Química Geral · Entropy, free energy and electrochemistry
Eletrólise, baterias e corrosão
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O que você vai aprender
- Predict the product at each electrode when a molten salt or an aqueous solution is electrolysed, including when water or an active-metal anode reacts instead of the ions.
- Use Q = It and F = 96 485 C/mol to convert between current, time, moles of electrons and the amount of product, in either direction.
- Set up electroplating and refining cells and calculate the mass, thickness or time of a metal deposit.
- Describe how alkaline, lead–acid, lithium-ion and hydrogen–oxygen fuel cells work, and calculate charge, mass and efficiency for them.
- Explain the corrosion of iron as an electrochemical cell and choose coatings, sacrificial anodes or impressed currents to prevent it.
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Electrolytic cells
An electrolytic cell uses an external power supply to drive a redox reaction that would not run by itself (E°cell < 0, ΔG° > 0). The electrode joined to the negative terminal is the cathode: electrons are pushed onto it and something is reduced there. The electrode joined to the positive terminal is the anode: electrons are pulled out of something there, so oxidation happens. Anode means oxidation and cathode means reduction in every kind of cell; only the signs of the terminals differ from a galvanic cell. Inside the melt or solution the current is carried by ions: cations drift toward the cathode and anions toward the anode. Molten sodium chloride (drawn) contains nothing else that can react, so sodium forms at the cathode (Na⁺ + e⁻ → Na) and chlorine at the anode (2Cl⁻ → Cl₂ + 2e⁻). The smallest voltage that can drive a reaction equals the size of its negative E°cell, with E°cell = E°cathode − E°anode as for any cell. Splitting water, 2H₂O → 2H₂ + O₂, has E°cell = 0.00 − 1.23 = −1.23 V, so at least 1.23 V must be applied under standard conditions; real cells need more to overcome resistance and overpotential.
Electrolysis in water
In aqueous solution water competes with the dissolved ions at both electrodes. At the cathode, the species with the most positive reduction potential is reduced first. Ag⁺ (+0.80 V) and Cu²⁺ (+0.34 V) are reduced to the metal. Na⁺ (−2.71 V), K⁺ (−2.93 V), Mg²⁺ (−2.37 V) and Al³⁺ (−1.66 V) are much harder to reduce than water, so water gives hydrogen instead: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (−0.83 V at 1 M OH⁻, about −0.41 V at pH 7). At the anode, the species with the least positive reduction potential is oxidized first. Iodide (I₂/I⁻ +0.54 V) is oxidized before water (O₂ + 4H⁺ + 4e⁻ → 2H₂O, +1.23 V at 1 M H⁺, about +0.82 V at pH 7). Sulfate, nitrate and fluoride are not oxidized, so water gives O₂ and H⁺ there. Electrolysis of copper(II) sulfate solution (drawn) between inert platinum electrodes therefore gives copper on the cathode and oxygen at the anode, and the solution turns acidic as its blue colour fades: 2Cu²⁺ + 2H₂O → 2Cu + O₂ + 4H⁺. Two corrections to this E° reasoning: O₂ evolution needs a large overpotential, which is why concentrated chloride solutions give Cl₂ in the chlor-alkali process, and an anode made of a reactive metal such as copper is itself oxidized and dissolves instead of producing oxygen.
Faraday's law: counting electrons
The charge passed is Q = I × t, in coulombs when the current is in amperes and the time in seconds. The Faraday constant, F = 96 485 C/mol, is the charge of one mole of electrons (Avogadro's number times the charge of one electron), so Q/F is the number of moles of electrons. The half-reaction turns moles of electrons into moles of product: Mⁿ⁺ + ne⁻ → M needs n mol of electrons per mole of metal, H₂ and Cl₂ need 2 per mole, and O₂ needs 4 per mole. Worked example: 4.00 A for 30.0 min through aqueous NiSO₄. Q = 4.00 A × 1800 s = 7200 C; moles of electrons = 7200 / 96 485 = 0.0746 mol; moles of Ni = 0.0746 / 2 = 0.0373 mol; mass = 0.0373 × 58.69 = 2.19 g. Run the same chain backwards to find a time (t = n(e⁻) × F / I) or a molar mass. Cells connected in series carry the same current for the same time, so each receives the same moles of electrons. Battery capacity is a charge too: 1 A·h = 3600 C and 1 mA·h = 3.6 C. Unless a problem states otherwise, assume 100% current efficiency, meaning every electron makes the product of interest.
Electroplating and refining
To plate an object, make it the cathode by joining it to the negative terminal, and dip it in a solution of the plating metal's ions. If the anode is a bar of the same metal, the anode dissolves as fast as metal deposits, so the ion concentration of the bath stays constant. Count electrons from the oxidation state of the metal in the bath, not from its most familiar ion: copper from Cu²⁺ needs 2 per atom, gold from the dicyanoaurate(I) ion, Au(CN)₂⁻, needs 1, and chromium plated from chromic acid (chromium +6) needs 6. For a coating of known thickness, mass = density × area × thickness, with the thickness converted to centimetres when the density is in g/cm³ (1 µm = 1 × 10⁻⁴ cm). Copper is refined in the same kind of cell: an impure copper anode dissolves and pure copper deposits on the cathode. Impurities more active than copper, such as zinc and iron, are oxidized into solution but stay there as ions because they are harder to reduce than Cu²⁺. Less active metals such as silver and gold are not oxidized at all and fall below the anode as anode mud.
