IONIC THEORY
Ionic theory as proposed by Arrhenius states that when an ionic compound is dissolved in water or melted, some or all its particles dissociate (break up) into free moving charged particles called ions. This dissociation into ions is called ionization.
These free ions move randomly in all directions inside the solution. as seen in fig. I
But the ions lose their freedom as soon as an electric current is passed through the solution and become orderly, surrounding themselves around the opposite pole/electrode as they begin to take electrons in and out of the solution through the electrodes.
Electrolysis is defined as the chemical decomposition of a compound by the passage of electricity through the solution of the compound or its molten form.
Terms commonly used in Electrolysis
i. ELECTROLYTE: An electrolyte is a compound which allows the passage of electricity through its solution or its molten state and is decomposed in the process.
Examples of electrolytes include dilute Acids and Alkalis and all electrovalent compounds like NaCl.
Electrolytes are grouped in two
1. Strong Electrolytes: These are compounds which ionize completely in solutions.
They usually have large amounts of ions in solution and hence are good conductors.Examples are all sodium and potassium salts, minerals acids, and caustic alkalis.
NaCl(aq) →Na+(aq) + Cl-(aq)
Weak Electrolytes: These are compounds that ionize partially in solution
There is slight dissociation of the ions in dilute solutions, and so they contain less ions in solution. Examples include organic acids, aqueous ammonia, etc.
CH3COOH(aq) →CH3COO-(aq) + H+(aq)
Non-Electrolytes: These are compounds which do not conduct electricity at all, whether molten or in solution
Non-electrolytes are mostly covalent compounds and only exist as molecules. Examples include vegetable oils, organic solvents like alcohols, benzene sugar solution
Conductor and Non-conductor
Conductors: These are metals which allow the passage of electricity through them.
Examples include all metals in general Silver is the best conductor followed by copper and ionic solutions
Non-conductor (Insulators): These are substances that do not conduct or allow electricity to pass through them. Examples include wood, paper, air, rubber, plastic.
Electrodes: these are wires rods or plates through which an electric current enters or leaves the electrolytes
2. Anodes: This is the positive electrode through which electrons leave electrolytes. It is the electrode where oxidation occurs
3. Cathode: This is the negative electrode through which electrons enter the electrolyte. It is also the electrode where reduction takes place
4. Cations: - these are positively charged ions. They migrate to the cathode (negative electrode) during electrolysis.
5. Anions: these are negatively charged ions. They migrate to the anode (positive electrode) during electrolysis.
5. Electrolytic Cell: This is a vessel or container containing two electrons connected to a battery and an electrolyte. It is used for Electrolysis.
When electrolysis is carried out on the solution of an ionic compound. There are usually two cations and two anions which migrate to the cathode and anode respectively but only one of each ion is preferentially discharged at the electrodes.
The following factors determine which ion gets discharged at each electrode.
1. Position of ions in the electrochemical series.
2. Concentration of ion in the electrolyte
3. Nature of the electrodes.
1. Position of the ion in the electrochemical Series: For positive ions, the metals that are lower down the series(less electropositive) are preferentially discharged to the ions higher up the series(more electropositive)
For negative ions, the less electronegative elements tend to lose their electrons more readily more electronegative ions will have, the greater tendency to accept electrons and so less electropositive ions are preferentially discharged..
2. Concentration of the ion in the electrolyte: - The ion with the higher concentration is preferentially discharged from solution. However, the influence of concentration is not effective when there is a wide gap between the two competing ions positions in the electrochemical series, while it is minimal if the ions are widely separated in their positions.
3. Nature Of the Electrodes: Some electrodes have strong affinity for certain ions and so select these elements out even when their positions in the electrochemical series and their concentrations do not warrant it. For example mercury has strong affinity for sodium metal ( forms the substance known as sodium amalgam Na/Hg) and so allows sodium to be discharged even if dilute solutions of sodium salts are used. Platinum (Pt) and Carbon (graphite) on the other hand have no affinity for any elements, hence they are described as inert electrodes( or passive electrodes).
Examples of Electrolysis
1 Electrolysis of Acidified Water (water containing drops of H2SO4)
The Hoffman Voltameter is used, both the anode and cathode are platinum foil.
The ions present in the electrolyte are:
Cations Anions
H2SO, → 2H+(g) + SO42-
H2O → H+(aq) + OH-(aq)
At the Cathode: H+ ion migrate to the cathode and take up electrons to form neutral hydrogen atoms.
H+(aq) + e-→ H(g)
The hydrogen atoms then combine to form hydrogen gas molecule
H(g) + H(g)→ H2(g)
Overall equation
2H+(aq) + 2e- →H2(g)
At the anode: Both SO42- and OH- ions migrate to the anode where OH- ions being lower in the electrochemical series is preferentially discharged and lose its electrons to the anode to become a neutral - OH group.
