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Every redox reaction involves the transfer of electrons from one species to another. Electrochemistry studies what happens when this electron transfer is separated and made to occur through an external circuit. Rusting of iron, batteries, electroplating and fuel cells are all applications of the same basic principle. These Redox Reactions and Electrochemistry JEE notes cover oxidation numbers, balancing, electrochemical cells, EMF, the Nernst equation, electrolysis, conductance, batteries and corrosion for quick JEE revision.
Redox Reactions and Electrochemistry JEE Notes: Oxidation Numbers, Balancing and Reaction Types
Oxidation, Reduction and Redox Reactions
Oxidation is the loss of electrons and results in an increase in oxidation number.
Reduction is the gain of electrons and results in a decrease in oxidation number.
Oxidation and reduction always occur together because electrons lost by one species must be accepted by another species. Such reactions are called redox reactions.
| Term | What Happens | Role |
|---|---|---|
| Oxidising agent | Gets reduced by gaining electrons | Oxidises another species |
| Reducing agent | Gets oxidised by losing electrons | Reduces another species |
JEE Tip: Remember the mnemonic OIL RIG — Oxidation Is Loss and Reduction Is Gain of electrons.
Oxidation Numbers
Oxidation number is the hypothetical charge assigned to an atom by assuming all bonds are completely ionic. It helps track electron transfer during redox reactions.
| Situation | Oxidation Number |
|---|---|
| Free element (Na, O₂, P₄, S₈) | 0 |
| Monoatomic ion | Equal to ionic charge |
| Hydrogen | +1 generally, −1 in metal hydrides |
| Oxygen | −2 generally, −1 in peroxides, −1/2 in superoxides, +2 in OF₂ |
| Fluorine | −1 always |
| Alkali metals | +1 |
| Alkaline earth metals | +2 |
| Neutral compound | Sum of oxidation numbers = 0 |
| Polyatomic ion | Sum of oxidation numbers = charge on ion |
Worked Example: Oxidation Number of Mn in KMnO₄
Let oxidation number of Mn be $$x$$.
Potassium has oxidation number +1 and oxygen has oxidation number −2.
$$+1+x+4(-2)=0$$
$$1+x-8=0$$
$$x=+7$$
Therefore, oxidation number of Mn in KMnO₄ is +7.
Worked Example: Oxidation Number of Sulphur in Na₂S₂O₃
Let oxidation number of sulphur be $$x$$.
$$2(+1)+2x+3(-2)=0$$
$$2+2x-6=0$$
$$2x=4$$
$$x=+2$$
The oxidation number obtained is the average value because the two sulphur atoms exist in different environments.
Balancing Redox Reactions by Half Reaction Method
Normal chemical balancing only conserves atoms, but redox balancing must conserve both atoms and charge. The half-reaction method separates oxidation and reduction processes and balances them individually.
Steps for Half Reaction Method
- Identify oxidation and reduction half reactions.
- Balance all atoms except oxygen and hydrogen.
- Balance oxygen using $$H_2O$$.
- Balance hydrogen using $$H^+$$ in acidic medium or $$OH^-$$ in basic medium.
- Balance charge by adding electrons.
- Multiply equations to make electrons equal.
- Add both half reactions together.
Worked Example: Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Reduction half reaction:
$$MnO_4^- \rightarrow Mn^{2+}$$
Add water to balance oxygen:
$$MnO_4^- \rightarrow Mn^{2+}+4H_2O$$
Add hydrogen ions:
$$MnO_4^-+8H^+\rightarrow Mn^{2+}+4H_2O$$
Add electrons to balance charge:
$$MnO_4^-+8H^++5e^- \rightarrow Mn^{2+}+4H_2O$$
Oxidation half reaction:
$$Fe^{2+}\rightarrow Fe^{3+}+e^-$$
Multiply the oxidation reaction by 5 and add:
$$MnO_4^-+8H^++5Fe^{2+}\rightarrow Mn^{2+}+4H_2O+5Fe^{3+}$$
Types of Redox Reactions
| Type | Description |
|---|---|
| Combination | Two substances combine to form one product |
| Decomposition | One compound breaks into simpler substances |
| Displacement | One element replaces another element |
| Disproportionation | The same element undergoes oxidation and reduction |
| Comproportionation | Two oxidation states combine to form an intermediate state |
Example of disproportionation:
$$2H_2O_2\rightarrow2H_2O+O_2$$
Oxygen in hydrogen peroxide undergoes both oxidation and reduction in the same reaction.