Batteries and fuel cells
A battery is one or more galvanic cells, so its anode is the negative terminal. A primary cell is discarded when spent; a secondary cell is recharged by an external supply that drives the cell reaction backwards. Alkaline cell (about 1.5 V): zinc powder in a KOH paste is the anode (Zn + 2OH⁻ → ZnO + H₂O + 2e⁻) and MnO₂ is reduced at the cathode, manganese going from +4 to +3. Lead–acid cell: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l). From PbO₂/PbSO₄ (+1.69 V) and PbSO₄/Pb (−0.36 V), E° = 1.69 − (−0.36) = 2.05 V per cell, and a 12-V car battery has six cells in series. Discharge coats both plates with PbSO₄ and uses up sulfuric acid, so the density of the electrolyte falls; charging reverses both changes. Lithium-ion cell (about 3.7 V): on discharge Li⁺ leaves lithiated graphite (LiC₆ → C₆ + Li⁺ + e⁻), crosses a nonaqueous electrolyte and enters the metal oxide (CoO₂ + Li⁺ + e⁻ → LiCoO₂), while the electrons travel through the external circuit; charging moves Li⁺ back into the graphite. A fuel cell is fed its reactants continuously instead of storing them. In the hydrogen–oxygen fuel cell, H₂ is oxidized at the anode (2H₂ → 4H⁺ + 4e⁻), O₂ is reduced at the cathode (O₂ + 4H⁺ + 4e⁻ → 2H₂O) and E° = 1.23 V. The largest electrical work a cell can deliver is −ΔG° = nFE°, so the maximum efficiency of a fuel cell is ΔG°/ΔH° for its reaction.
Corrosion and its prevention
Rusting is a galvanic cell on the surface of the iron. At anodic regions iron is oxidized, Fe → Fe²⁺ + 2e⁻ (Fe²⁺/Fe −0.44 V); the electrons flow through the metal to cathodic regions, where oxygen is reduced: O₂ + 2H₂O + 4e⁻ → 4OH⁻ (+0.40 V), or O₂ + 4H⁺ + 4e⁻ → 2H₂O in acid. Oxygen then oxidizes the Fe²⁺ (drawn) to iron(III), which precipitates as hydrated iron(III) oxide, Fe₂O₃·xH₂O, the rust. Both water and oxygen are needed, and dissolved salt speeds corrosion by making the water a better electrolyte. Under a drop of water the oxygen-poor centre becomes the anode and the well-aerated edge becomes the cathode, so pits form where oxygen is scarce. Prevention works either by keeping water and oxygen away (paint, oil, a tin coating) or by making the iron a cathode. Galvanized steel is coated with zinc (Zn²⁺/Zn −0.76 V), which is oxidized in preference to iron even where the coating is scratched; a scratched tin coating (Sn²⁺/Sn −0.14 V) makes things worse, because the exposed iron becomes the anode. A sacrificial anode of magnesium or zinc wired to a ship's hull or a buried pipe corrodes in place of the steel, and an impressed current from a DC supply, with the steel joined to the negative terminal, does the same job without consuming an active metal. Aluminium (−1.66 V) is more easily oxidized than iron, yet it lasts in air because a thin, tightly bound layer of Al₂O₃ passivates the surface.
What to carry forward
Name the electrodes by the reaction (oxidation at the anode, reduction at the cathode) and decide what reacts by comparing reduction potentials with those of water, remembering overpotential and reactive-metal anodes. For every quantity, follow one chain: Q = It, moles of electrons = Q/F with F = 96 485 C/mol, then the half-reaction's electron count gives moles of product. Batteries and fuel cells are galvanic cells run for useful work; corrosion is an unwanted galvanic cell, stopped by barriers or by making the iron the cathode.
Problemas de prática (40)
Praticar é grátis com uma conta.
Preveja o produto 5
Electroplating and refining
- Where the impurities go in copper refiningPredict the fate of more and less active metal impurities during electrolytic refining from their reduction potentials.
Batteries and fuel cells
- Inside a discharging lead–acid batteryPredict how the electrodes and the electrolyte of a lead–acid battery change as it discharges.
- What moves when a Li-ion cell dischargesPredict the directions of ion and electron flow in a lithium-ion cell during discharge.
Electrolytic cells and their products
- Electrolysis of molten magnesium chloridePredict the cathode and anode products when a molten binary salt is electrolysed.
- Electrolysis of silver nitrate solutionPredict the electrode products for an aqueous salt whose cation is reduced but whose anion cannot be oxidized.
Classificar 3
Batteries and fuel cells
- Electrodes of an alkaline cellIdentify the substance oxidized in a battery and the electrode and terminal where that happens.
