JEE Main & Advanced

Electrochemistry

Electrochemistry for JEE Main & Advanced

1
Module 1

Conductance and Galvanic Cells

Electrolytic Conductance, Kohlrausch's LawTopic 1

Electrochemistry: Study of chemical changes producing/accompanying electrical energy and vice versa.

Electrolytes: Conduct electricity in molten/aqueous state due to ions. Two types:

  • Strong: Completely dissociate (NaCl, HCl, NaOH, $H_2SO_4$)
  • Weak: Partially dissociate ($CH_3COOH, NH_4OH$)

Non-electrolytes: Do not conduct (e.g., glucose, urea).

Conductance: Reciprocal of resistance, $G = 1/R$. Unit: siemens (S) or $\Omega^{-1}$.

Specific Conductance (Conductivity, $\kappa$ "kappa"): $$\kappa = \frac{1}{\rho} = \frac{l}{RA}$$

  • $\kappa$ unit: S/m or S·cm⁻¹
  • Conductance of 1 cm cube of solution

Cell Constant ($l/A$): Geometric factor of conductivity cell; constant for a given cell.

Molar Conductivity ($\Lambda_m$): Conductance of all ions from 1 mole of electrolyte: $$\Lambda_m = \frac{\kappa \times 1000}{c}$$ where $c$ = concentration in mol/L. Unit: S·cm²/mol.

Equivalent Conductivity ($\Lambda_{eq}$): Conductance for 1 equivalent. $\Lambda_{eq} = \kappa \times 1000/N$.

Variation with concentration:

  • $\kappa$ decreases with dilution (fewer ions per cm³)
  • $\Lambda_m$ increases with dilution (more dissociation in weak electrolytes; more freedom of ion movement in strong)

Strong vs Weak Electrolyte Behavior:

For strong electrolytes, Debye-Hückel-Onsager: $\Lambda_m = \Lambda_m^\ominus - A\sqrt{c}$ (linear with $\sqrt c$).

  • Extrapolation to $c = 0$ gives $\Lambda_m^\ominus$ (limiting molar conductivity, at infinite dilution).

For weak electrolytes: $\Lambda_m$ increases sharply at low $c$ (more dissociation). Cannot find $\Lambda_m^\ominus$ by extrapolation; use Kohlrausch's law.

Kohlrausch's Law of Independent Ion Migration: At infinite dilution, each ion contributes independently to $\Lambda_m^\ominus$: $$\Lambda_m^\ominus = \nu_+ \lambda_+^\ominus + \nu_- \lambda_-^\ominus$$

where $\nu$ = number of cations/anions, $\lambda^\ominus$ = limiting ionic conductivity.

Application: Find $\Lambda_m^\ominus$ for weak electrolytes: $\Lambda_m^\ominus(CH_3COOH) = \Lambda_m^\ominus(CH_3COONa) + \Lambda_m^\ominus(HCl) - \Lambda_m^\ominus(NaCl)$.

Degree of Dissociation ($\alpha$) of Weak Electrolyte: $$\alpha = \frac{\Lambda_m}{\Lambda_m^\ominus}$$

For acid HA: $K_a = c\alpha^2/(1-\alpha)$.

Worked Examples
1

Resistance of $0.01$ M KCl in a cell is $1500\,\Omega$. Conductivity at this conc = $1.413 \times 10^{-3}$ S/cm. Find cell constant.

Show solution

$\kappa = G/(l/A) = 1/(R \cdot l/A) \implies l/A = \kappa \cdot R$ $l/A = 1.413 \times 10^{-3} \times 1500 = 2.12$ cm⁻¹.

Final Answer: Cell constant = $2.12$ cm⁻¹.

2

$\Lambda_m^\ominus(NH_4Cl) = 130$, $\Lambda_m^\ominus(NaOH) = 248$, $\Lambda_m^\ominus(NaCl) = 126$. Find $\Lambda_m^\ominus(NH_4OH)$.

Show solution

By Kohlrausch: $\Lambda_m^\ominus(NH_4OH) = \Lambda_m^\ominus(NH_4Cl) + \Lambda_m^\ominus(NaOH) - \Lambda_m^\ominus(NaCl)$ $= 130 + 248 - 126 = 252$ S·cm²/mol.

Final Answer: $\Lambda_m^\ominus(NH_4OH) = 252$ S·cm²/mol.

✎ Self-Check — 5 questions0 / 5
Q1.

Specific conductivity $\kappa$ with dilution:

Q2.

Molar conductivity with dilution (strong electrolyte):

Q3.