Electrochemical Cells, EMF and Electrochemical Series
Galvanic Cells
A galvanic cell converts chemical energy into electrical energy using a spontaneous redox reaction.
A galvanic cell contains two half-cells connected by an external wire and a salt bridge.
| Component | Function |
|---|---|
| Anode | Oxidation occurs here; negative terminal |
| Cathode | Reduction occurs here; positive terminal |
| Salt bridge | Maintains electrical neutrality and completes circuit |
| Electron flow | From anode to cathode |
JEE Tip: Remember An Ox for oxidation at anode and Red Cat for reduction at cathode. Oxidation always occurs at the anode in both galvanic and electrolytic cells.
Cell Notation
The standard representation of a cell is:
$$\text{Anode | Anode solution || Cathode solution | Cathode}$$
For Daniell cell:
$$Zn(s)|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu(s)$$
Half reactions:
- Anode: $$Zn\rightarrow Zn^{2+}+2e^-$$
- Cathode: $$Cu^{2+}+2e^-\rightarrow Cu$$
Overall reaction:
$$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$$
EMF and Standard Electrode Potential
The electromotive force (EMF) of a cell is the potential difference between the cathode and anode when no current is drawn from the cell. It represents the driving force behind the cell reaction.
The standard electrode potential of a half-cell is measured relative to the Standard Hydrogen Electrode (SHE), whose standard potential is defined as zero.
Standard conditions include:
- Concentration = 1 M
- Pressure = 1 bar
- Temperature = 298 K
$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$
A positive value of $$E^\circ_{cell}$$ indicates that the reaction is spontaneous.
Worked Example: EMF of Daniell Cell
Given:
- $$E^\circ(Cu^{2+}/Cu)=+0.34V$$
- $$E^\circ(Zn^{2+}/Zn)=-0.76V$$
Copper has a higher reduction potential, so it acts as the cathode.
$$E^\circ_{cell}=0.34-(-0.76)$$
$$E^\circ_{cell}=1.10V$$
Since the value is positive, the reaction is spontaneous.
Electrochemical Series
The electrochemical series arranges elements according to their standard reduction potentials.
It helps predict:
- Spontaneity of redox reactions.
- Oxidising and reducing strength.
- Ability of metals to displace hydrogen or other metals.
| Half Reaction | Standard Reduction Potential (V) |
|---|---|
| $$Li^+ + e^- \rightarrow Li$$ | -3.05 |
| $$Zn^{2+}+2e^- \rightarrow Zn$$ | -0.76 |
| $$Fe^{2+}+2e^- \rightarrow Fe$$ | -0.44 |
| $$2H^++2e^- \rightarrow H_2$$ | 0.00 |
| $$Cu^{2+}+2e^- \rightarrow Cu$$ | +0.34 |
| $$Ag^++e^- \rightarrow Ag$$ | +0.80 |
| $$F_2+2e^- \rightarrow 2F^-$$ | +2.87 |
JEE Tip: More positive reduction potential means stronger oxidising ability, while more negative reduction potential means stronger reducing ability.
Worked Example: Displacement Reaction
Will iron displace copper from copper sulphate solution?
Given:
- $$E^\circ(Fe^{2+}/Fe)=-0.44V$$
- $$E^\circ(Cu^{2+}/Cu)=+0.34V$$
Iron has lower reduction potential, so it acts as the reducing agent.
$$E^\circ_{cell}=0.34-(-0.44)$$
$$E^\circ_{cell}=0.78V$$
Since $$E^\circ_{cell}>0$$, iron displaces copper.
