EAPCET Chemistry · Intermediate second year

Electrochemistry and Chemical Kinetics

This chapter combines electrochemistry — cells, electrode potentials and electrolysis — with chemical kinetics — rates, orders and activation energy.

Subject
Chemistry
Taught in
Intermediate second year
Chemistry in the paper
40 of 160 questions

Key formulas and results

  • Nernst equation at 298 K: E=E∘−0.0591nlog⁡QE = E^\circ - \frac{0.0591}{n} \log Q
  • ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ
  • Faraday’s first law: m=ZItm = ZIt
  • First-order rate constant k=2.303tlog⁡aa−xk = \frac{2.303}{t} \log\frac{a}{a - x}; half-life t1/2=0.693kt_{1/2} = \frac{0.693}{k}
  • Arrhenius equation: k=Ae−Ea/RTk = A e^{-E_a/RT}
  • Standard EMF and equilibrium constant at 298 K: Ecell∘=0.0591nlog⁡KE^\circ_{\text{cell}} = \frac{0.0591}{n} \log K
  • Conductance and cell constant: G=1RG = \frac{1}{R}, κ=G×lA\kappa = G \times \frac{l}{A}, resistivity ρ=1κ\rho = \frac{1}{\kappa}
  • Molar conductivity: Λm=κ×1000M\Lambda_m = \frac{\kappa \times 1000}{M} in S cm2 mol−1\text{S cm}^2\,\text{mol}^{-1}, with κ\kappa in S cm−1\text{S cm}^{-1}
  • Degree of dissociation of a weak electrolyte: α=ΛmΛm∘\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}
  • Kohlrausch’s law: Λm∘=λcation∘+λanion∘\Lambda^\circ_m = \lambda^\circ_{\text{cation}} + \lambda^\circ_{\text{anion}}; for strong electrolytes Λm=Λm∘−bC\Lambda_m = \Lambda^\circ_m - b\sqrt{C}
  • Mass deposited in electrolysis: w=M×I×tn×96500w = \frac{M \times I \times t}{n \times 96500}
  • Half-life and order: t1/2∝1an−1t_{1/2} \propto \frac{1}{a^{n-1}}; zero order t1/2=a02kt_{1/2} = \frac{a_0}{2k}, second order t1/2=1ka0t_{1/2} = \frac{1}{k a_0}
  • Arrhenius equation at two temperatures: log⁡k2k1=Ea2.303R×T2−T1T1T2\log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \times \frac{T_2 - T_1}{T_1T_2}

What this chapter covers

  • Galvanic cells and electrode potentials
  • Nernst equation
  • Conductance and Kohlrausch’s law
  • Electrolysis and Faraday’s laws
  • Rate laws and order
  • Activation energy and the Arrhenius equation

How it is usually asked

  • Cell emf
  • Mass deposited in electrolysis
  • Molar conductivity
  • Order, rate constant and half-life
  • Activation energy

Understanding Electrochemistry and Chemical Kinetics

In a galvanic cell, a spontaneous redox reaction produces electricity: oxidation at the anode (negative) and reduction at the cathode (positive). The cell emf is E°cell = E°cathode − E°anode, using standard reduction potentials. A positive E°cell means ΔG° = −nFE° is negative, so the reaction is spontaneous. The Nernst equation adjusts emf for non-standard concentrations, and at equilibrium E = 0, giving E° = (0.0591/n) log K at 298 K.

Conductance of a solution rises with ion concentration, but molar conductivity rises on dilution. For strong electrolytes it rises only slightly (Λm falls linearly with √c); for weak electrolytes it rises steeply, so Λ°m is found with Kohlrausch’s law of independent migration of ions. In electrolysis, Faraday’s laws link charge to products: one faraday, 96 500 C, is the charge of one mole of electrons, and mass deposited = (charge/96 500) × (molar mass/n).

Chemical kinetics measures how fast reactions go. The rate law, found by experiment, gives the order, which need not match the stoichiometric coefficients. A first-order reaction has a constant half-life, t½ = 0.693/k, independent of starting concentration; a zero-order reaction has t½ = [A]₀/2k. Rate rises with temperature because more molecules exceed the activation energy, as described by the Arrhenius equation; a catalyst lowers Ea without changing ΔH.

Worked example

EAPCET-style question

For the Daniell cell Zn∣Zn2+(0.1 M)∥Cu2+(0.001 M)∣Cu\text{Zn} \mid \text{Zn}^{2+}(0.1\ \text{M}) \parallel \text{Cu}^{2+}(0.001\ \text{M}) \mid \text{Cu} with Ecell∘=1.10 VE^\circ_{\text{cell}} = 1.10\ \text{V}, the emf at 298 K is:

  1. (1)0.98 V
  2. (2)1.04 V
  3. (3)1.10 V
  4. (4)1.16 V

Solution

  1. Cell reaction: Zn+CuX2+→ZnX2++Cu\ce{Zn + Cu^2+ -> Zn^2+ + Cu}, with n=2n = 2.
  2. Nernst: E=E∘−0.0591nlog⁡[ZnX2+][CuX2+]E = E^\circ - \frac{0.0591}{n} \log\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}.
  3. [ZnX2+][CuX2+]=0.10.001=100\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} = \frac{0.1}{0.001} = 100, so log⁡100=2\log 100 = 2.
  4. E=1.10−0.05912×2=1.10−0.059=1.041 V≈1.04 VE = 1.10 - \frac{0.0591}{2} \times 2 = 1.10 - 0.059 = 1.041\ \text{V} \approx 1.04\ \text{V}.
  5. 1.16 V comes from inverting QQ, and 0.98 V from taking n=1n = 1.