Kohlrausch's law applies at:

Q4.

Unit of molar conductivity:

Q5.

Degree of dissociation $\alpha$ from conductivity:

Electrochemical Cells, EMF, Standard Electrode PotentialTopic 2

Electrochemical Cells: Devices for interconversion of chemical and electrical energy.

Two Types:

TypeProcessEnergy Conversion
Galvanic (Voltaic)Spontaneous redoxChemical → Electrical
ElectrolyticNon-spontaneous (driven by external EMF)Electrical → Chemical

Galvanic Cell Components:

  • Anode: Oxidation (loss of e⁻); negative terminal in galvanic cell
  • Cathode: Reduction (gain of e⁻); positive terminal in galvanic cell
  • Salt bridge: Maintains electrical neutrality; contains KCl or KNO₃ in agar
  • External circuit: Wire connecting electrodes

Daniell Cell:

  • Anode: Zn rod in ZnSO₄ solution; Zn → Zn²⁺ + 2e⁻
  • Cathode: Cu rod in CuSO₄ solution; Cu²⁺ + 2e⁻ → Cu
  • Overall: $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$
  • $E^\ominus_{cell} = 1.10$ V

Cell Notation: Anode | Anode soln || Cathode soln | Cathode. Daniell: $\text{Zn(s)} | \text{Zn}^{2+}(1M) || \text{Cu}^{2+}(1M) | \text{Cu(s)}$

EMF (Electromotive Force): Potential difference between electrodes when no current flows; max work obtainable. $$E_{cell} = E_{cathode} - E_{anode}$$

Always: $E_{cathode} > E_{anode}$ for spontaneous cell, so $E_{cell} > 0$.

Standard Electrode Potential ($E^\ominus$): Potential of electrode at $1$ M concentration, $1$ bar, $298$ K, relative to SHE.

Standard Hydrogen Electrode (SHE): Reference; $E^\ominus = 0$ V.

  • Half-reaction: $2H^+ + 2e^- \rightleftharpoons H_2$; $H_2$ gas at $1$ bar, $H^+$ at $1$ M

Electrochemical Series: Half-cells arranged by decreasing $E^\ominus$:

Half-Reaction$E^\ominus$ (V)
$F_2 + 2e^- \to 2F^-$$+2.87$
$MnO_4^- + 8H^+ + 5e^- \to Mn^{2+}$$+1.51$
$Cl_2 + 2e^- \to 2Cl^-$$+1.36$
$Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}$$+1.33$
$O_2 + 4H^+ + 4e^- \to 2H_2O$$+1.23$
$Br_2 + 2e^- \to 2Br^-$$+1.09$
$Ag^+ + e^- \to Ag$$+0.80$
$Cu^{2+} + 2e^- \to Cu$$+0.34$
$2H^+ + 2e^- \to H_2$$0.00$
$Pb^{2+} + 2e^- \to Pb$$-0.13$
$Sn^{2+} + 2e^- \to Sn$$-0.14$
$Fe^{2+} + 2e^- \to Fe$$-0.44$
$Zn^{2+} + 2e^- \to Zn$$-0.76$
$Al^{3+} + 3e^- \to Al$$-1.66$
$Mg^{2+} + 2e^- \to Mg$$-2.37$
$Na^+ + e^- \to Na$$-2.71$
$K^+ + e^- \to K$$-2.93$
$Li^+ + e^- \to Li$$-3.04$

Higher $E^\ominus$: Better oxidizing agent (gets reduced). Lower $E^\ominus$ (more negative): Better reducing agent (gets oxidized; better at giving electrons).

Free Energy and EMF: $$\Delta G^\ominus = -nFE^\ominus_{cell}$$

  • $n$: moles of electrons transferred
  • $F$ = $96500$ C/mol (Faraday constant)
Worked Examples
1

Calculate $E^\ominus_{cell}$ for Daniell cell from $E^\ominus(Cu^{2+}/Cu) = +0.34$ V, $E^\ominus(Zn^{2+}/Zn) = -0.76$ V.

Show solution

Cu²⁺ is reduced (higher $E^\ominus$, cathode), Zn is oxidized (anode): $E^\ominus_{cell} = E^\ominus_{cathode} - E^\ominus_{anode} = 0.34 - (-0.76) = 1.10$ V.

Final Answer: $E^\ominus_{cell} = 1.10$ V.

2

Find $\Delta G^\ominus$ for Daniell cell.

