Electrochemistry: Electrolysis, Electroplating and Batteries

The correct answer is D, 20 A. The working rule for charging a storage battery is that the charging current should be about one-tenth of the ampere-hour capacity of the battery, which is called the C by 10 rate. For a battery rated 200 ampere-hours, one-tenth of 200 gives 20 amperes, and at that rate a fully discharged battery takes roughly ten to twelve hours to charge. A rate much higher than this heats the plates, boils the electrolyte and shortens the life of the cell, while a much lower rate leaves the battery undercharged and allows sulphation. A, 12 A, and B, 15 A, are wrong because they are below the one-tenth value and are simply arbitrary figures for a 200 ampere-hour cell. C, 10 A, is the trap: it is one-twentieth of the rating, and candidates who remember the words ten per cent but apply them to the wrong number mark it.

The correct answer is C, 12 V. When cells are joined in parallel all their positive terminals are connected together and all their negative terminals together, so each cell is across the same two points and the voltage of the combination is the voltage of one cell, that is 12 volts. What the parallel arrangement does add is the current capacity: six identical cells in parallel can supply six times the current, or the same current for six times as long, and the internal resistance of the group falls to one-sixth of that of a single cell. B, 72 V, is wrong because adding the voltages is what happens in a series connection, where the positive of one cell is joined to the negative of the next. A, 6 V, and D, 2 V, are wrong because no method of grouping identical cells reduces the terminal voltage below that of one cell; 2 V happens to be the voltage of a single lead-acid cell, of which six in series make such a 12 volt battery.

The correct answer is B, the anode. The anode is the electrode joined to the positive terminal of the supply, so negatively charged ions, the anions, travel to it and give up electrons there; loss of electrons is oxidation. A is wrong because the cathode is joined to the negative terminal, attracts the positively charged cations and gives them electrons, so reduction, not oxidation, occurs there. The rule holds in a galvanic cell as well, where the anode is the negative terminal but is still the seat of oxidation, which is the point examiners test. C is wrong because the electrolyte is only the medium in which the ions move; the reactions happen where the ions meet the electrodes. D is wrong because a salt bridge belongs to a galvanic cell with two half-cells and merely completes the circuit while keeping the solutions electrically neutral; it is not a site of oxidation.

The correct answer is B, 96,500 coulombs. One faraday is the charge carried by one mole of electrons, obtained by multiplying the charge on a single electron by the Avogadro number, and it comes to about 96,500 coulombs; passing one faraday through an electrolyte liberates one gram equivalent of the substance at an electrode. A is wrong because 9,650 coulombs is one-tenth of the correct value and is a careless-reading trap. C is wrong because the figure 1.6 belongs to the charge on one electron, which is 1.6 into ten raised to the power minus nineteen coulombs, not to the faraday. D is wrong because 6.023 into ten raised to the power twenty-three is the Avogadro number, a count of particles and not a quantity of charge, though the two are linked in the very calculation that gives the faraday.

The correct answer is B, the cathode. The metal ions in the plating bath carry a positive charge, so they move towards the negative electrode and are reduced to metal there; the article must therefore be the cathode for the coating to build up on it. A is wrong because the anode is a plate of the pure plating metal, which dissolves into the solution as the plating proceeds and so keeps the concentration of the bath steady, which is why the bath does not have to be renewed often. C is wrong because the electrolyte is a solution of a salt of the plating metal, such as a copper salt for copper plating, and is not an electrode at all. D is wrong because the polarity cannot be chosen freely: reversing it would strip metal from the article instead of depositing it, which is in fact how electro-polishing and anodising work.

The correct answer is C, chlorine. In the chlor-alkali process a concentrated solution of sodium chloride is electrolysed; chloride ions are oxidised at the anode and come off as chlorine gas, hydrogen is set free at the cathode, and sodium hydroxide remains in the solution, which is why the industry yields three useful products at once. A is wrong because hydrogen appears at the cathode, not the anode. B is wrong because oxygen is the anode product when water or a dilute solution of an acid or of a salt of a very reactive metal is electrolysed; in concentrated brine the chloride ion is discharged in preference to water. D is wrong because the solution contains no carbonate or carbon source at all, so carbon dioxide cannot be formed. Remember the trio: chlorine, hydrogen and caustic soda from brine.

