Electrolytic Cells and Electrolysis
Electrical energy plays a significant role in great quantities of chemical reactions. The field of science that engages the production of electricity from the energy released during spontaneous chemical reactions and the use of electrical energy to accomplish non-spontaneous chemical changes is called electrochemistry.
The basis of these types of processes is redox reactions, which we have learned in the previous class. A large number of chemical and biological reactions are redox reactions. They are used to obtain energy for domestic, transport, or industrial purposes, to burn fuel for the digestion of food in animals; Several industrial processes for photosynthesis to obtain energy from the Sun, extracting metals from their ores and manufacturing important chemicals, operating dry and wet batteries, fuel cells, etc.
Reactions carry through electrochemically perchance energy efficient and less polluting. Thereupon, the study of electrochemistry is important for the creation of new environmentally friendly technologies. Therefore, electrochemistry is a very massive and interdisciplinary subject. Redox reactions keep us alive.
Electrolysis is the procedure of decomposition of an electrolyte by the passage of electricity throughout its aqueous solution or the molten state of an electric current. This cell is utilized to perform electrolysis which is electrolyte cells.
Water, for example, can be electrolyzed (with the help of an electrolytic cell) to produce gaseous oxygen and hydrogen. This is accomplished by utilising the flow of electrons (into the reaction environment) to overcome the non-spontaneous redox reaction’s activation energy barrier.
The following are the three major components of electrolytic cells:
- The cathode (which is negatively charged for electrolytic cells)
- The anode is a type of anode that is (which is positively charged for electrolytic cells)
- Electrolyte: The electrolyte serves as a conduit for electrons to flow between the cathode and the anode. Water (containing dissolved ions) and molten sodium chloride are common electrolytes in electrolytic cells.
Working of an Electrolytic Cell
- The process of electrolysis is done by taking a solution of electrolyte in a suitable vessel. The vessel is called an electrolytic tank. It is made of either glass or a material that is a poor conductor of electricity. Two metal rods or plates are suspended in an electrolytic solution. These are connected to the terminals of the battery with the help of metal wires. These metallic rods or plates allow the flow of electric current and are called electrodes. The electrode associated with the positive terminal of the battery is termed the anode while the electrode connected to the negative terminal of the battery is called the cathode.
- When an electrolyte is dissolved in water, it splits into negative and positive ions. Positively charged ions are termed cations and negatively charged ions are termed as anions. On passing an electric current throughout the solution. The ions are enticed by the oppositely charged electrodes. For that reason, cations move towards the cathode while anions move towards the anode. This movement of ions in the solution is known as electrolytic or ionic conduction and is the flow of current through the solution.
- On reaching the anode, the ions give up their electrons (which are expelled from the solution). On the contrary, cations take up the electron the cathode. Therefore, the cation and anion are again discharged at the electrode and converted into neutral particles. This is known as primary change. The primary products may be aggregated in this way or they may be further transformed into molecules or compounds. These are known as secondary products and the change is termed secondary change.
- The process of electrolysis was explained by the principle of ionization. According to ionic theory, electrolytes exist as ions in solution and the function of electricity is simply to direct these ions towards their respective electricity. Electrolytes can be electrolyzed only in the dissolved or molten state.
- One of the straightforward electrolytic cells consists of two copper rods dipped in an aqueous solution of copper sulphate. When an electric current (DC voltage) is applied to the two electrodes, Cu2+ ions discharge at the cat (negatively charged electrode) and the following reaction occurs-
Cu2+ (aq) + 2e– ⇒ Cu(s)
- Copper metal is deposited at the cathode. In mode, copper is converted to Cu2+ by the reaction
Cu(s) ⇒ Cu2+(aq) + 2e–
- Thus, copper is dissolved at the anode. In terms of oxidation and reduction, we can say that copper is oxidized at the anode while it is reduced at the cathode. They are the foundation of an important industrial process in which impure copper is converted into high purity copper with many other metals such as Na, Mg, Al, etc., They are also generated on a massive scale by electrochemical reduction of their respective cations where no suitable chemical reducing agents are available for reduction. This is one of the important methods in metallurgy.
- By way of illustration, sodium, and magnesium metals are produced by the electrolysis of their fused chlorides, and aluminium is produced by the electrolysis of aluminium oxide in the presence of ergoline (Na3AIF6).
The cell which converts electrical energy into chemical energy is called an electrolytic cell. This oxidation occurs at the anode and is a positive plate while reduction occurs at the cathode and is a negative plate. In electrolytic cells, electrical energy is used to perform non-spontaneous chemical reactions and the process that takes place in an electrolytic cell is called electrolysis.