Show solution

$n = 2$ (Zn loses 2 e⁻). $\Delta G^\ominus = -nFE^\ominus = -2 \times 96500 \times 1.10 = -212,300$ J/mol $= -212.3$ kJ/mol.

Final Answer: $\Delta G^\ominus = -212.3$ kJ/mol (spontaneous).

✎ Self-Check — 5 questions0 / 5
Q1.

SHE has $E^\ominus$:

Q2.

In Daniell cell, anode is:

Q3.

Cell with positive $E^\ominus_{cell}$:

Q4.

Strongest oxidizing agent in electrochemical series:

Q5.

$\Delta G^\ominus = -nFE^\ominus$ relates:

2
Module 2

Nernst Equation and Applications

Nernst Equation, Cell EMF CalculationsTopic 1

Nernst Equation: Relates cell EMF to concentration (non-standard conditions).

For: $aA + bB \to cC + dD$: $$E_{cell} = E^\ominus_{cell} - \frac{RT}{nF}\ln\frac{[C]^c[D]^d}{[A]^a[B]^b}$$

At $25°$C ($T = 298$ K), using base-10 log and $R = 8.314$: $$E_{cell} = E^\ominus_{cell} - \frac{0.0591}{n}\log\frac{[C]^c[D]^d}{[A]^a[B]^b}$$

Note: Activities of pure solids and pure liquids = 1.

For a single electrode (e.g., $M^{n+} + ne^- \to M$): $$E_{M^{n+}/M} = E^\ominus_{M^{n+}/M} + \frac{0.0591}{n}\log[M^{n+}]$$

(Positive sign because we use reduction potential and concentration of the oxidized form increases potential.)

Equilibrium Constant Relation: At equilibrium, $E_{cell} = 0$: $$E^\ominus_{cell} = \frac{0.0591}{n}\log K_c$$

So $K_c = 10^{nE^\ominus/0.0591}$.

Effects of Concentration:

  • $E_{cell}$ varies if concentrations are not 1 M
  • Diluting cathode side decreases $E_{cell}$
  • Diluting anode side increases $E_{cell}$

Concentration Cells: Same electrode in two different concentrations. $E^\ominus_{cell} = 0$. $$E_{cell} = \frac{0.0591}{n}\log\frac{[M^{n+}]_{conc}}{[M^{n+}]_{dilute}}$$

Worked Examples
1

Daniell cell with $[Zn^{2+}] = 0.1$ M, $[Cu^{2+}] = 0.5$ M. Find $E_{cell}$.

Show solution

$E^\ominus = 1.10$ V, $n = 2$. $E_{cell} = 1.10 - (0.0591/2)\log([Zn^{2+}]/[Cu^{2+}])$ $= 1.10 - 0.0296 \log(0.1/0.5) = 1.10 - 0.0296 \log(0.2)$ $= 1.10 - 0.0296 \times (-0.699) = 1.10 + 0.0207 = 1.121$ V.

Final Answer: $E_{cell} = 1.121$ V.

2

$E^\ominus_{cell} = 0.59$ V for $n = 2$. Find $K_c$.

Show solution

$\log K_c = nE^\ominus/0.0591 = 2 \times 0.59/0.0591 = 19.97$. $K_c \approx 10^{20}$.

Final Answer: $K_c \approx 10^{20}$.

✎ Self-Check — 5 questions0 / 5
Q1.

Nernst equation at $25°$C:

Q2.

At equilibrium, $E_{cell}$ is:

Q3.

Concentration cell has:

Q4.

Increasing $[Cu^{2+}]$ in cathode side of Daniell cell:

Q5.

Relation $\log K = nE^\ominus/0.0591$ uses:

Electrolysis, Faraday's Laws, Corrosion, BatteriesTopic 2

Electrolysis: Non-spontaneous chemical change driven by passage of electric current.

Electrolytic Cell:

  • Anode (positive, connected to + terminal of battery): Oxidation
  • Cathode (negative): Reduction

Products of Electrolysis:

For solution of NaCl (brine):

  • Anode: $2Cl^- \to Cl_2 + 2e^-$ (not $O_2$ because of overpotential)
  • Cathode: $2H_2O + 2e^- \to H_2 + 2OH^-$ (Na⁺ has lower $E^\ominus$ than H₂O)

For molten NaCl:

  • Anode: $2Cl^- \to Cl_2 + 2e^-$
  • Cathode: $Na^+ + e^- \to Na$

For copper sulfate (with Cu electrodes — refining):

  • Anode: $Cu \to Cu^{2+} + 2e^-$
  • Cathode: $Cu^{2+} + 2e^- \to Cu$

(Used in electrorefining of copper.)