Nernst Equation, Gibbs Energy and Equilibrium Constant
Nernst Equation
Standard electrode potentials are measured under standard conditions. The Nernst equation calculates cell potential under non-standard conditions.
$$E_{cell}=E^\circ_{cell}-\frac{RT}{nF}\ln Q$$
At 298 K:
$$E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q$$
Where:
- $$n$$ = number of electrons transferred
- $$F$$ = Faraday constant = $$96500\,C/mol$$
- $$Q$$ = reaction quotient
Important: Pure solids and liquids are not included in the reaction quotient.
Worked Example: Nernst Equation
Calculate the EMF of:
$$Zn|Zn^{2+}(0.1M)||Cu^{2+}(0.01M)|Cu$$
For Daniell cell:
$$E^\circ_{cell}=1.10V$$
$$n=2$$
$$Q=\frac{[Zn^{2+}]}{[Cu^{2+}]}$$
$$Q=\frac{0.1}{0.01}=10$$
$$E_{cell}=1.10-\frac{0.0591}{2}\log10$$
$$E_{cell}=1.10-0.02955$$
$$E_{cell}=1.07V$$
Equilibrium Constant from EMF
At equilibrium:
$$E_{cell}=0$$
The reaction quotient becomes the equilibrium constant:
$$Q=K$$
$$\log K=\frac{nE^\circ_{cell}}{0.0591}$$
A large positive cell potential means a large equilibrium constant and a reaction strongly favoured towards products.
Gibbs Energy and Cell Potential
$$\Delta G=-nFE_{cell}$$
Under standard conditions:
$$\Delta G^\circ=-nFE^\circ_{cell}$$
| Cell Potential | Gibbs Energy | Nature |
|---|---|---|
| $$E^\circ_{cell}>0$$ | $$\Delta G^\circ<0$$ | Spontaneous |
| $$E^\circ_{cell}<0$$ | $$\Delta G^\circ>0$$ | Non-spontaneous |
Worked Example: Gibbs Energy
Calculate $$\Delta G^\circ$$ for a cell with:
$$E^\circ_{cell}=1.10V$$
$$n=2$$
$$\Delta G^\circ=-(2)(96500)(1.10)$$
$$\Delta G^\circ=-212300J$$
$$\Delta G^\circ=-212.3kJ$$
Electrolysis and Faraday's Laws
Electrolytic Cells
Electrolysis uses electrical energy to force a non-spontaneous reaction to occur.
| Feature | Galvanic Cell | Electrolytic Cell |
|---|---|---|
| Reaction | Spontaneous | Non-spontaneous |
| Energy Conversion | Chemical to electrical | Electrical to chemical |
| Anode | Negative | Positive |
| Cathode | Positive | Negative |
Faraday's First Law of Electrolysis
The mass deposited at an electrode is directly proportional to the quantity of electricity passed.
$$m=\frac{MIt}{nF}=\frac{MQ}{nF}$$
Where:
- $$m$$ = mass deposited
- $$M$$ = molar mass
- $$I$$ = current
- $$t$$ = time
- $$Q$$ = charge passed
- $$F$$ = Faraday constant
Faraday's Second Law
When the same quantity of electricity passes through different electrolytes, the deposited masses are proportional to their equivalent weights.
$$\frac{m_1}{m_2}=\frac{E_1}{E_2}$$
Equivalent weight:
$$E=\frac{M}{n}$$
Conductance, Molar Conductivity and Kohlrausch's Law
Conductance
The ability of a solution to conduct electricity is called conductance. It is the reciprocal of resistance.
$$G=\frac{1}{R}$$
The SI unit of conductance is siemens (S).
