Content Writer
Electrochemistry is the branch of chemistry which deals with a combination of reactors and electrodes. As the name suggests, it studies about the production of electricity from chemical reactions.
- Electrochemistry deals with the relationship between electrical energy and chemical changes.
- The reaction involves the movement of electrons through the electronic conducting phase.
- In electrolysis, any reaction is based on the electrical potential difference.
- Some cases result in the generation of potential differences from a battery or cell.
- This process of generation of potential difference is called an electrochemical reaction.
- Production of chemical change, which is also known as electrolysis, is the first category of electrochemical reaction.
- Generation of electricity by redox reaction is the second category of electrochemical reaction.
- Disposable batteries in our phones are a real-life example of electrochemistry.
Key Terms: Electrochemistry, Potential Difference, Electrochemical Reaction, Electrolysis, Nernst Equation, Standard Electrode Potential, Temperature, Salt Bridge, Metals, Chemical Reactions, Electrode
What is Electrochemistry?
[Click Here for Sample Questions]
Electrochemistry is a branch of chemistry that studies the generation of electricity by the process of spontaneous chemical reactions. It also deals with the applications of electrical energy to non-spontaneous chemical reactions.
- Electrochemistry is the study of the movement of electrons in oxidation and reduction reactions.
- The reaction is carried out at a polarized electrode surface.
- Each electronic plate is oxidized or reduced at a particular potential difference.
- It is used to examine the phenomena resulting from combined chemical and electrical effects.
- The process of oxidation and reduction is carried out on separate surfaces.
- This process of oxidation and reaction process taking place simultaneously is called a redox reaction.
- Batteries used in flashlights and the use of calculators are some examples of electrochemistry.
Read More:
Electrochemical Cell
[Click Here for Sample Questions]
An electrochemical cell is an equipment that is able to generate electrical energy through chemical reactions. In more detailed terms, it is a device that produces a potential difference between electrodes via chemical reactions.
- It comprises mainly two electron conductors.
- The electron conductors are separated through an ionic conductor.
- An electrochemical cell is composed of two half-cells.
- Oxidation occurs in one-half of the cell, while reduction occurs in the other half of the cell.
- A reverse reaction is possible in such a cell when a non-spontaneous chemical reaction takes place.
An electrochemical cell comprises the following components:
Types of Electrochemical Cells
[Click Here for Sample Questions]
There are two types of electrochemical cells which are as follows:
Galvanic Cell
Galvanic cells also known as voltaic cell in which the chemical energy is converted or transformed into electrical energy. The redox reaction taking place in this kind of cells are spontaneous in nature.
- In these cells, the Electrons are originated from species undergoing oxidation.
- The anode in galvanic cell is negatively charged while the cathode is positively charged.
Electrolytic Cell
In electrolytic cells, the electrical energy is transformed into chemical energy. These type of cells have reactions which non-spontaneous. In these cells, the electrons are originated from some external source.
- The anode in the electrolytic cell is positively charged while the cathode is negatively charged.
Electrode Potential
[Click Here for Previous Year Questions]
Electrode Potential is termed as the tendency of an element to either gain or lose electrons in order to become negatively or positively charged. Depending upon the type of reaction that has taken place, the electrode potential can be called oxidation or reduction potential.
- The oxidation and reduction potentials are opposite in sign but are equal in magnitude.
- Here, the values of E are not additive as it is not thermodynamic in nature.
- The oxidation reaction is carried out at the anode.
- The reduction reaction is carried at the cathode.
Standard Electrode Potential
Standard Electrode Potential is denoted as Eocell which is measured in comparison to Standard Hydrogen Electrode. It is defined as the standard EMF value of a cell. It is the measurement of the potential for equilibrium.
- The potential difference between the electrode and electrolyte is known as the potential of electrode.
- At unit concentration, this electrode potential is called standard electrode potential.
