Electrochemistry
Electrochemistry for JEE Main & Advanced
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: $K_a = c\alpha^2/(1-\alpha)$.
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.
$\Lambda_m^\ominus(NH_4Cl) = 130$, $\Lambda_m^\ominus(NaOH) = 248$, $\Lambda_m^\ominus(NaCl) = 126$. 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.
Specific conductivity $\kappa$ with dilution:
Molar conductivity with dilution (strong electrolyte):
Kohlrausch's law applies at:
Unit of molar conductivity:
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:
| Type | Process | Energy Conversion |
|---|---|---|
| Galvanic (Voltaic) | Spontaneous redox | Chemical → Electrical |
| Electrolytic | Non-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)
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.
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).
SHE has $E^\ominus$:
In Daniell cell, anode is:
Cell with positive $E^\ominus_{cell}$:
Strongest oxidizing agent in electrochemical series:
$\Delta G^\ominus = -nFE^\ominus$ relates:
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. $E^\ominus_{cell} = 0$. $$E_{cell} = \frac{0.0591}{n}\log\frac{[M^{n+}]_{conc}}{[M^{n+}]_{dilute}}$$
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.
$E^\ominus_{cell} = 0.59$ 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: $K_c \approx 10^{20}$.
Nernst equation at $25°$C:
At equilibrium, $E_{cell}$ is:
Concentration cell has:
Increasing $[Cu^{2+}]$ in cathode side of Daniell cell:
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):
| Cell | Anode | Cathode | Electrolyte | Voltage |
|---|---|---|---|---|
| Dry cell (Leclanché) | Zn (case) | C (graphite) rod in $MnO_2$ + C powder | $NH_4Cl$ + $ZnCl_2$ paste | $1.5$ V |
| Alkaline cell | Zn powder | $MnO_2$ + C | KOH | $1.5$ V |
| Mercury cell | Zn-Hg amalgam | HgO | KOH (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):
| Cell | Anode | Cathode | Electrolyte | Voltage |
|---|---|---|---|---|
| Lead-acid | Pb | $PbO_2$ | $38\%\,H_2SO_4$ | $2$ V/cell |
| Ni-Cd | Cd | $NiO(OH)$ | 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!
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.
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.
Faraday's first law: mass deposited proportional to:
In electrolysis of brine, gas at cathode:
Rusting of iron requires:
Galvanising is coating iron with:
Lead-acid battery generates voltage approximately:
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