Specific Conductance
Specific conductance or conductivity represents the conductance of a solution kept between electrodes of unit area separated by unit distance.
$$\kappa=G\frac{l}{A}$$
Where:
- $$\kappa$$ = specific conductivity
- $$l$$ = distance between electrodes
- $$A$$ = area of electrodes
Cell Constant
The ratio of distance between electrodes to the area of electrodes is called cell constant.
$$\text{Cell constant}=\frac{l}{A}$$
Therefore:
$$\kappa=G\times\frac{l}{A}$$
Molar Conductivity
Molar conductivity is the conductance of all ions produced by one mole of an electrolyte dissolved in solution.
$$\Lambda_m=\frac{\kappa\times1000}{C}$$
Where:
- $$\Lambda_m$$ = molar conductivity
- $$C$$ = concentration in mol/L
Effect of Dilution on Conductivity
| Quantity | Effect on Dilution |
|---|---|
| Specific conductivity (κ) | Decreases |
| Molar conductivity (Λm) | Increases |
Specific conductivity decreases because the number of ions per unit volume decreases. Molar conductivity increases because ions move more freely at lower concentration.
| Electrolyte | Behaviour on Dilution |
|---|---|
| Strong electrolyte | Molar conductivity increases slowly |
| Weak electrolyte | Molar conductivity increases sharply |
Kohlrausch's Law
Kohlrausch's law states that at infinite dilution, each ion contributes independently to the total molar conductivity.
$$\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ$$
Applications of Kohlrausch's law:
- Finding limiting molar conductivity of weak electrolytes.
- Calculating degree of dissociation.
- Finding dissociation constant of weak electrolytes.
Degree of Dissociation
$$\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}$$
For weak electrolytes:
$$K_a=\frac{C\alpha^2}{1-\alpha}$$
Worked Example: Kohlrausch's Law
Find limiting molar conductivity of acetic acid using:
- $$\Lambda_m^\circ(CH_3COONa)=91.0$$
- $$\Lambda_m^\circ(HCl)=426.2$$
- $$\Lambda_m^\circ(NaCl)=126.5$$
Using Kohlrausch's law:
$$\Lambda_m^\circ(CH_3COOH)=\Lambda_m^\circ(CH_3COONa)+\Lambda_m^\circ(HCl)-\Lambda_m^\circ(NaCl)$$
$$=91.0+426.2-126.5$$
$$=390.7\,S\,cm^2/mol$$
Batteries and Corrosion
Types of Batteries
Batteries are practical applications of galvanic cells used to convert chemical energy into electrical energy.
| Battery Type | Features |
|---|---|
| Primary Battery | Non-rechargeable and irreversible reactions |
| Dry Cell | Zn anode, MnO₂ cathode, approximately 1.5 V |
| Mercury Cell | Zn and HgO electrodes, constant voltage |
| Secondary Battery | Rechargeable battery with reversible reactions |
| Lead Acid Battery | Used in automobiles, approximately 2 V per cell |
| Fuel Cell | Reactants supplied continuously from outside |
Lead Acid Battery
During discharge:
Anode:
$$Pb+SO_4^{2-}\rightarrow PbSO_4+2e^-$$
Cathode:
$$PbO_2+SO_4^{2-}+4H^++2e^-\rightarrow PbSO_4+2H_2O$$
Overall reaction:
$$Pb+PbO_2+2H_2SO_4\rightarrow2PbSO_4+2H_2O$$
Corrosion and Rusting of Iron
Corrosion is the gradual deterioration of metals due to electrochemical reactions with the environment.
Rusting of iron requires both oxygen and water.
Anodic reaction:
$$Fe\rightarrow Fe^{2+}+2e^-$$
Cathodic reaction:
$$O_2+4H^++4e^-\rightarrow2H_2O$$
The final rust formed is hydrated iron(III) oxide:
$$Fe_2O_3\cdot xH_2O$$
Prevention of Corrosion
- Galvanisation: Coating iron with zinc, which acts as a sacrificial metal.
- Painting: Prevents contact with oxygen and moisture.
- Electroplating: Coating with metals like chromium or nickel.
- Cathodic protection: Connecting iron with a more reactive metal like magnesium.
- Alloying: Producing corrosion-resistant alloys like stainless steel.
