What is Electrolysis in chemistry?
What is Electrolysis in chemistry?
Electrolysis is the gradual breakdown or decomposition of a compound by the passage of electricity through it.
To understand very well what is Electrolysis in Chemistry, we need to really read down and understand everything in this post.
By decomposition, we mean dissociating into ions. So decomposition into ioans by the passage of electricity it is important to also remember that electrolysis the compound that is broken into ions. It’s actually called the Electrolyte.
I’ll continue by describing certain terms associated with electrolysis.
What is an electrolyte?
An electrolyte is a compound that conducts electricity in molten or aqueous form.
There are basic terms we use in describing an electrolyte.
We use molten electrolyte, concentrated or aqueous forms in describing and explaining the mechanisms of electrolysis..
When the electrolyte is in molten form, there is no water
What is an electrode?
An electrode or electrodes are conductors in form of wires, plates or rods through electrons enter or leave the electrolyte.
Platinum and graphite(allotrope of carbon) are best used as electrodes because inert electrodes .
These elements are used because they conduct electricity and they are unreactive so they won’t interfere with the mechanisms of electrolysis.
What is an electrolytic cell?
An electrolytic cell is a cell that converts electrical energy to chemical cell.
The electrolytic cell is the cell used to demonstrate electrolysis.
The electrolytic cell is made of two electrodes inserted in a solution of electrolyte with the electrodes connected by a wire to the battery for direct current supply.
Hint
Conduction of electricity in a compound is by the movement of ions.
For example electrolytes conduct electricity via movement of ions.
Conduction of electricity in an electrode or metals by the movement of electrons.
For example why we use graphite and platinum in addition to being being inert is that they also conduct electricity.
Mechanism of electrolysis
I will like to stress here that the mechanism of electrolysis influences the all we can learn in this course because it is very important. .
I want to start by saying that there are two reactions that happen in the electrolytic cell.
The two reactions are reduction and oxidation.
Reduction happen at the cathode while oxidation happens at the anode.
Hint
In the electrolytic cell, the anode is positively charged while the cathode is negatively charged.
In the electrochemical cell, the anode is negatively charged while the cathode is positively charged.
However reduction occurs at the cathode while oxidation occurs at the anode in both electrolytic and electrochemical cell.
So the mechanism of electrolysis of electrolysis like I said earlier before involves a redox reaction.
Redox in the sense that reduction and oxidation occur simultaneously
I will like you to note these two pneumonic;
AOCR = Anode is fir oxidation Cathode is for reduction
OILRIG = oxidation is loss of electrons, reduction is gain of electrons.
Reduction
Reduction is loss of electrons and increase in oxidation number.
For example;
Ca2+ +2e-===Ca
You can see that there is gain of electrons and decrease in oxidation number.
Oxidation
Oxidation is the loss of electrons and increase in oxidation number.
For example
2Cl- -2e- ===Cl2
2Cl- ===Cl2 +2e-
So let’s dive into the electrolysis of some compounds now and how to predict the products formed at cathode and anode.
Take note of these terms;
Molten form( no water)
Concentrated aqueous ( little water)
Dilute aqueous ( excess water)
Factors considered in the discharge of ions
There are three factors we consider in the discharge of ions.
-
Position of the ions in the electrochemical series
-
Nature of electrodes ( is it inert or reactive?)
-
Concentration of ions
Mechanism of electrolysis involves migration and discharge of ions.
The ions that are discharged are the ions lower in the series .
Electrolysis of concentrated aqueous compounds
In concentrated forms , there is little water and consequently
We have two ions coming from water, hydrogen ion(H+) and hydroxide ion *OH-)
Example 1: Electrolysis of Brine( concentrated sodium chloride)
So here there are two sets of ions
Electrolyte NaCl===Na+ + Cl-
Little water H2O ==H+ + OH-
Migration of ions
The two positive ions migrate to negative electrode ( cathode) while the two negative ions migrate to positive electrode (anode)
Discharge of ions
Here the position of the ions in the electrochemical series is considered.
The ions lower in the series are discharged.
At the Cathode (Reduction occurs)
Since the cathode is negatively charged, it will attract sodium ion(Na+) and hydrogen ion(H+).
But the lower ion in the series is hydrogen ion, so it will be discharged.
2H++2e- ===H2
At the anode ( oxidation occurs)
Since the anode is positively charged, the two negative ions Cl- and OH-.will migrate to the anode but one will be discharged .
Remember that chloride ion is concentrated and also close to the reference ion so it will be discharged in preference to hydroxide ion.
2Cl- -2e- ===Cl2
2Cl-===Cl2 +2e-
Example 2: Electrolysis of caustic soda( concentrated sodium hydroxide)
Also in this compound, there are two sets of ions
Electrolyte NaOH===Na+ + OH-
Little water H2O ==H+ + OH-
Migration of ions
The two positive ions migrate to negative electrode ( cathode) while the two negative ions migrate to positive electrode (anode)
Discharge of ions
Here the position of the ions in the electrochemical series is considered.
The ions lower in the series are discharged.
At the Cathode (Reduction occurs)
Since the cathode is negatively charged, it will attract sodium ion(Na+) and hydrogen ion(H+).
But the lower ion in the series is hydrogen ion, so it will be discharged.
2H++2e- ===H2
At the anode ( oxidation occurs)
Since the anode is positively charged, the two negative ions OH- and OH-.will migrate to the anode , so same ion will be discharged .
OH- -e- ===HOH
OH +OH ===H2O + O
O+ O ===O2
Overall equation
4OH- ==2H2O + O2 + 4e-
Please take note that when the ions are concentrated and are quite close to the reference ions (H+ and OH-), the concentrated ions will be discharged instead.
