Electrochemistry and Kinetics Numerical Review

Introduction

Electrochemistry and kinetics problems share a hidden structure: each asks you to choose a basis and convert everything onto it. Electrolysis runs on moles of electrons, cell potentials adjust through concentrations, and rate problems run on time. Mixed questions deliberately combine these, for example by plating out metal and then asking for a new electrode potential. This review gathers the key relations from earlier pages and practises choosing the right basis for each step.

Core explanation

Electron basis. Whenever charge, current or electrical energy appears, convert to moles of electrons first: n(e⁻) = It/F. From there, divide by z to reach moles of product, or multiply by E to reach energy (w = n(e⁻)FE). The same electron count links ΔG = −nFE to the extent of reaction: n is electrons per mole of reaction, and the extent tells you how many moles of reaction occur.

Concentration basis. Cell potentials depend on concentrations through the Nernst equation, E = E° − (0.0592/n) log Q at 298 K. After an electrolysis or a cell discharge, recompute concentrations from the moles consumed and the solution volume before substituting. For conductivity, the basis switches to mol m⁻³ so that Λm = κ/c comes out in S m² mol⁻¹.

Time basis. Rate constants carry reciprocal time units, and current-based charge requires seconds. Before any calculation, convert every time to the unit demanded by the equation in use: seconds for Q = It; the unit of k for ln([A]₀/[A]) = kt.

Key relations at a glance.

Quantity Relation Common trap --- --- --- Cell potential E°cell = E°(cathode) − E°(anode) Multiplying E° by coefficients Gibbs energy ΔG = −nFE Forgetting n Composition effect E = E° − (0.0592/n) log Q Including solids in Q Electrolysis n(e⁻) = It/F Time in minutes Molar conductivity Λm = κ/c c in mol dm⁻³ First-order decay ln([A]₀/[A]) = kt Mixing time units Arrhenius ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) Using °C

Linking steps. In a combined problem, the output of one step becomes the input of the next. Plating copper reduces [Cu²⁺], which lowers the Cu²⁺/Cu potential; a temperature change alters k through Arrhenius, which then changes the half-life. Keep unrounded intermediate values and round only the final answers.

Consistency checks. Verify with an independent route where possible: electrical work should equal both ξ(−ΔG) and QE; a first-order time should match a whole number of half-lives when the fraction is a power of one half; a potential should move in the direction Le Chatelier's principle predicts.

Step-by-step reasoning

1. Read the whole problem and list the quantities asked for. 2. Choose the basis for each step: electrons, concentration or time. 3. Convert all data to that basis with SI-consistent units. 4. Carry the result of each step forward without early rounding. 5. Check each answer by a second route or by a physical-sense test.

Visual explanation

Draw a flow chart with three hubs: "mol e⁻", "concentration" and "time". Arrows labelled It/F, ÷z, ×V, Nernst and ln-ratio connect the data given to the hubs and the hubs to the answers. Every mixed problem becomes a path through this chart.

Real-world analogy

Travelling across several countries requires changing money at each border. Mixing currencies in one sum gives nonsense; converting everything into one currency at each stage keeps the accounts right. Electrons, concentrations and times are the currencies of these problems.

Real-world example

A copper refinery tracks both the mass of copper deposited (an electron-basis calculation) and the falling Cu²⁺ concentration of the electrolyte (a concentration-basis calculation), because the cell voltage and the purity of the deposit both depend on keeping that concentration within a working range.

Why?

Why emphasise the basis rather than memorising combined formulae? Combined formulae hide the physical steps, so errors in electron number or units pass unnoticed. Working through a clear basis at each stage makes every conversion visible and checkable.

Common misconception

"After electrolysis the electrode potential stays at its standard value." Removing ions changes their concentration, and the Nernst equation shows that the potential shifts. Only when every species remains at standard concentration does E equal E°.

Worked example

Question: 250 cm³ of 0.100 mol dm⁻³ CuSO₄ is electrolysed with copper electrodes arranged so that copper is only deposited, at 0.750 A for 20.0 min. Find the mass of copper deposited and the Cu²⁺/Cu electrode potential afterwards (E° = +0.34 V, 298 K).

Reasoning: Q = 0.750 × 1200 = 900 C; n(e⁻) = 900 ÷ 96 485 = 9.33 × 10⁻³ mol; n(Cu) = 4.66 × 10⁻³ mol; mass = 4.66 × 10⁻³ × 63.55 = 0.296 g. Initial Cu²⁺ = 0.0250 mol; remaining = 0.0250 − 0.00466 = 0.0203 mol; concentration = 0.0203 ÷ 0.250 = 0.0813 mol dm⁻³. E = 0.34 − (0.0592/2) log(1/0.0813) = 0.34 − 0.032 = 0.31 V.

Answer: 0.296 g of copper; E ≈ +0.31 V.

Quick check

1. Which basis should you convert to first in a question that gives current, time and asks for the volume of hydrogen released? Answer: The electron basis: find n(e⁻) = It/F, then n(H₂) = n(e⁻)/2, then use the gas equation.

Exam focus

Multi-step questions award method marks for each linked stage, so set out every step with a label (charge, electrons, product, concentration, potential). State the basis explicitly and convert units at the start. Round only final answers, usually to three significant figures.

Advanced insight

Electrochemistry and kinetics meet directly in electrode kinetics: the current at an electrode is a reaction rate, and the overpotential needed to drive it follows an exponential dependence similar to the Arrhenius equation (the Butler–Volmer relation). This is why real electrolysis requires more voltage than E°cell predicts, and why catalytic electrode materials matter.

Summary

Mixed problems become routine once each step is assigned a basis: moles of electrons for charge and energy, concentration for Nernst and conductivity calculations, and consistent time units for rate laws and Q = It. Carry unrounded values between steps and verify answers by an independent route or a physical-sense check.

Practice questions

1. A current of 0.500 A flows for 1930 s through dilute acid. What amount of hydrogen is produced? Answer: Q = 965 C; n(e⁻) = 0.0100 mol; n(H₂) = 0.00500 mol. 2. A cell with n = 2 and E = 0.95 V delivers 0.0300 mol of reaction. Find the electrical work. Answer: w = nξFE = 2 × 0.0300 × 96 485 × 0.95 ≈ 5.5 kJ. 3. A first-order reaction has t½ = 40 min at 300 K and Ea = 60 kJ mol⁻¹. Estimate t½ at 310 K. Answer: ln(k₂/k₁) = (60 000/8.314) × (10/93 000) ≈ 0.776, so k rises by about 2.17 and t½ ≈ 40/2.17 ≈ 18 min. 4. Explain why a Nernst calculation after electrolysis needs the solution volume. Answer: Faraday's law gives moles removed, but the Nernst equation needs concentrations, so the remaining moles must be divided by the volume.