Table of contents | |
What is an Electrolytic Cell? | |
Working of an Electrolytic Cell | |
What is Electrolysis? | |
Products of Electrolysis | |
Some Important Questions |
An electrolytic cell can be defined as an electrochemical device that uses electrical energy to facilitate a non-spontaneous redox reaction.
The three primary components of electrolytic cells are:
- Cathode
(which is negatively charged for electrolytic cells)- Anode
(which is positively charged for electrolytic cells)- Electrolyte
(The electrolyte provides the medium for the exchange of electrons between the cathode and the anode. Commonly used electrolytes in electrolytic cells include water (containing dissolved ions) and molten sodium chloride.
Components of Electrolytic Cell
An Electrolytic Cell is an electrochemical device that uses an external source of electrical energy to drive a non-spontaneous chemical reaction. It's the opposite of a galvanic or voltaic cell, which spontaneously generates electrical energy from a chemical reaction.
Here's how an electrolytic cell works:
Working of an Electrolytic Cell
Anode | Positive | Loss of electron or oxidation takes place | Positive Current enters |
Cathode | Negative | Gain of electron or reduction takes place | Current leaves |
Reaction at Cathode: Na+ + e– → Na
Reaction at Anode: 2Cl– → Cl2 + 2e–
Cell Reaction: 2NaCl → 2Na + Cl–
The decomposition of electrolyte solution by passage of electric current, resulting in deposition of metals or liberation of gases at electrodes is known as electrolysis.
Electrolysis
Key characteristics of electrolysis include:
These laws describe the relationship between the amount of substance deposited or produced during an electrolysis process and the quantity of electricity (electric charge) passed through the system.
The amount of substance deposited or liberated at an electrode is directly proportional to the amount of charge passed (utilized) through the solution.
W ∝ Q W = ZQ |
where
when Q = 1 coulomb, then W = Z
Thus, weight deposited by a 1-coulomb charge is called electrochemical equivalent.
Let 1-ampere current is passed till 't' seconds.
Then, Q = It => W = ZIt
W = ZIt |
1 Faraday = 96500 coulomb = Charge of one mole electrons
One faraday is the charge required to liberate or deposit one gm equivalent of a substance at the corresponding electrode.
Let 'E' is equivalent weight then 'E' gm will be liberated by 96500 coulombs.
1 Coulomb will liberate gm ; By definition, Z =
W =
When gas is evolved at an electrode, then the above formula changes as,
V =
where V = volume of liberated gas, Ve = equivalent volume of gas.
Equivalent volume may be defined as:
The volume of gas liberated by 96500 coulombs at STP.
Hence, According to Faraday's first law of electrolysis:
Where:
When the same amount of charge is passed through different electrolyte solutions connected in series then the weight of substances deposited or dissolved at anode or cathode is in the ratio of their equivalent weights. i.e.
w1/w2 = E1/E2 |
Q.1. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is:
a) 9.65 x 104 s
b) 28.95 x 104 s
c) 19.3 x 104 s
d) 38.6 x 104 s
Solution:
Mass of H2 = m = 0.01 mol x 2 g mol-1
Equivalent weight of H2 = E = Atomic weight / valence = 1 g mol-1 / 1 = 1 g mol-1
Current in ampere = I = 10 milliamperes = 10 x 10-3 amperes.
Conclusion: The correct option is 'c'.
Q.2. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of electrodes (relative atomic mass of Na = 23, Hg = 200; 1 Faraday constant = 96500 Coulombs mol-1) :
The total number of moles of chlorine gas evolved is:
a) 3.0
b) 2.0
c) 0.5
d) 1.0
Solution:
First we have to calculate the amount of NaCl present in 4.0 molar solution.
The number of moles of NaCl in 500 mL of 4.0 molar solution = Molarity x Volume (in L) = 4.0 x 500 x 10-3 = 2.0 moles.
The following reaction occurs during electrolysis of aqueous solution of NaCl.Dissociation of NaCl: 2 NaCl(aq) ------> 2Na+(aq) + 2Cl-(aq)
At cathode: 2H2O + 2e- -------> H2 + 2OH-
At anode: 2Cl-(aq) ---------> Cl2(g) + 2e-
Complete reaction: 2NaCl(aq) + 2H2O ------>H2 + Cl2(g) + 2Na+(aq) + 2OH-(aq)
That means, two moles of NaCl gives one mole of Cl2 gas.
