CBSE Class 12 Chemistry Notes: Electrochemistry – Important Question Answers

  1. Calculate charge (in coloumbs) on 1 gram ion of O2.

CBSE Class 12 Chemistry Notes: Electrochemistry    Important Question Answers Image by AglaSemAns. ⇒ 1 gram ion O2-
⇒ 1 mol O2-
⇒ 1 mol × (2e-)
⇒ 2 mol O-
⇒ 2 (1 mole e-)
⇒ 2 × I F
⇒ 2 × 96500 C
⇒ 193000 C
= 193 × 105 C 5

  1. How long a current of 1.5 A has to be passed through a solution of AgNO3 to coat a metal surface of 40 cm2 using a 0.005 mm thich layer? Density of silver is 10.5g cm3

Ans. Volume of silver coated = A × b
Mass of Ag depainted = A × b × d
(A × b × d/At wt) × 1 × 96500 = I × t
(40 × 5 × 10-4 × 10-5/108.5) × 1 × 96500 = 1.5 × t (sec)

  1. A 100 watt 110 volt incondescent lamp is connuted in series with on electrolyte cell containing codmium sulphate solution. What weight of cadmium will be deposited by the current flowing for 10 hours.

Ans. P = 100 Watt
V = 110 Volts
P = V × I
I = (P/V) = 100/110 = 10/11
10 hrs = 10 × 3600 = 36 × 103 sec
(mass/Al wt) × N.F × 96500 = It
M = 19060

  1. An aqueous solution of NaCl on electrolysis gives H2(g)t and NaOH as given:
    2Cl(aq) + 2H2O → 2OH-(aq) + H2(g) + Cl
    A direct current of 25A (efficincy 62%) is passed the 20L of NaCl solution (20% by wt). How long will it take to produce 1 kg Cl2?
    (assume no LOSS due to evaporation).

Ans.
CBSE Class 12 Chemistry Notes: Electrochemistry    Important Question Answers Image by AglaSem

  1. A current of 1.70 A is passed through 300 mL of 0.160 M of ZnSO4 for 230 seconds with a current efficiency of 90 % Find out the molarity of Zn2+ after the deposition of Zn.

Ans. I = 1.7A × 0.9
t = 230S
mass × VF × 96500 = T× +

CBSE Class 12 Chemistry Notes: Electrochemistry    Important Question Answers Image by AglaSem

x m (x < 0.16)
x = n
n × 2 × 96500 = 0.9 × 17 × 230
x = (n original – n deposit)/0.3
n (deposit) = (0.9 × 17 × 230)/2 × 96500
x = (0.10 × 0.3 – (0.9 × 17 × 230)/2 × 96500)/0.3

  1. NH4ClO4, which is used in the solid fuel in the booster rockets on the space shuttle is produced commercially from NaClO4, which is produced commercially by electrolysis of NaCl solution
    NaCl + 4H2O → NaCl4 + 4H2
    How making faradays are used to produced 1.00 kg of NaClO4?

Ans. N factor for NaClO4 is 8
4H2O + Cl- → Cl-O4 + 8H+ + 8e-
(1000/122.5) × 8 = 64 F

  1. How the ELECTROLYSIS of malten NaCl and aqueuos NaCl are “different”?

Ans.

  • The products obtained on the electrodes depend on their oxidation and reduction potentials.
  • In molten salt, mostly the anions and cations of the salt, only undergo the change, where as with aqueous soln; water molecules also compete with the anions a cations for oxidation and reduction on electrodes.
  • On anode that speices which has higher oxidation potential undergo the charge, where as cathode that species which has higher reduction potential undergo change.
  • In many cases over potential for a species deviate it from it’s standard oxidation and reduction potentials.
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