The equation used to determine the difference of two half-cell is:
Eocell = Eored, Cathode – EoRed, Anode
- Where, Eocell = Standard cell potential
- Eored, Cathode = Standard reduction potential for reduction
- EoRed, Anode = Standard reduction potential for oxidation
Cell Potential or Emf of a Cell
[Click Here for Previous Year Questions]
Cell Potential of a cell is denoted as Ecell. It is measured in Volts (J/C). In general terms, the cell potential is a way to measure the quantity of potential difference that exists between the two half cells of a battery or electrochemical cell.
- For an electrochemical cell, the cell potential is measured using a voltmeter.
- Due to the presence of electric current, the electron moves from high potential to lower potential.
- The electrode of higher potential is called the anode.
- On the other hand, electrodes of lower potential are called cathodes.
Electromotive force (EMF)
[Click Here for Previous Year Questions]
The electromotive force is a type of force that is produced either by an electrochemical cell or by changing the magnetic field. The work done on a unit electric charge is termed electromotive force. The symbol for electromotive force is ε, and its unit is Volt.
- The formula for electromotive force is:
ε = V + Ir
- Where, ε = electromotive force
- V = voltage of the cell
- I = current across the circuit
- r = internal resistance of the cell
Standard Hydrogen Electrode (SHE)
Standard Hydrogen electrode (SHE) is used as a reference electrode. At temperature of 298K, the standard electrode potential is 0. The equation of reaction for redox half-cell of SHE is:
2H+ (aq) + 2e– → H2 (g)
Electrochemical Series
[Click Here for Sample Questions]
An electrochemical series is a form of a list that describes the arrangement of electrons in the order of their increasing electrode potential values. It is also known as an activity series.
- Electrochemical series are basically half-cell potential values.
- It will calculate the potential difference of various electrodes.
- The series is based on the ease of oxidation.
- Alkali metals are more oxidised than lower metals.
- Copper and gold, which are used in the creation of jewellery, are precious metals.
Nernst Equation
[Click Here for Sample Questions]
Nernst equation is responsible for providing the relation between the cell potential of an electrochemical cell, the standard cell potential, temperature, and the reaction quotient. Under non-standard conditions, it determines the cell potentials of electrochemical cells.
- It is often used to compute the cell potential of an electrochemical cell.
- The emf is measured at any given temperature, pressure, and reactant concentration.
The Nernst equation for single electrode potential is given by:
Ecell = E0 – [RT/nF] ln Q
- Where, Ecell = cell potential of the cell
- E0 = cell potential under standard conditions
- R = universal gas constant
- T = temperature
- n = number of electrons transferred in the redox reaction
- F = Faraday constant
- Q = reaction quotient
- The relationship between the Nernst equation, equilibrium constant, and Gibbs energy change is illustrated below:
ΔG° = – nFE°cell
- The relationship between ΔG° and equilibrium constant is: ΔG°= -RTlnQk
The video below explains this:
Nernst Equation Detailed Video Explanation:
Resistance
[Click Here for Previous Year Questions]
Resistance is defined as the process where electric current flows through a bulb or any other conductor. The conductor or the transmitter offers some impediment to the flow and this impediment is called electrical resistance and is signified by R.
- The SI unit of resistance is ohms.
- According to the ohm’s law, the relationship between current flowing through any conductor and the potential difference across it is given by:
V ∝ I V = IR
- Where,
- V = potential difference measured across the conductor (in volts)
- I = current through the conductor (in amperes)
- R = constant of proportionality called Resistance (in ohms)
- The electrical resistance of a circuit is determined by the ratio between the voltage applied to the current flowing through it.
Specific Conductance / Conductivity
[Click Here for Sample Questions]
Specific conductivity (otherwise called conductivity) is the measure of the capacity of that material to conduct electricity. It is characterized as the corresponding of obstruction of that material. SI unit of conductance is S (Siemens).
- It is addressed by the image "k".