Answer: 1.04 V

Shortcuts and results to remember

  • First-order reactions: t75%=2 t1/2t_{75\%} = 2\,t_{1/2}, t87.5%=3 t1/2t_{87.5\%} = 3\,t_{1/2}, t99.9%≈10 t1/2t_{99.9\%} \approx 10\,t_{1/2} and t99%=2 t90%t_{99\%} = 2\,t_{90\%}; zero-order reactions: t75%=1.5 t1/2t_{75\%} = 1.5\,t_{1/2} and t100%=2 t1/2t_{100\%} = 2\,t_{1/2}.
  • Units of the rate constant for order nn are (mol L−1)1−n s−1(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}: zero order mol L−1 s−1\text{mol L}^{-1}\,\text{s}^{-1}, first order s−1\text{s}^{-1}, second order L mol−1 s−1\text{L mol}^{-1}\,\text{s}^{-1}.
  • Temperature coefficient: k2k1=μΔT/10\frac{k_2}{k_1} = \mu^{\Delta T/10}, where μ\mu is usually about 2 (range 1.5 to 3), so a 10 °C rise roughly doubles the rate.
  • A metal–metal ion concentration cell has Ecell∘=0E^\circ_{\text{cell}} = 0, so E=0.0591nlog⁡CcathodeCanodeE = \frac{0.0591}{n} \log\frac{C_{\text{cathode}}}{C_{\text{anode}}}, and it works only when the cathode side is more concentrated.
  • Molecularity is always a whole number from 1 to 3 and applies to elementary steps only; order can be zero or fractional and must be found by experiment.

More practice questions

Practice question 1

A first-order reaction is 75% complete in 60 minutes. Its half-life is

  1. (1)15 min
  2. (2)30 min
  3. (3)45 min
  4. (4)60 min

Solution

  1. 75% completion leaves one quarter of the reactant
  2. a→a2→a4a \to \frac{a}{2} \to \frac{a}{4} takes two half-lives
  3. So t75%=2 t1/2=60t_{75\%} = 2\,t_{1/2} = 60 min
  4. t1/2=30t_{1/2} = 30 min

Answer: 30 min

Practice question 2

A current of 2 A is passed through CuSOX4\ce{CuSO4} solution for 965 s. The mass of copper deposited (Cu = 63.5) is

  1. (1)0.3175 g
  2. (2)0.635 g
  3. (3)1.27 g
  4. (4)6.35 g

Solution

  1. Charge Q=I×t=2×965=1930 CQ = I \times t = 2 \times 965 = 1930\ \text{C}
  2. CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}, so n=2n = 2
  3. w=M×Qn×96500=63.5×19302×96500w = \frac{M \times Q}{n \times 96500} = \frac{63.5 \times 1930}{2 \times 96500}
  4. w=0.635 gw = 0.635\ \text{g}

Answer: 0.635 g

Common mistakes to avoid

  • Multiplying E° by the stoichiometric coefficient when balancing half-reactions — electrode potential is an intensive property and is never scaled.
  • Putting product over reactant the wrong way in the Nernst equation — Q is [products]/[reactants], so for Zn | Cu it is [Zn²⁺]/[Cu²⁺].
  • Assuming every half-life depends on initial concentration — for first order it is constant, which is how you identify first-order data.
  • Equating order with stoichiometric coefficients — order is found experimentally and can be zero or fractional.

How to prepare it

The units of the rate constant tell you the order of the reaction — a quick way to check options.

Once the formulas feel familiar, test them the way the exam does: practise Electrochemistry and Chemical Kinetics questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Electrochemistry and Chemical Kinetics chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Electrochemistry and Chemical Kinetics part of the EAPCET Chemistry syllabus?
Yes. Electrochemistry and Chemical Kinetics is taught in Intermediate second year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Electrochemistry and Chemical Kinetics for EAPCET?
The units of the rate constant tell you the order of the reaction — a quick way to check options.
How many Chemistry questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Electrochemistry and Chemical Kinetics?
Multiplying E° by the stoichiometric coefficient when balancing half-reactions — electrode potential is an intensive property and is never scaled. Putting product over reactant the wrong way in the Nernst equation — Q is [products]/[reactants], so for Zn | Cu it is [Zn²⁺]/[Cu²⁺]. Assuming every half-life depends on initial concentration — for first order it is constant, which is how you identify first-order data. Equating order with stoichiometric coefficients — order is found experimentally and can be zero or fractional.