- Which cell is not rechargeable?Classify common batteries as primary (single-use) or secondary (rechargeable) cells.
Corrosion and protection
- The two halves of rustingAssign the anode and cathode half-reactions of iron corroding in moist air.
Comparar 4
Electrolytic cells and their products
- Which ion deposits first?Order competing cathode reductions, including the reduction of water, by their reduction potentials.
- Potassium iodide: melt versus solutionCompare the electrolysis products of a molten salt with those of its aqueous solution using reduction potentials.
Faraday's law
- Charge needed per gram of metalRank metals by the charge needed to deposit one gram, combining the electrons per ion with the molar mass.
Corrosion and protection
- Scratched galvanized steel versus scratched tin plateCompare how zinc and tin coatings behave once scratched, using reduction potentials to decide which metal becomes the anode.
Calcular 15
Faraday's law
- Charge passed by a steady currentCalculate the charge passed by a constant current from Q = It with the time in seconds.
- Sodium from a Downs cellCalculate the mass of an alkali metal produced by an industrial molten-salt electrolysis.
- Moles of electrons deliveredConvert a current and a time in hours into moles of electrons with the Faraday constant.
- Volume of oxygen from water electrolysisCombine Faraday's law with the ideal gas law to find the volume of oxygen evolved at the anode.
- Two cells in seriesUse equal moles of electrons in series cells to relate the masses of two metals deposited by the same charge.
Electroplating and refining
- Chromium from chromic acidUse the oxidation state of a metal in an oxoanion bath to find the electrons per atom and the mass plated.
- Gold plating from a cyanide bathFind the electrons per atom from the metal's oxidation state in a complex ion and use it to calculate the mass plated.
- Time to nickel-plate a partCalculate the time needed to plate a given mass of a divalent metal at a fixed current.
- Time to build a silver layerConvert a coating thickness and area into a mass of metal and then into the plating time needed.
Batteries and fuel cells
- Hydrogen used by a fuel cellUse Faraday's law to calculate the mass of fuel a fuel cell consumes to deliver a given current for a given time.
- Maximum efficiency of a hydrogen fuel cellCalculate the thermodynamic maximum efficiency of a fuel cell as the ratio of ΔG° to ΔH°.
- Minimum voltage to recharge a car batteryCalculate the minimum charging voltage of a multi-cell battery from the standard potentials of its half-reactions.
- Lithium behind a phone battery ratingConvert a battery capacity in milliampere-hours into a charge and then into the mass of lithium that must move between the electrodes.
Electrolytic cells and their products
- Minimum voltage to electrolyse HICalculate the minimum applied voltage for an electrolysis from the E°cell of the reaction actually driven.
Corrosion and protection
- Magnesium used by a sacrificial anodeCalculate how much of a sacrificial anode is consumed by a steady protective current over a long period.
Encontre o erro 1
- Electrolysis with copper electrodesFind the error in a product prediction that ignores oxidation of an active-metal anode.
Interprete as evidências 3
Electrolytic cells and their products
- Reading a copper sulfate electrolysisInterpret the observations of an aqueous electrolysis to choose the balanced overall reaction that explains all of them.
Corrosion and protection
- Four tubes of iron nailsInterpret a controlled experiment to identify which conditions are needed for iron to rust and which speed it up.
- A drop of salt water on ironInterpret indicator colours under a water drop to locate the anodic and cathodic regions of differential-aeration corrosion.
Trabalhe de trás para frente 1
- Identify a metal from its depositWork backwards from the mass deposited and the charge passed to the molar mass of an unknown metal.
Planeje uma síntese 3
Corrosion and protection
- Protecting a buried pipeline with a power supplyDesign an impressed-current system that makes a steel structure the cathode.
- Choosing a sacrificial anodeChoose a metal for cathodic protection of steel by comparing reduction potentials.
Electroplating and refining
- Setting up a silver-plating cellChoose the electrode connections and electrolyte needed to electroplate an object with a given metal.
Explique por quê 5
Corrosion and protection
- Why aluminium does not crumble awayExplain why aluminium resists corrosion although its reduction potential shows it is more easily oxidized than iron.
Faraday's law
- What the Faraday constant measuresExplain the Faraday constant as the charge of one mole of electrons and connect it to Avogadro's number and the electron charge.
Electrolytic cells and their products
- How charge crosses a molten saltExplain how current is carried through an electrolyte by the migration of cations and anions toward the electrodes of opposite sign.
- Why add sodium sulfate to water?Explain the role of an unreactive electrolyte in the electrolysis of water and name the gas formed at each electrode.
Batteries and fuel cells
- Why lithium-ion cells contain no waterExplain why a high-voltage lithium cell needs a nonaqueous electrolyte by comparing its electrode potentials with the decomposition of water.
Travou em eletrólise, baterias e corrosão? Escaneie o problema.
Tire uma foto da reação da sua lista ou do livro e desenhe o mecanismo com cada seta corrigida. 1 scan grátis agora, sem cadastro; mais 1 com uma conta gratuita.
Depois, Scan Standard: /mês