Since there are 2.0 moles of NaCl present in the solution, one mole of Cl2 gas will be evolved at anode upon complete electrolysis.
Conclusion: The correct option is 'd'.
Electrolysis of molten sodium chloride (NaCl) is a chemical process that involves the decomposition of sodium chloride into its constituent elements, sodium (Na) and chlorine (Cl), using electricity. Here's how the electrolysis of molten sodium chloride works:
Na+ accumulates at negatively charged electrode and Cl- ions will surround positively charged electrode
Overall Reaction: The net result of the electrolysis is the decomposition of sodium chloride into sodium metal and chlorine gas:
2NaCl(l) → 2Na(l) + Cl₂(g)
Standard Potential (SRP) Values: To determine which ion is discharged preferentially, the concept of standard potential is used. Standard reduction potential (SRP) values for cations are compared, with the cation having a higher SRP value being discharged preferentially, provided the ions are at a concentration of 1 M.
Standard Oxidation Potential (SOP) Values: For anions, standard oxidation potential (SOP) values are compared. Anions with higher SOP values are preferentially discharged, again assuming a 1 M concentration for each ion. The standard potential values are determined under specific conditions: 1 M concentration, 1 atm pressure for gases, and a temperature of 25°C.
Answer: H2, Cl2
Solution:
Q.2. What is the product formed at the cathode in the electrolysis of aqueous CuSO4?
a) Copper metal
b) Oxygen gas
c) Hydrogen gas
d) Sulphur
Answer: a
Solution: In the electrolysis of aqueous CuSO4, Cu2+, SO42+, H+ and OH– are the ions formed after dissociation. Copper ions have much higher reduction potential than water. Hence, these ions are easily reduced and deposited as Cu at the cathode
1. Products of electrolysis depend on the material being electrolyzed. In other words, the nature of electrolyte governs the process of electrolysis. The process is fast for a strong electrolyte whereas for a weak electrolyte an extra potential better known as overpotential is required.
Products of electrolysis depend on upon the value of this over potential too.
2. Products of electrolysis depend on the nature of electrodes too. That is, in the case of the inert electrode (say gold, platinum), it doesn’t participate in the reaction whereas if the electrode used is reactive in nature it takes part in the reaction.
3. Various oxidising and reducing species present in the electrolytic cell do affect the products of electrolysis.
4. The products of electrolysis depend on standard electrode potentials of the different oxidizing and reducing species present in the electrolytic cell.
5. In the case of multiple reactions, the product of electrolysis depends on the standard electrode potential of various reactions taking place. For example, electrolysis of an aqueous solution of sodium chloride. Out of the multiple reduction reactions taking place, the reduction reaction which has the highest value of standard electrode potential takes place at the cathode. Similarly, out of the multiple oxidation reactions, the oxidation reaction which has the lowest value of standard electrode potential takes place at the anode.
The SRP Values at 25º C for some of the reduction half-reaction are given in the table below.
S. NO. | Reduction half cell reaction | Eo in volts at 250 |
1. | F2 + 2e- →→ 2F- | + 2.65 |
2. | S2O82- + 2e- → 2SO42- | + 2.01 |
3. | Co3+ + e- → Co2+ | + 1.82 |
4. | PbO2 + 4H+ + SO42- + 2e- → PbSO4 + 2H2O | + 1.65 |
5. | MnO4- + 8H+ + 5e- → Mn2+ + 4H2O | + 1.52 |
6. | Au3+ + 3e- → Au | + 1.50 |