Redox Reactions and Electrochemistry Formula Sheet
| Concept | Formula |
|---|---|
| Cell EMF | $$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$ |
| Nernst Equation | $$E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q$$ |
| Gibbs Energy | $$\Delta G=-nFE_{cell}$$ |
| Faraday's Law | $$m=\frac{MIt}{nF}$$ |
| Charge | $$Q=It$$ |
| Conductance | $$G=\frac{1}{R}$$ |
| Molar Conductivity | $$\Lambda_m=\frac{\kappa\times1000}{C}$$ |
| Kohlrausch's Law | $$\Lambda_m^\circ=\nu_+\lambda_+^\circ+\nu_-\lambda_-^\circ$$ |
JEE Important Points, Common Mistakes and Quick Revision
Points JEE Repeatedly Tests
- Oxidation involves loss of electrons, while reduction involves gain of electrons.
- The oxidising agent gets reduced, whereas the reducing agent gets oxidised.
- Oxidation number helps identify whether an element has gained or lost electrons.
- Oxygen generally has oxidation number −2, except in peroxides, superoxides and OF₂.
- Hydrogen generally has oxidation number +1, except in metal hydrides where it is −1.
- Fluorine always has oxidation number −1 because it is the most electronegative element.
- In a galvanic cell, oxidation occurs at the negative anode and reduction occurs at the positive cathode.
- In an electrolytic cell, oxidation still occurs at the anode and reduction at the cathode, but the signs of electrodes are reversed.
- A positive standard cell potential indicates a spontaneous reaction.
- The electrochemical series helps compare oxidising and reducing strength.
- In Nernst equation problems, always determine the value of electrons transferred from the balanced overall reaction.
- Only gaseous and dissolved species appear in reaction quotient; pure solids and liquids are ignored.
- During electrolysis, the amount of substance deposited depends on charge passed and equivalent weight.
- Specific conductivity decreases on dilution, while molar conductivity increases.
- Kohlrausch's law is especially useful for finding limiting molar conductivity of weak electrolytes.
- Rusting requires both oxygen and moisture and can be prevented by galvanisation and cathodic protection.
Common Mistakes to Avoid
- Using oxidation potentials instead of reduction potentials: Standard cell potential is calculated using reduction potentials only.
- Taking incorrect value of n: The number of electrons must be taken from the balanced overall reaction, not a single half reaction.
- Including solids in reaction quotient: Pure solids and liquids do not appear in the Nernst equation expression.
- Inverting the reaction quotient: Always write products divided by reactants according to the balanced reaction.
- Confusing conductance and conductivity: Conductance depends on resistance, while conductivity depends on cell dimensions.
- Assuming conductivity and molar conductivity behave similarly: They show opposite trends during dilution.
- Using Faraday's law without unit conversion: Time must always be converted into seconds.
- Assuming anode is always negative: The sign depends on whether the cell is galvanic or electrolytic.
- Reading electrochemical series incorrectly: More positive reduction potential means stronger oxidising ability.
- Ignoring coefficients while balancing redox reactions: Electron balance is essential for correct balancing.
Quick Revision Notes for Redox Reactions and Electrochemistry
- Redox reactions involve simultaneous oxidation and reduction.
- Oxidation number changes indicate electron transfer.
- Half reaction method balances atoms as well as charge.
- Galvanic cells convert chemical energy into electrical energy.
- Electrolytic cells use electrical energy to drive non-spontaneous reactions.
- Anode is the site of oxidation and cathode is the site of reduction.
- Cell potential determines whether a reaction is spontaneous.
- Nernst equation calculates cell potential under non-standard conditions.
- Gibbs energy connects electrochemical potential with spontaneity.
- Faraday's laws explain the quantitative relationship between electricity and chemical deposition.
- Molar conductivity increases with dilution because ion mobility improves.
- Kohlrausch's law allows calculation of limiting molar conductivity.
- Batteries are practical applications of electrochemical cells.
- Corrosion is an electrochemical process that can be controlled using protective methods.
Exam focus: The most frequently tested areas from this chapter are oxidation number calculation, redox balancing, electrochemical series, Nernst equation, Faraday's laws and conductivity. Practising numerical applications of these concepts is essential for JEE preparation.