OH-(aq) → OH + e-
The neutral –OH group combines in pairs to form one molecule of water and one atom of oxygen.
OH + OH → H2O(l) + O(g)
The oxygen atoms then combine with another free oxygen atom to form an oxygen gas molecule.
O(g) + O(g)→O2(g)
Overall equation
4OH-(aq)→2H2O(l) + O2(g) + 4e-
Note: At the end of the electrolysis the solution becomes more concentrated or more acidic as the components of water (H2 and O2) are removed
2. Electrolysis of concentrated NaCl (brine)
The Hoffman’s voltameter is also used for this electrolysis with a platinum or carbon cathode, but the anode must be carbon.
This is because chlorine can attack the carbon rod/anode by The ions present are
Cations Anions
NaCl(aq) → Na+(aq) Cl-(aq)
H2O → H+(aq) OH-(aq)
At the Cathode: Both Na+ and H+ ion move to the cathode. Although Na+ ion is higher in concentration, H+ ion is preferentially discharged because it is lower (position) in the electrochemical series.
H+ takes up electrodes to form neutral hydrogen atoms.
H+(aq) + e–→ H(g)
The hydrogen atoms then combine in pairs to form diatomic hydrogen gas molecule
H(g) + H(g)→H2(g).
Overall equation
2H+(aq) + 2e-→ H2(g)
Thus, hydrogen gas is obtained at the cathode.
At the anode: both Cl- and OH- ions migrate to the anode where Cl- ions are preferentially discharged. This is because it is higher in concentration than OH- ion and the two ions are close to each other in the series.
Cl- (aq)→Cl(g) + e-
The chlorine atoms combine to give the molecules.
Cl(g) + Cl(g)→Cl2
Overall equation
2Cl-(aq)→ Cl2(g) + 2e-
Chlorine gas is obtained at the anode.
3.Electrolysis Of Copper (II) Tetraoxosulphate (VI) Solution Using Different Anode:
With carbon or platinum electrodes.
Cations Anions
CuSO4 → Cu2+(g) + SO42-(aq)
H2O → H+(aq) + OH-(aq)
At the Cathode: Cu2+ and H+ ion migrate to the cathode where Cu2+ being lower than and less electropositive than H+ is preferentially discharged as metallic copper on the cathode.
Cu2+(aq) + 2e-→ Cu(s)
At the anode: SO42- and OH- ions migrate to the anode where OH- ions lose their electrons to become a neutral - OH group.
OH-(aq) → OH + e-
OH + OH → H2O(l) + O(g)
O(g) + O(g)→O2(g)
Overall equation
4OH-(aq) →2H2O(l) + O2(g) + 4e-
Electrolysis of CuSO4 using different electrodes
1. Using Pt. or C- electrodes: - Pt. and C-electrodes as we know are inert or passive electrodes and do not determine the product of the electrolysis
Uses Of Electrolysis
1. Extraction of metals like Na, K, Mg, Ca and Al.
2. Purification of metals copper,
3. Electroplating
4. Preparation of some elements like Cl2 as sodium hydroxide, hydrogen and chlorine from electrolysis of brine using cathode.
FARADAY’S LAWS OF ELECTROLYSIS
Faraday’s First Law Of Electrolysis: State that the mass (m) of an element discharged during electrolysis is directly proportional to the quantity of electricity (Q) that passed through the electrolyte
Mathematically
M α Q
Q = It
M α It removing the sign of proportionality we have
M = ZIt
Where Z is a constant known as the electrochemical equivalent of the substance.
M = Mass of substance in gram
Q = Quantity of electricity in coulombs
I = Current in ampere
t = Time in seconds
Verification Of Faraday’s First Law Of Electrolysis
Method
1. Fill a beaker with a 1M solution of copper (II) tetraoxosulphate (VI) solution.
2. Weigh two clean copper plates and place them in the beaker
3. Connect the circuit as shown in the diagram above, and place a variable resistor adjusted to maintain a current of 2A for 10 minutes.
4. Remove the cathode after 10mins washed with water dry and then reweigh
5. Replace the cathode and pass the same for 15 minutes
6. Repeat the process in step 4 and 5 cleaning, obtaining three more readings at 20mins, 25 and 30mins
Tabulate the result as shown in the table below
Current (A) | Time (S) | Quantity of Electricity(Q) | Mass of Copper |
2A | 10min | 2 x 10x60 | M1 |
15mins | 2 x 15 x 60 | M2 | |
20mins | 2x20x60 | M3 | |
25mis | 2x25x60 | M4 | |
30mins | 2x30x60 | M5 |
Plot a graph of mass(M) deposited against quantity of electricity(Q)
Repeat the entire process above, but this time varying the quantity of electricity while keeping the time constant.