Electrolysis of NaCl Solution
Na+, Cl–, H+, and OH– ions are present in the aqueous solution of NaCl due to the ionisation of water and the presence of NaCl. Water has a low ion population because it is a weak electrolyte. When a potential difference between the two electrodes is reached during a reaction, Na+ and H+ ions migrate towards the negative electrode, the cathode, while Cl– and OH– ions move towards the positive electrode, the anode.
Between H+ and Na+ ions at the cathode, H+ ions are discharged, but Cl– ions are discharged in preference to OH– ions on the same ground.
NaCl ⇌ Na+ + Cl–
H2O ⇌ H+ + OH–
- H+ + e– → H
- 2H → H2
- Cl– → Cl + e–
- 2Cl → Cl2
As a result, the Na+ and OH– ions stay undisturbed in the aqueous solution, resulting in NaOH crystals when the solution is evaporated.
Faraday laws of Electrolysis
Faraday’s First Law of Electrolysis
The amount of chemical reaction manifested by passing current to any electrode during electrolysis is proportional to the amount of electricity flowing through the electrolyte (in solution or molten state).
w ∝ Q
Q = ZQ
w = Zit
- Q = quantity of electricity
- t = current (A),
- t = time (sec),
- Z=constant proportionality called electrochemical equivalent.
Faraday’s Second Law of Electrolysis
The amount of dissociated substances that can pass through the same amount of electricity using an electrolytic solution is proportional to their chemical equivalent mass, i.e.-
W1/E1 = W2/E2 = W3/ E3 …….
- W1 = mass of substance 1 deposited
- E1 = equivalent weight
Applications of Electrolysis
- Caustic soda is prefabricated by electrolysis of sodium chloride solution
- manufacture of O2 and H2
- Therapeutic supplication
Problem 1. How many coulombs are needed for 40.5 g of aluminium to react when the electrode is:
Al3+ + 3e– ⇒ Al
1 mol of Al requires 3 mol of electrons or 3 × 96500 C
1 mol of Al = 27g
27g of Al require =3 × 96500 C
40.5g of Al require =(3*96500C × 40.5)/27 = 434,250 C
Problem 2. In the electrolysis of acidic water, it is desired to obtain hydrogen at 1cc sec at the STP position. What should be the current pass?
2H+ + 2e– ⇒ H2
1 mol of H2 or 22400 cc of H2 at STP requires = 2 × 96500 C
1cc of H2 at NTP requires = (2×96500)/22400 = 8.616 C
Now, Q = I × t
I = Q/t = 8.616/1s
= 8.616 ampere
Problem 3. How many moles of mercury will be produced by galvanic isolation 1.0 M Hg(NO3)2 solutions with a current of 2.00 A for 3 hours?
Hg2+ + 2e– ⇒ Hg
Quality of electricity passed = I × t(sec)
= 2.0 A × 3.0 × 60 × 60 = 21600 C
2 × 96500 C of electricity produce mercury = 1 mol
21600 C of electricity will produce mercury = 1× 21600 / (2×96500) = 0.112mol
Problem 4. A solution of CuSO is electrolyzed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Current strength (I) = 1.5 A
Time (t) = 10 min = 10 x 60 = 600s
amount of electricity passed = I x t = (1.5 A) x (600 s) = 900 C (A s = C)
Copper is deposited as: Cu +2e7 → Cu(s)
2 mol of electrons or 2 x 96500 C of current deposit copper = 63.56 g
900 C of current will deposit copper = 63.56/(2 ×96500) = 0.296 g
Problem 5. Calculate how long it will take to deposit 1.0 g of chromium when a current of 1.25 A flows through a solution of chromium (III) sulphate. (Molar mass of Cr=62).
Cr3++3e– → Cr(s)
3 mol of electricity are needed to deposit 1 mol of Cr,
52 g of Cr require current = 3 x 96500 C
1g of Cr will require current = (3 x 96500)/52 =5567.3 C
number of coulombs = Current x t
Time (s) required = No. of coulombs / Current
Time (s) required = 5567.3 C / 1.25 (Ampere)
= 4453.8s or 1.24hr
Problem 6. How many hours does it take to reduce 3 mol of to Fe with 2.0 A current? (F= 96500 C) Solution: Reduction of Fe³+ to Fe²+.
Fe3+ +e– →Fe²+
Reduction of 1 mol of Fe3+ requires = 96500 C
Reduction of 3 mol of Fe³+ require = 3 × 96500 C = 2.895 x 10°C
Quantity of electricity = Current × Time
2.895 x 10 = 2 x Time
Time = 2.895×105 = 14475×105 s
Time= (14475 x 105 )/(60 x 60) = 40.21 hours
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