Faraday's Laws of Electrolysis:

1st Law: Amount of substance liberated at electrode $\propto$ quantity of charge: $$w = ZIt = ZQ$$ where $Z$ = electrochemical equivalent (g/C), $I$ = current (A), $t$ = time (s), $Q$ = charge.

2nd Law: When same charge passes through different electrolytes, masses of substances deposited are in ratio of equivalent weights: $$\frac{w_1}{w_2} = \frac{E_1}{E_2}$$

Faraday constant: $F = 96500$ C/mol $= eN_A$. Charge of 1 mole of electrons.

Mass deposited: $w = E \cdot Q/F = E \cdot I \cdot t/F$ $w = (M/n) \cdot I t/96500$, where $M$ = molar mass, $n$ = valence change.

Corrosion: Oxidation of metals in environment.

Rusting of Iron: Electrochemical process.

  • Anode: $Fe \to Fe^{2+} + 2e^-$
  • Cathode: $O_2 + 4H^+ + 4e^- \to 2H_2O$ (or $O_2 + 2H_2O + 4e^- \to 4OH^-$)
  • Further: $Fe^{2+}$ oxidized to $Fe^{3+}$; with $H_2O$ → $Fe_2O_3 \cdot xH_2O$ (rust, hydrated $Fe(III)$ oxide)

Conditions: Moisture + air (especially with $CO_2$ or salts dissolved).

Prevention:

  • Painting, galvanising (Zn coating), tin plating, electroplating
  • Cathodic protection: Connect Fe to a more active metal (Zn, Mg) that corrodes preferentially (sacrificial anode)
  • Anodising (Al with $Al_2O_3$ layer)
  • Alloying (stainless steel: Fe + Cr + Ni)

Batteries (Commercial Galvanic Cells):

1. Primary cells (non-rechargeable):

CellAnodeCathodeElectrolyteVoltage
Dry cell (Leclanché)Zn (case)C (graphite) rod in $MnO_2$ + C powder$NH_4Cl$ + $ZnCl_2$ paste$1.5$ V
Alkaline cellZn powder$MnO_2$ + CKOH$1.5$ V
Mercury cellZn-Hg amalgamHgOKOH (paste)$1.35$ V

Reactions (dry cell):

  • Anode: $Zn \to Zn^{2+} + 2e^-$
  • Cathode: $2MnO_2 + 2NH_4^+ + 2e^- \to Mn_2O_3 + 2NH_3 + H_2O$

2. Secondary cells (rechargeable):

CellAnodeCathodeElectrolyteVoltage
Lead-acidPb$PbO_2$$38\%\,H_2SO_4$$2$ V/cell
Ni-CdCd$NiO(OH)$KOH$1.4$ V

Lead acid cell (car battery):

  • Anode: $Pb + SO_4^{2-} \to PbSO_4 + 2e^-$
  • Cathode: $PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \to PbSO_4 + 2H_2O$
  • Reverses on charging

Fuel Cells: Convert chemical energy from fuel directly to electricity. $H_2$-$O_2$ fuel cell:

  • Anode: $H_2 + 2OH^- \to 2H_2O + 2e^-$
  • Cathode: $O_2 + 2H_2O + 4e^- \to 4OH^-$
  • Used in space programs (Apollo missions); produces drinkable water!
Worked Examples
1

A current of $5$ A is passed for $30$ min through CuSO₄ solution. Mass of Cu deposited? ($M_{Cu} = 63.5, n = 2$)

Show solution

$Q = It = 5 \times 30 \times 60 = 9000$ C. $w = (M/n)(Q/F) = (63.5/2)(9000/96500) = 31.75 \times 0.0933 = 2.96$ g.

Final Answer: $w \approx 2.96$ g.

2

Galvanising involves coating iron with which metal? Why is it effective?

Show solution

Iron is coated with zinc. Zn is more active (more negative $E^\ominus$: $-0.76$ V vs Fe $-0.44$ V), so it oxidizes preferentially (sacrificial anode), protecting iron. Even if coating scratched, Zn corrodes first.

Final Answer: Zinc; acts as sacrificial anode by being more electropositive.

✎ Self-Check — 5 questions0 / 5
Q1.

Faraday's first law: mass deposited proportional to:

Q2.

In electrolysis of brine, gas at cathode:

Q3.

Rusting of iron requires:

Q4.

Galvanising is coating iron with:

Q5.

Lead-acid battery generates voltage approximately:

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