Conclusion: The mass of copper deposited at the cathode is directly proportional to the quantity of electricity passed.
Faraday’s Second Law Of Electrolysis: State that when the same quantity of electricity is passed through solutions of different electrolytes, the relative number of moles of the elements discharged at each electrode is inversely proportional to the charges on the ions of each of the element
According to Faraday the minimum quantity of electricity required to liberate one mole of a univalent ion during electrolysis is equal to 1 Faraday and
1 Faraday = 96500 coulombs
Mass = Quantity of Electricity
Molar mass Faraday
Verification Of Faraday’s Second Law
Method
1. Fill two beakers up to ⅔ of their volumes with 1M of copper (II) tetraoxosulphate (VI) solution and 1M solution of silver trioxonitrate (V) solution.
2. Weigh and Place two clean plates of copper and Silver electrodes in their respective solutions
3. Connect a battery and complete the circuit as shown above, attach a variable resistor adjusted to maintain a steady current of 0.5A. Allow the current to pass through the solution for 25 minutes.
4. The cathode is removed, washed with water dried and then reweighed to obtain the masses of copper and silver deposited
5. The ratio of the number of moles of copper and silver deposited is then calculated.
6. The process is repeated to obtain at least three more readings for accuracy.
Amounts n (Number of moles) = Mass of element deposited
Its relative atomic mass.
Observation: On passing the same quantity of electricity through the solutions, the ratio of the number of moles of copper and silver deposited is 1:2.
This ratio is inversely proportional to the ratio of the charges on the ions, Cu2+ and Ag+ , or the number of moles of electrons required to liberate 1 mole each of the ions.
Cu2+(aq) + 2e- → Cu(S) and Ag+(aq) + e- → Ag(s)
2 moles 63.5g 1 mole 108g
Conclusion: When the same quantity of electricity is passed through a solution of different electrolytes the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of each of the elements respectively.
CALCULATION BASED ON THE FIRST LAW OF ELECTROLYSIS
1. In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited in 30mins is 0.66g. Calculate the error in the ammeter reading.[electrochemical equivalent of copper =0.00033gC-1]
Solution
M = ZIt
M = 0.66g, Z = 0.00033gC-1, t = 30mins = 1800 seconds
I = M/Zt
I = 0.66/0.00033 x 1800
I = 0.66/0.594
I = 1.11A
The error in the ammeter reading is 1.11 – 1 = 0.11A
2. Calculated the time in minutes, required to plate a substance of total surface area 300cm2, a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm3 and one coulombs liberates 0.00033g copper.
Solution
Given that area = 300cm2 , thickness = 0.6mm = 0.06cm
Mass = 0.00033g, density = 8.8g/cm3
Density = mass/volume
Mass = density x volume
Mass = 8.8 x 300 x 0.06
Mass = 158.4g
From M = Zit
t = m/ZI
t = 158.4/2 x 0.00033
t = 240000secs
t= 4000mins
3. At what time must a current of 5Amp pass through a solution of zinc sulphate to deposited 1g of zinc. Electrochemical equivalent (e.c.e) = 0.0003387
Solution
Given current = 5Amp
e.c.e = 0.0003387g/c
Mass = 1g
M = ZIt
t = M
ZI
time = 1
0.0003387 x 25
|
time = 1,180secs
time = 19mins 40 seconds
4. In an electrolysis experiment, a cathode of mass 5g is found to weigh 5.01g, after a current of 5A flows for 50 seconds. What is the electrochemical equivalent for the deposited substance?
Solution
M = 5.01 – 5 = 0.01g, t = 50sec, I = 5A
M = ZIt
Z = m/It
Z = 0.01/50 x 5
Z = 0.01/250
Z = 4 x 10-5g/c
5. The electrochemical equivalent of silver is 0.0012g/c. if 0.36g of silver is to be deposited by electrolysis on a surface by passing a steady current for 5.0 minutes. Calculate the value of the current.
Solution
Z= 0.0012g/c, m = 0.36g, t = 5mins = 300secs,
M = ZIt
I = m/Zt
I = 0.36/0.0012 x 300
I = 0.36/0.36
I = 1A.
CALCULATION BASED ON THE SECOND LAW OF ELECTROLYSIS
6. A current of 4.5A is passed through a solution of gold salt for 1 hour 45 minutes. Calculate
(i) The mass of gold deposited
(ii) The number of moles of gold deposited
(iii) If the same current is used, find the time taken for 5.5g of gold to be deposited (Au = 197, 1 Faraday = 96500c)
Solution
(i) Au+ + e-→ Au
197g 1F 197g
Mass = Quantity of Electricity
Molar mass Faraday
M Q
Mm F
Mass = ?