- Consequently, by definition,
G = 1/R
R= ρl/A
K= 1/ρ
G =kl/A
- Where, K = conductivity,
- ρ = resistivity of the material
- G= conductance
- R= opposition
- l = length
- A= space of cross-segment
Molar Conductivity
Molar Conductivity is defined as the conductivity of a solution of an electrolyte which is separated by molar concentration of electrolyte. It gauges the efficiency with which an electrolyte can conduct electricity in the solution. Its SI Unit is S.m2.mol-1.
The expression for the molar conductivity is given by the formula:
∧m = K/C
- Where, K= specific conductivity
- C = concentration in mole per liter
Variation of molar conductivity along with concentration
The molar conductivity for strong electrolytes as well as for weak electrolytes increases with the dilution i.e. decrease in the concertation. It is the conductance of all the ions which are produced by one mole of electrolyte.
- For strong electrolytes, molar conductivity increases sharply along with increasing concentration.
- For weak electrolytes, the molar conductivity gradually increases with an increase in concentration.
Kohlrausch law of Independent Migration of Ions
[Click Here for Previous Year Questions]
Kohlrausch law states that at infinite dilution, each ion migrates independently of its co-ion and makes its own contribution to the total molar conductivity of an electrolyte.
- A scientist named Friedrich Kohlrausch performed experiments several times.
- He came out with a graph that plots the molar conductivity versus the square root of the concentration of a solution,
- According to the law at infinite dilution, the total molar conductivity is the algebraic sum of molar conductivities of cation and anion.
∧ = λ+0 + λ-0
- Where, ∧ = molar conductivity of a solution
- λ+0 = molar conductivity of a cation
- λ-0 = molar conductivity of a anion
- The molar conductivity at infinite dilution is given by expression:
∧ = mλ+0 + nλ-0
Faraday’s Law of electrolysis
[Click Here for Sample Questions]
Michael Faraday was the scientist who gave the two laws of electrolysis to explain the quantitative aspects of electrolysis. The two laws of electrolysis are the first law of electrolysis and the second law of electrolysis.
- The quantity of electricity required for oxidation-reduction relies upon the stoichiometry of the electrode reaction.
Faraday’s First Law of Electrolysis
Faraday’s first law of electrolysis is the primary law that states that, during electrolysis, the quantity of chemical reaction that occurs at any electrode under the effect of electrical energy is directly proportional to the amount of electricity passed through the electrolyte.
Faraday’s Second Law of Electrolysis
According to the second law of electrolysis, during the process of electrolysis, when a similar amount of electricity passes through the electrolytic solution, various substances liberated are proportional to their chemical equivalent weights.
Types of electrodes
[Click Here for Previous Year Questions]
Electrode is defined as any substance which is a good conductor of electricity and usually connects non-metallic parts of the circuit. The main use of electrodes is to generate electrical current.
- It passes that current through the non-metallic objects to alter the current in different ways.
There are mainly two types of electrode which are named as reactive and inert electrodes:
Inert
The inert type of the electrode does not participate in any reactions. Commonly used inert electrodes are platinum, gold, graphite and rhodium.
Reactive
A reactive type of electrode is known to take part in the reactions and dissolve into the electrolyte. Zinc, copper, lead, and silver are commonly used reactive electrodes.
Batteries
[Click Here for Previous Year Questions]
Batteries are referred to as a parallel combination of electrochemical cells. There are mainly two categories into which batteries are classified: primary and secondary batteries.
Primary Batteries
Primary Batteries are kinds of cells/batteries have high density and get discharged slowly. There is no fluid inside these types of cells, so they are known as dry cells.
- They are easy to use, acquire high internal resistance.
- It undergoes a chemical reaction which is irreversible.
Secondary Batteries
Secondary batteries consist of high energy density and are made up of molten salts and wet cells. The internal resistance of secondary cells is low, and their chemical reaction is reversible. It is quite complicated to use as compared to primary cells.