Electrolysis of Dilute aqueous solutions
Electrolysis of Dilute aqueous solutions implies that there is excess water.
So there is competition for discharge of ions between the two positive and negative ions.
Let’s dive into some examples of electrolysis of Dilute aqueous solutions.
Examples 1: Electrolysis of dilute Copper II sulphate
Here there is excess water so the ions from the salt are no longer concentrated.
Please note that the difference between concentrated and Dilute solutions depends on the level or amount of water added.
In the Electrolysis of Dilute Copper Ii sulphate;
CuSO4 ===Cu2+ SO42-
H2O===H+ + OH-
At the cathode Cathode ( Reduction occurs here)
Electrolysis of Molten compounds
Example 1 : Electrolysis of molten Lead II Bromide (PbBr2)
In molten, there is no water so we only focus on the ions .
Mechanism
PbBr2 ===Pb2+ + 2Br-
At the Cathode (Reduction occurs here)
Pb2+ + 2e- ===Pb
At th Anode (Oxidation occurs here)
2Br- -2e- ===Br2
2Br- ===Br2 +2e-
Example 2: Electrolysis of fused calcium chloride
CaCl2===Ca2+ 2Cl-
At the cathode Cathode ( Reduction occurs here)
Ca2+ + 2e- ===Ca
At the Anode (Oxidation occurs here)
2Cl- -2e- ===Cl2
2Cl- ===Cl2 +2e-
Example 3: Electrolysis of molten aluminium oxide
Al2O3 ===Al3+ + O2-
Cathode (-ve): Gain of electrons
At the Cathode (reduction occurs here)
Al3+ + 3e- ===Al
At the Anode (Oxidation occurs here)
O2- -4e- ===O2
2O2- ===Ol2 +4e-
In conclusion of these mechanism of electrolysis of electrolytes without water, I
It is important to reemphasize that there is no hydrogen ion and hydroxide ion and as a result only the ions involved are discharged.
So the ions that are migrated are discharged.
How do you calculate mass produced in electrolysis?
Now we have two laws of electrolysis;
How do you calculate more produced in electrolysis? First law of electrolysis and second law of electrolysis.
First law of Electrolysis
The Faradays First law of Electrolysis
The faraday’s first law of electrolysis states that the mass of an element discharged during electrolysis is directly proportional to the quantity of electricity passed through it.
Explanation of first law of electrolysis
First law of electrolysis is basically on the fact that the quantity of electricity passed determines the mass or volume of elements discharged
Example 1
Calculate the mass of an element copper deposited during the electrolysis of CuSO4 by passing 3456.8C .
Solution
Cu2+ +2e- ====Cu
2F ==Cu
2 ×96500C ====Cu
193000C =====63.5g
3456.8C =======X
X = 3456.8 x 63.5 / 3456.8
Example 2
If 3.5A of current was passed for 2 hours during the electrolysis of Dilute aqueous silver nitrate . What mass of silver would be deposited .
Solution
Ag+ + e- =====Ag
1F====108g
Q=it
Q= 3.5 x2 x 60x 60
Q= 25200C
96500C====108g
25200C====X
X = 25200 x 108/96500
X= 28.2g
Example 3
How long will it take to deposit 14.5g of copper if a current of 3.5A of electricity.
Solution
Cu2+ + 2e- ====Cu
2F ======63.5g
2 x 96500C ====63.5g
193000C ======63.5g
X =========14.5g
X = 193000 x 14.5/ 63.5
X= 44071C
Q =it
t = Q/I
t= 44071/3.5
t=12591s
Faraday’s second law of electrolysis
Faraday’s second law of electrolysis states that if the same quantity of electricity is passed through two or more electrolytes connected in series, then the number of moles of ions discharged is inversely proportional to the charges on the ions.
Explanation of Faraday’s second law of electrolysis
Faradays second law of electrolysis explains that for serial connection of electrolytes, the same quantity of electricity is passed through all of them.
Example 1
If there are two electrolytes dilute CuSO4 and molten sodium chloride connected in series. If 23.5g of copper was deposited.
Calculate;
a. Quantity of electricity
b.Mass of sodium deposited
Solution
Start with the given
Cu2+ +2e- ======Cu
2F ======63.5g
193000C=====63.5g
X ========= 23.5g
X = 193000 x23.5 /63.5
X= 71425C
So apply the quantity of electricity to get the mass of sodium deposited
Na+ + e- ====Na
F =====Na
96500C=====23
71425C =====X
X= 71425 X 23/96500
X= 17g of sodium deposited
Example 2
Three electrolytic cells acidified water, concentrated potassium chloride and concentrated copper II tetraoxosulphate VI
If 32g of hydrogen was liberated by the same quantity of electricity, calculate
a.Quantity of electricity passed
b.volume of chlorine liberated
c.Volume of oxygen liberated
Solution
So, the first step is to calculate the quantity of electricity.
2H+ +2e- ====H2
2F ======2g
193000C====2g
X==≈======32g
X= 193000 X 32/2
X= 3088000
b. 2Cl- =======Cl2 +2e-
1 mole of Cl2 ======2F
22.4dm3======193000
X ===== 3088000
X= 22.4 X 3088000/193000
×= 358.4dm3
c.4OH-====2H2O ÷ O2 +4e
4F=====1 mole of O2
4 x96500C=====22.4fm3
386000C====22.4dm3
3088000=== X
X= 3088000x 22.4/386000
XX537.6 dm3