The correct answer is B, the lead-acid cell. Its reaction is reversible, so passing a current through it in the opposite direction converts the products back into lead and lead dioxide and the cell can be used again; such a cell is called a secondary or storage cell, and the car battery is the common example. A is wrong because the dry cell of the Leclanche type is a primary cell: once the zinc container and the manganese dioxide are consumed the cell is thrown away, and attempting to recharge it can make it leak. C is wrong because the Daniell cell, with its zinc and copper electrodes in their own salt solutions, is a classic primary cell of the laboratory. D is wrong because a mercury button cell of the kind used in watches is also a primary cell. Nickel-cadmium and lithium-ion cells are the other important secondary cells.

The correct answer is C, zinc. In the Leclanche dry cell the outer container itself is made of zinc and acts as the anode, being oxidised to zinc ions as the cell works, which is why an old cell often leaks when the container is eaten through. A is wrong because copper has no part in this cell; it is the cathode metal of the Daniell cell, a different arrangement. B is wrong because the graphite rod in the centre of the dry cell is the cathode, an inert conductor that carries the current away from the reduction reaction rather than taking part in it. D is wrong because manganese dioxide is the depolariser packed around the graphite rod: it is reduced and absorbs the hydrogen that would otherwise collect and stop the cell, but it is not the anode. The electrolyte is a moist paste of ammonium chloride and zinc chloride, and the cell gives about 1.5 volts.

The correct answer is A, dilute sulphuric acid. The lead-acid cell has a spongy lead negative plate and a lead dioxide positive plate dipped in dilute sulphuric acid, and as the cell discharges both plates turn to lead sulphate while the acid becomes weaker, which is why its state of charge can be judged from the specific gravity of the acid. B is wrong because concentrated nitric acid is a strong oxidising agent that would attack the lead plates and the container instead of working as an electrolyte. C is wrong because the moist paste of ammonium chloride belongs to the Leclanche dry cell, a primary cell that cannot be recharged. D is wrong because potassium hydroxide is the alkaline electrolyte of the nickel-cadmium and nickel-iron cells, which is a different family of secondary cells. Each lead-acid cell gives about two volts, so six in series make a twelve volt battery.

The correct answer is B, cryolite. Pure alumina melts at a very high temperature, so it is dissolved in molten cryolite, which is sodium aluminium fluoride, usually with a little fluorspar added; this brings the working temperature down sharply and makes the melt a good conductor. Aluminium collects at the carbon-lined cathode at the bottom of the cell and oxygen is set free at the carbon anodes, which slowly burn away and must be replaced. A is wrong because bauxite is the ore from which alumina is first purified by the Bayer process; the ore itself is not electrolysed. C is wrong because molten sodium chloride is electrolysed in Down's process to obtain sodium metal and chlorine. D is wrong because limestone is a flux of the blast furnace in iron making and has no role in this cell.

The correct answer is C, zinc. Galvanisation is the coating of iron or steel with zinc, usually by dipping the article in molten zinc, and it protects the iron twice over: the layer keeps out air and moisture, and because zinc is the more reactive metal it corrodes in place of the iron even where the coating is scratched, which is sacrificial protection. A is wrong because the coating of tin on iron, as on the sheet of food cans, is called tinning; it protects only so long as it is unbroken, since tin is less reactive than iron. B is wrong because chromium plating, done by electrolysis, is used on taps and car fittings mainly for a bright and hard surface. D is wrong because nickel plating is likewise an electroplating process used on machine parts and cheap jewellery, often as an undercoat below chromium.

The correct answer is B, water. In this cell hydrogen is fed to the anode and oxidised, oxygen is fed to the cathode and reduced, and the two combine to give water, so the cell produces electricity for as long as the gases are supplied and leaves no polluting exhaust. Cells of this kind gave both power and drinking water on the Apollo spacecraft and are now used in buses, trains and backup power units. A is wrong because no carbon enters the reaction; carbon dioxide would appear only if a hydrocarbon or an alcohol were the fuel. C is wrong because hydrogen peroxide is not the product of the overall cell reaction, although peroxide species can form as unwanted intermediates at a poor catalyst. D is wrong because ammonia contains nitrogen, which is not a reactant here, though ammonia is studied separately as a possible carrier of hydrogen fuel.