Corrosion
[Click Here for Sample Questions]
Corrosion is defined as the natural process of conversion of a refined metal into a more chemically stable form such as oxide, hydroxide, or sulfide. In general terms, corrosion is the gradual destruction of material through the chemical or/and electrochemical reaction with their environment.
Factors Affecting Corrosion
Factors affecting corrosion include:
- Exposure to air containing gases like CO2, SO2, SO3 etc.
- Exposure to moisture especially salt water
- Presence of impurities like salt
- An increase in temperature increases corrosion.
- Nature of the primary layer of oxide
- Presence of corrosive in the environment: acids can easily accelerate the process of corrosion.
Mechanism of corrosion
The formation of rust on the surface of iron occurs through the following steps.
- At the anodic region: Iron in contact with water forms anode and gets oxidized to Fe2+.
Anodic region: Fe(s) →Fe2+ (aq) + 2e– E0 Fe2+ /Fe = –0.44 V
- The electrons released moves to the other portion of the iron sheet.
- This portion of the iron sheet serves as cathode.
- At the cathodic region: The surface, oxygen in the presence of H+ ions get reduced to form H2O.
Cathodic region:
O2(g) + 4H+(aq) + 4e– →2H2O(l) EoO2/H+ H2O = + 1.23 V
The overall reaction of the local cell is equal to the sum of the cathodic and anodic reactions.
Overall cell reaction: 2Fe(s) + O2(g) + 4H+(aq) →2Fe2+(aq) + 2H2O(l) E0 cell = 1.67 V
Important Topics for JEE MainAs per JEE Main 2024 Session 1, the important topic included in the chapter Electrochemistry is as follows:
|
Things to Remember
- Electrochemistry is the branch of chemistry that studies the relationship between electric potential differences.
- It also checks for the changes that take place in the chemical reactions.
- Galvanization protects the iron from rusting.
- Redox reaction takes place with the help of external voltage.
- Conductance is defined as the process by which electric current passes through the conductor.
Also Read:
Previous Year Questions
- The change in potential of the half-cell Cu2+|CuCu2+|Cu, when aqueous Cu2+Cu2+ solution is diluted 100100 times at 298K298K ? (2.303RTF=0.06)
- Which one of the following has the highest molar conductivity?
- Corrosion of iron is essentially an electrochemical phenomenon where the cell reaction are
- When a solution of CuSO4 is electrolysed for 15 minutes with a current of 1 amp, then the mass of Cu deposited at the cathode will be
- A hypothetical electrochemical cell is shown below :- TA|A+(xM)||B+(yM)|⊕BAT|A+(xM)||B+(yM)|B⊕ The emf measured is +0.20V. The cell reaction is:
- The e.m.f. of a cell, whose half-cells are given below, is: Mg2++2e−→Mg(s);Mg2++2e−→Mg(s); E∘=−2.37VE∘=−2.37V Cu2++2e−→Cu(s);Cu2++2e−→Cu(s); E∘=+0.34V
- Which of the following processes occurs in the electrolysis of an aqueous solution of nickel chloride at nickel anode?
- Rusting of iron is a chemical reaction. The reaction is:
- When a molten ionic hydride is electrolysed :
- What amount of Cl2 gas liberated at anode, if 1 ampere current is passed for 30 minute from NaCl solution?
- Kohlrausch's law states that at
- Then the standard e.m.f. of the cell with the reaction Fe2++Zn⟶Zn2++FeFe2++Zn⟶Zn2++Fe is
- The EMF of hte cell reaction Fe2++Zn=Zn2++FeFe2++Zn=Zn2++Fe is
- The formal potential of Fe3+/Fe2+Fe3+/Fe2+ in a sulphuric acid and phosphoric acid mixture (E∘=+0.61V)(E∘=+0.61V) is much lower than the standard potential (E∘=+0.77V).(E∘=+0.77V). This is due to
- What amount of electricity can deposit 1 mole of Al metal at cathode when passed through molten AlCl3
- 1C electricity deposits
- The value of the dissociation constant of the acid at the given concentration at 25∘C is
- The potential of a hydrogen electrode at pH=10 is
- During galvanization of iron, which metal is used for coating iron surface ?