7. | Cl2 + 2e- → 2Cl- | + 1.36 |
8. | Cr2O2-7 + 14 H+ + 6e- → 2Cr3+ + 7H2O | + 1.33 |
9. | O2 + 4H+ + 4e- → 2H2O | + 1.229 |
10. | Br2 + 2e- → 2Br- | + 1.07 |
11. | NO3- + 4H+ + 3e- → NO + 2H2O | + 0.96 |
12. | 2Hg2+ + 2e- → Hg22+ | + 0.92 |
13. | Cu2+ + I- + e- → CuI | + 0.86 |
14. | Ag+ + e- → Ag | + 0.799 |
15. | Hg22+ + 2e- → 2Hg | + 0.79 |
16. | Fe3+ + e- → Fe2+ | + 0.77 |
17. | I2 + 2e- → 2I- | + 0.535 |
18. | Cu+ + e- → Cu | + 0.53 |
19. | Cu2+ + 2e- → Cu | + 0.34 |
20. | Hg2Cl2 + 2e- → 2Hg + 2Cl- | + 0.27 |
21. | AgCl + e- → Ag + Cl- | + 0.222 |
22. | Cu2+ + e- → Cu+ | + 0.15 |
23. | Sn4+ + 2e- → Sn2+ | + 0.13 |
24. | 2H+ + 2e- → H2 | + 0.00 |
25. | Fe3+ + 3e- → Fe | - 0.036 |
26. | Pb2+ + 2e- → Pb | - 0.126 |
27. | Sn2+ + 2e- → Sn | - 0.14 |
28. | Agl + e- → Ag + I- | - 0.151 |
29. | Ni2+ + 2e- → Ni | - 0.25 |
30. | Co2+ + 2e- → Co | - 0.28 |
31. | Cd2+ + 2e- → Cd | - 0.403 |
32. | Cr3+ + e- → Cr2+ | - 0.41 |
33. | Fe2+ + 2e- → Fe | - 0.44 |
34. | Cr3+ + 3e- → Cr | - 0.74 |
35. | Zn2+ + 2e- → Zn | - 0.762 |
36. | 2H2O + 2e- → H2 + 2OH- | - 0.828 |
37. | Mn2+ + 2e- → Mn | - 1.18 |
38. | Al3+ + 3e- → Al | - 1.66 |
39. | H2 + 2e- → 2H- | - 2.25 |
40. | Mg2+ + 2e- → Mg | - 2.37 |
41. | Na+ + e- → Na | - 2.71 |
42. | Ca2+ + e- → Ca | - 2.87 |
43. | Ba2+ + 2e- → Ba | - 2.90 |
44. | Cs+ + e- → Cs | - 2.92 |
45. | K+ + e- → K | - 2.93 |
46. | Li+ + e- → Li | - 3.03 |
Q.1. What is the product formed at the cathode in the electrolysis of molten NaCl?
a) Chlorine gas
b) Sodium metal
c) Hydrogen gas
d) Oxygen gas
Answer: b
Explanation: In the electrolysis of NaCl, if the electrolyte is molten NaCl, then the only ions formed after dissociation are Na+ and Cl– ions. The cathode being a negatively charged electrode attracts the positive Na+ ions and neutralizes it to form Sodium metal.
Q.2. What is the product formed at the cathode in the electrolysis of aqueous Na2SO4?
a) Sodium metal
b) Oxygen gas
c) Hydrogen gas
d) Sulphur
Answer: c
Explanation: Na2SO4 dissociates into Na+ and SO42- ions in the electrolysis of aqueous Na2SO4. Na+ has much lower reduction potential than water and hence Na+ ions are not reduced at the cathode. Instead, reduction of water occurs giving out hydrogen gas at the cathode.
Q.3. What is the product formed at the cathode in the electrolysis of aqueous CuSO4?
a) Copper metal
b) Oxygen gas
c) Hydrogen gas
d) Sulphur
Answer: a
Explanation: In the electrolysis of aqueous CuSO4, Cu2+, SO42+, H+ and OH– are the ions formed after dissociation. Copper ions have much higher reduction potential than water. Hence, these ions are easily reduced and deposited as Cu at the cathode.
Q.4. What are the two electrodes used in Daniell cell?
a) Pt and Cu
b) Al and Zn
c) Al and Pt
d) Zn and Cu
Answer: d
Explanation: The two electrodes that are used in a Daniell cell are zinc (as anode) and copper (as cathode) electrodes which are dipped in a solution containing its own ions, generally zinc sulphate and copper sulphate.
Q.5. What is the electrolyte used in the electroplating of gold?
a) Molten gold
b) [AgCN2]–
c) AuCN
d) AuCl3
Answer: c
Explanation: The electrolyte in electrolysis should contain the metal to be coated, gold in this case. AuCN is used because it is exceptionally stable and doesn’t resist the flow of Au+ ions from anode to cathode.
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1. What is an electrolytic cell? |
2. How does an electrolytic cell work? |
3. What is electrolysis? |
4. What are Faraday's Laws of Electrolysis? |
5. What are the products of electrolysis? |
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