Molar mass = 197g
Quantity of electricity = I x t =
I = 4.5A
t = 1 hour 45minutes = 105 minutes = 105 x 60 = 6300 seconds
Quantity of electricity = I x t = 4.5 x 6300 = 28,350C
Faraday = 96500F
Mass = 28,350
197 96500
Mass = 0.29378
197
Mass of Gold deposited = 57.88g
(ii) Number of mole = Mass
Molar mass
Number of mole = 57.88
197
The number of mole of Gold deposited = 0.30mol
(iii) Mass = Quantity of Electricity
Molar mass Faraday
M Q
Mm F
|
5.5 = Quantity of Electricity
197 96500
0.02792 = Quantity of Electricity
96500
Quantity of Electricity = 2694.28C. This is the quantity of electricity (Q) required for 5.5g of Au to be deposited.
Q = It
t = Q/I
t = 2694
4.5
t = 598.7
The time taken is 9.98 minutes
7. Calculate the current that must be passed into a solution of aluminium salt for 1hr.30minutes in order to deposited 1.5g of Aluminium (Al = 27)
Solution
Al3+ + 3e- →Al(s)
27g 3F 27g
Mass = Quantity of Electricity
Molar mass Faraday
M Q
Mm F
Mass = 1.5g, Molar mass = 27g/mol
Quantity of electricity = ?
Faraday = 3 x 96500 = 289500
1.5 = Quantity of electricity
27 289500
0.0556 = Quantity of electricity
289500
Quantity of electricity = 16096.2C
Quantity of electricity = I x t
I = Q/t
t = 1 hr.30mins = 90mins = 90 x 60 = 5400seconds
I = 16096.2
5400
The current (I) = 2.98Amperes
8. 0.222g of a divalent metal is deposited when a current of 0.45A is passed through a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. (1 Faraday = 96500 coulombs)
Solution
M2+ + 2e- →M
1mole 2F 1 mole of atoms.
Mass = Quantity of Electricity
Molar mass Faraday
M Q
Mm F
Mass = 0.222g, molar mass = ?
Quantity of electricity = It = 0.45 x 25 x 60 = 675C
Faraday = 2 x 96500 = 193000F
0.222 = 675
Mm 193000
0.222 = 0.003497
Mm
Molar mass = 0.222
0.0003497
|
The relative atomic mass of metal M is 63.5g/mol
Therefore metal M is copper.
9. A given quantity of electricity was passed through three electrolytic cells connected in series containing solutions of Silver trioxonitrate (V), Copper(II) tetraoxosulphate (VI) and Sodium Chloride respectively. If 10.5g of Copper are deposited in the second electrolytic cell. Calculate
(a) The mass of Silver deposited in the first cell.
(b) The Volume of Chloride liberated in the third cell at 180C and 760mmHg pressure. (Ag=108, Cu=63.5, 1Faraday=96500C, molar volume of gases at s.t.p =22.4dm3.)
Solution
In the second cell
Cu2+(aq) + 2e- →Cu(s)
63.5g 2F 63.5g
Mass = Quantity of Electricity
Molar mass Faraday
M = Q
Mm F
mass of copper = 10.5g, molar mass of copper 63.5g/mol, Faraday = 2 x 96500 = 193000C, Quantity of electricity = ?
10.5 = Quantity of electricity
63.5 193000
0.16535 = Quantity of electricity
193000
Quantity of electricity = 31913.4 Coulombs
(a) The mass of silver deposited in the first cell
Ag+(aq) + e-→Ag(s)
108g 1F 108g
Mass = Quantity of Electricity
Molar mass Faraday
M Q
Mm F
M = 31913.4
108 96500
M = 0.3307
108
Mass = 108 x 0.3307
Mass of silver deposited = 35.7g
(b) Volume of chlorine liberated in the third cell
2Cl- →Cl2 + 2e-
2F
Mass = Quantity of Electricity
Molar mass Faraday
M 31913.4
Mm 2 x 96500
Mole = 31913.4
193000
Mole = 0.1654mol
1 molar volume of gas = 22.4dm3 x mole
Volume of chlorine = 22400cm3 x 0.1654
Volume of chlorine at s.t.p = 3704cm3
Convert the volume at s.t.p to its volume at 180C and 740mmHg
P1V1 = P2V2
T1 T2
Where P1 = 760mmHg, P2 = 740mmHg
T1 = 273 K T2 = 18 + 273 = 291K
V1 = 3704cm3 V2 = ?
V2 = P1V1T2 760 x 3704 x 291
P2T1 740 x 273
2. Calculated the time in minutes, required to plate a substance of total surface area 300cm2, a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm3 and one coulombs liberates 0.00033g copper.
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