 4OH− (aq)

b) H+ (aq) + OH− (aq) → H2O (l)

c) 2H2 (g) + O2 (g) → 2H2O (g)

d) H+ + e− → 1/2 H2

17. The equivalent conductance of M/36 solution of a weak monobasic acid is 6 mho cm2 equivalent –1 and at infinite dilution is 400 mho cm2 equivalent –1. The dissociation constant of this acid is

a) 1.25 ×10−6

b) 6.25 ×10−6

c) 1.25 ×10−4

d) 6.25 ×10−5

Solution:

 α = Λ / Λo = 6/400

Ka =α2C

= 6/400 × 6/400 × 1/36

= 6.25 ×10−6

Option (b)

18. A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance ( κ =1.25 ×10−3 S cm−1 ) in the cell and the measured resistance was 800 W at 25º C . The cell constant is,

a) 10−1 c m−1

b) 101 c m−1

c) 1 c m−1

d) 5.7 ×10−12

Solution:

R = ρ. l/A

cell constant = R/ρ

= κ.R       (1/ ρ =κ )

= 1.25 × 10−3 Ω−1cm−1 × 800 Ω

= 1 cm−1

Option (c)

19. Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is 1 .85 ×10−5 S m−1 . Solubility product of the salt AB at 298K ( Λºm )AB = 14 × 10−3 S m2 mol−1 .

a) 5.7 ×10−12

b) 1.32 ×10−12

c) 7.5 ×10−12

d) 1.74 ×10−12

Solution: Option (d)

20. In the electrochemical cell: Zn | ZnSO4 (0.01M) || CuSO4 (1.0M) | Cu , the emf of this Daniel cell is E1. When the concentration of is changed to 1.0M and that CuSO4 changed to 0.01M, the emf changes to E2. From the above, which one is the relationship between E1 and E2?

a) E1 < E2

b) E1 > E2

c) E2 ≥ E1

d) E1 = E2

Solution:

Zn(s) → Zn2+ (aq) + 2e-

Cu2+ (aq)+2e- → Cu(s)

Zn(s) +Cu2+ (aq) → Zn2+ (aq) + Cu(s)

Ecell =Eºcell – (0.0591/2) log ( [zn2+]/ [Cu2+] )

E1 =Eºcell - 0.0591/2 log (10-2/1)

E1 =Eocell + 0.0591 ........(1)

E2 =Eºcell - 0.0591/2 log (1/10-2)

E2 =Eocell - 0.0591 .........(2)

E1>E2

Option (b)

21. Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

Then the species undergoing disproportionation is

a) Br2

b) BrO−4

c) BrO3-

d) HBrO

Solution:

( Ecell )A =-1.82+1.5=-0.32V

( Ecell )B =-1.5+1.595=+0.095V

( Ecell )c =-1.595+1.0652=-0.529V

The species undergoing disproportionation is HBrO

(Option D)

22. For the cell reaction

2Fe3+ (aq) + 2l−(aq) → 2Fe2+ (aq) + l2 (aq)

Eocell = 0.24V at 298K. The standard Gibbs energy (∆, Gº) of the cell reactions is :

a) -46.32 KJ mol−1

b) -23.16 KJ mol−1

c) 46.32 KJ mol−1

d) 23.16 KJ mol−1

Solution: Option (a)

23. A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time through copper sulphate solution

a) 31.75

b) 15.8

c) 7.5

d) 63.5

Solution: Option (b)

24. A gas X at 1 atm is bubbled through a solution containing a mixture of 1MY- and 1MZ- at 25oC . If the reduction potential of Z>Y>X, then

a) Y will oxidize X and not Z

b) Y will oxidize Z and not X

d) Y will oxidize both X and Z

d) Y will reduce both X and Z

Solution: Option (a)

25. Cell equation : A + 2B- → A2+ +2B;

A2+ + 2e− → A  ; Eº = + 0.34V and log10K = 15.6 at 300K for cell reactions find E for B+ + e − → B

a) 0.80

b) 1.26

c) -0.54

d) -10.94

Solution: Option (a)

PTA Question Oneword:

1. Assertion : A small piece of Zinc dissolved in dilute nitric acid but hydrogen gas not evolved.

Reason: HNO₃ is an oxidizing agent and this oxidizes hydrogen.

a) Both Assertion and Reason is true. Reason is a correct explanation for assertion.

b) Both Assertion and Reason is true. Reason is not correct explanation for the assertion.

c) Assertion is correct but Reason is wrong.

d) Both assertion and Reason are wrong.

Answer: c)

2. Which of the following statement is not correct with respect to electrolytic conductance?

a) Conductivity increases with the decreases in Viscosity

b) Conductivity increases with increase in temperature

c) Molar conductance of a solution decreases with increase in dilution

d) Conductance decrease with increase in temperature

Answer: c)

3. Which of the following is Secondary cell?

a) Laclanche cell

b) Lithium ion battery

c) Mercury button cell

d) both (a) and (c)

Answer: b)

4. The general representation of a fuel cell is

a) Fuel/Electrode/Electrolyte/Electrode/Oxidant

b) Oxidatant/Electrode/Electrolyte/Electrode/Fuel

c) Fuel/Electrode/Electrolyte/Electrode/Reductant

d) Oxidant/Electrode/Electrolyte/Electrode/Reductant

Answer: a)

5. Laptops have

a) Lead storage battery

b) fuel cell

c) Mercury button cell

d) Lithium ion-battery

Answer: d)