- Ratio of electrochemical equivalents of Cu and Cr is
Sample Questions
Ques. (A)The cell in which the following reaction occurs :
2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(s)
Has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
(Given : 1 F = 96,500 C mol–1)
(B)How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given : 1 F = 96,500 C mol–1)? (3 marks)
Ans: (a) We will apply the following equation to solve this question:
ΔG° = – nFE°cell
In the above reaction, the number electrons being transferred(n) is 2.
Hence, ΔG°=-2x96500x0.236 = -45548 J mol–1 = – 45.548 kJ mol–1
(b) Given that I = 0.5 A , t= 2 hours = 7200 seconds
Q= It = 7200x0.5 = 3600 C
Hence, no of electrons = \(\frac{Total charge}{Charge on 1 electron}\) = 3600/1.6x10-19 = 2.25x1022 electrons
Ques. Calculate emf and ΔG for the following cell
Mg(s)| Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)
Eo(Mg2+/Mg)= -2.37 V, Eo(Cu2+/Cu)= 0.34 V ?(5 marks)
Ans. The cell reactions are as follows:
Cathode: Cu2++2e- → Cu(s)
Anode: Mg → Mg2+ + 2e-
Overall : Cu2++ Mg → Cu(s)+ Mg2+
As evident from the reactions, n=2
From nernst equation
Ecell = Eocell – \(\frac{2.303RT}{nF} log \frac{[Mg^{2+}]}{[Cu^{2+}]}\)
Eocell = Eocathode - Eoanode = 0.34+2.37 = 2.71 V
[Mg2+]= 0.001 M , [Cu2+ ]= 0.0001 M
Putting values in the above equation,
Eocell = 2.68 V
ΔG° = – nFE°cell = -2x96500x2.68 = – 517,240 J mol–1 = – 517.24 kJ mol–1
Ques. Answer the following questions: (A) ExWhat is meant by ‘limiting molar conductivity
(B) Express the relation between conductivity and molar conductivity of a solution held in a cell? (2 marks)
Ans. (A) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λm.
(B) Λm = K/C = Conductivity/ Concentration
Ques. What is the effect of catalyst on:
(A) Gibbs energy (ΔG) and (B) activation energy of a reaction? (2 marks)
Ans. (A) There will be no effect of the catalyst on Gibbs energy.
(B) The catalyst provides an alternative pathway by decreasing the activation energy of a reaction.
Ques. What is the effect of adding a catalyst on (a) Activation energy (Ea), and (b) Gibbs energy (AG) of a reaction? (2 marks)
Ans. (a) On adding a catalyst in a reaction, the activation energy reduces and rate of reaction is fastened. (b) A catalyst does not alter Gibbs energy (AG) of a reaction.
Ques. What is the difference between anode and cathode? (4 marks)
Ans. The difference between anode and cathode are as follows:
Anode | Cathode |
---|---|
Anode is an electrode where oxidation takes place. | Cathode is an electrode where reduction takes place. |
It is positively charged in electrolytic cell | It is negatively charged in electrolytic cell |
It holds negative polarity in galvanic cells. | It holds positive polarity in galvanic cell |
In electrolytic cell, an anode is responsible to attract anions | In an electrolytic cell, cathode is responsible for attracting cations. |
Ques. (A)The cell in which the following reaction occurs :
2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(s)
Has E°cell = 0.336 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
(Given : 1 F = 96,500 C mol–1)
(B)How many electrons flow through a metallic wire if a current of 0.8 A is passed for 2 hours? (Given : 1 F = 96,500 C mol–1)? (3 marks)
Ans: (a) We will apply the following equation to solve this question:
ΔG° = – nFE°cell
In the above reaction, the number electrons being transferred(n) is 2.
Hence, ΔG°=-2x96500x0.336 = -64,848 J mol–1 = – 64.84 kJ mol–1
(b) Given that I = 0.8 A , t= 2 hours = 7200 seconds
Q= It = 7200 x 0.8 = 5760 C
Hence, no of electrons = \(\frac{Total charge}{Charge on 1 electron}\) = 5760/1.6x10-19 = 3.6 x1022 electrons
Ques. Calculate emf and ΔG for the following cell
Mg(s)| Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)
Eo(Mg2+/Mg)= -3.37 V, Eo(Cu2+/Cu)= 0.74 V ? (5 marks)
Ans. The cell reactions are as follows:
Cathode: Cu2++2e- → Cu(s)
Anode: Mg → Mg2+ + 2e-
Overall : Cu2++ Mg → Cu(s)+ Mg2+
As evident from the reactions, n=2
From nernst equation
Ecell = Eocell – \(\frac{2.303RT}{nF} log \frac{[Mg^{2+}]}{[Cu^{2+}]}\)
Eocell = Eocathode - Eoanode = 0.74+3.37 = 4.11 V
[Mg2+]= 0.001 M , [Cu2+ ]= 0.0001 M
Putting values in the above equation,
Eocell = 4.11 V
ΔG° = – nFE°cell = -2 x 96500 x 4.11 = – 793,230 J mol–1 = – 793.230 kJ mol–1
Ques. The emf of a cell corresponding to the reaction, Zn + 2H+ (aq) → Zn+2 (0.1M) + H2(g) 1 atm is 0.40 v at 25 0 C. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. (E0cell = – 0.86 v )? (5 marks)
Ans. Eocell = 0 – (-0.86) = 0.86 v
Applying the Nernst equation,
Ecell = Eocell – 0.0591 / 2 log [zN+2][H2] / [H]+
0.40 = 0.76 – 0.0591 / 2 log 0.1 x 1 / [H+]2
log 0.1 / [H+]2 = 2 x 0.7009 / 0.40
log 0.1 – log [H+]2 = 3.5045
2 pH = 3.5045 – log 0.1
pH = 4.5045 / 2 = 2.25
Ques. The solution of metal of atomic mass X was electrolysed for 1 hour with a current of 0.35 ampere. The mass of the metal deposited was 0.395 g. Find the metal X if its valency is 2? (3 marks)
Ans.Given, I = 0.35 ampere, t= 1 hr = 60 x 60 = 3600s
Q = I ✕ t
Q = 0.35 ✕ 3600
= 1260 coulombs
Therefore, 1260 coulombs of electricity deposit = 0.395
96500 coulomb of electricity deposit = 0.395 X 96500 / 1260 = 29.86 g
Valency of metal = atomic mass / equivalent mass
Atomic mass of metal X = 29.86 X 2 = 59.73g
Therefore, the metal X is copper.
Ques. (A)The cell in which the following reaction occurs :
2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(s)
Has E°cell = 0.446 V at 298 K. Calculate the standard Gibbs energy of the cell reaction.
(Given : 1 F = 96,500 C mol–1)
(B)How many electrons flow through a metallic wire if a current of 0.9 A is passed for 4 hours? (Given : 1 F = 96,500 C mol–1)? (3 marks)
Ans: (a) We will apply the following equation to solve this question:
ΔG° = – nFE°cell
In the above reaction, the number electrons being transferred(n) is 2.
Hence, ΔG°=-2x96500x0.446 = -86078 J mol–1 = – 86.07 kJ mol–1
(b) Given that I = 0.9 A , t= 4 hours = 14400 seconds
Q= It = 14400 x 0.9 = 12960 C
Hence, no of electrons = \(\frac{Total charge}{Charge on 1 electron}\) = 12960/1.6x10-19 = 8.1 x1022 electrons
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:
Comments