Electrolysis
At a Young Scientist exhibition, a student wants to silver-plate a teaspoon. Nothing happens if you simply dip the spoon into silver nitrate solution. So how can a battery make the silver ions stick to the spoon? Electrolysis is also how aluminium and magnesium are made and how water is split into hydrogen and oxygen.
An electrolytic cell: a forced reaction
In a galvanic cell, a spontaneous reaction produces an electric current. In electrolysis it is the other way round: a direct current from an external source forces a redox reaction that would not happen by itself. If you put two inert electrodes into acidified water and connect them to a battery, bubbles of hydrogen form at one electrode and bubbles of oxygen at the other.
The electrolysis of a salt is carried out in an electrolytic cell connected to an external direct current source. a) Justify the polarity (positive or negative sign) of…
The electrolysis of a salt is carried out in an electrolytic cell connected to an external direct current source. a) Justify the polarity (positive or negative sign) of the anode and the cathode in an electrolytic cell, comparing it with the sign convention of a galvanic cell. b) State at which of the two electrodes oxidation occurs and at which reduction occurs, explaining towards which electrode the cations and anions present in the electrolyte migrate.
Gas volumes and extension: Faraday's laws
In an electroplating process, a constant direct current of $3.50\text{ A}$ is passed through a cell containing an aqueous solution of silver nitrate, $\mathrm{AgNO_3}$,…
In an electroplating process, a constant direct current of $3.50\text{ A}$ is passed through a cell containing an aqueous solution of silver nitrate, $\mathrm{AgNO_3}$, for a time of $45.0\text{ min}$. At the cathode, the silver ions are reduced according to the half-equation: $\mathrm{Ag^+}(aq) + \mathrm{e^-} \rightarrow \mathrm{Ag}(s)$. Data: Faraday constant, $F = 96\,485\text{ C}\times\text{mol}^{-1}$; molar mass of silver, $M(\mathrm{Ag}) = 107.9\text{ g}\times\text{mol}^{-1}$. Calculate: a) The total electric charge that has passed through the electrolytic cell. b) The mass of pure silver metal deposited at the cathode, assuming a current efficiency of $100\,\%$.
Molten salts
In a molten ionic compound (no water) only the compound's ions are present, so the cation is reduced at the cathode and the anion oxidised at the anode. This is how reactive metals such as sodium, magnesium and aluminium are made. For molten $\mathrm{MgCl_2}$: cathode $\mathrm{Mg^{2+} + 2e^- \rightarrow Mg}$; anode $\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$. In brine, by contrast, water is discharged instead of $\mathrm{Na^+}$ (see below).
In an industrial plant, magnesium metal is obtained by the electrolysis of molten magnesium chloride, $\mathrm{MgCl_2}(l)$, at $750\,^{\circ}\text{C}$ using inert…
In an industrial plant, magnesium metal is obtained by the electrolysis of molten magnesium chloride, $\mathrm{MgCl_2}(l)$, at $750\,^{\circ}\text{C}$ using inert graphite electrodes. The half-equations of the process are: at the cathode, $\mathrm{Mg^{2+}} + 2\,\mathrm{e^-} \rightarrow \mathrm{Mg}(l)$; at the anode, $2\,\mathrm{Cl^-} \rightarrow \mathrm{Cl_2}(g) + 2\,\mathrm{e^-}$. If the cell operates with a constant current of $5000\text{ A}$ for $4.00\text{ h}$, determine: a) The mass of magnesium metal obtained at the cathode. b) The moles of chlorine gas, $\mathrm{Cl_2}$, formed simultaneously at the anode. Data: $F = 96\,485\text{ C}\times\text{mol}^{-1}$; $M(\mathrm{Mg}) = 24.31\text{ g}\times\text{mol}^{-1}$. Assume a current efficiency of $100\,\%$.
Electrolysis of aqueous solutions
Molten sodium chloride gives shiny sodium metal. Dissolve the same salt in water and pass a current, and you get hydrogen gas and never any sodium. Why does the water get in the way?
In solution, water can also be oxidised or reduced, so there are two candidates at each electrode. The one that is easier to discharge wins.
To coat a metal object by electrodeposition, the experimental set-up shown in the following diagram is used: . In the cell, a copper(II) sulfate solution,…
To coat a metal object by electrodeposition, the experimental set-up shown in the following diagram is used:
. In the cell, a copper(II) sulfate solution, $\mathrm{CuSO_4}$, acts as the electrolyte; the anode consists of a pure copper plate connected to the positive terminal of the source, and the cathode is the object to be copper-plated, connected to the negative terminal. A direct current of $I = 2.00\text{ A}$ is passed for a time of $t = 30.0\text{ min}$. Given that the efficiency of the electric current in the cathodic process is $90.0\,\%$, calculate: a) The amount of electric charge actually used in the electrodeposition. b) The mass of copper deposited on the surface of the object. Data: $M(\mathrm{Cu}) = 63.5\text{ g}\times\text{mol}^{-1}$; $F = 96\,485\text{ C}\times\text{mol}^{-1}$; half-equation: $\mathrm{Cu^{2+}}(aq) + 2\,\mathrm{e^-} \rightarrow \mathrm{Cu}(s)$.
An aqueous $1.0\text{ M}$ solution of sodium sulfate, $\mathrm{Na_2SO_4}$, is electrolysed using inert platinum electrodes at $25\,^{\circ}\text{C}$. The following…
An aqueous $1.0\text{ M}$ solution of sodium sulfate, $\mathrm{Na_2SO_4}$, is electrolysed using inert platinum electrodes at $25\,^{\circ}\text{C}$. The following standard reduction potentials are known: $E^{\circ}(\mathrm{Na^+}/\mathrm{Na}) = -2.71\text{ V}$; $E^{\circ}(\mathrm{S_2O_8^{2-}}/\mathrm{SO_4^{2-}}) = +2.01\text{ V}$; for water: $2\,\mathrm{H_2O}(l) + 2\,\mathrm{e^-} \rightarrow \mathrm{H_2}(g) + 2\,\mathrm{OH^-}(aq)$, $E = -0.41\text{ V}$ (at $\mathrm{pH} = 7$); $\mathrm{O_2}(g) + 4\,\mathrm{H^+}(aq) + 4\,\mathrm{e^-} \rightarrow 2\,\mathrm{H_2O}(l)$, $E = +0.82\text{ V}$ (at $\mathrm{pH} = 7$). a) Justify thermodynamically which substance is reduced at the cathode and which is oxidised at the anode, writing the corresponding half-equations and the overall reaction. b) Calculate the total combined volume of gases released, measured at $1.00\text{ atm}$ and $298\text{ K}$, after passing a current of $1.50\text{ A}$ for $2.00\text{ h}$. Data: $R = 0.082\text{ atm}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$; $F = 96\,485\text{ C}\times\text{mol}^{-1}$. Use the ideal model without overpotentials and assume 100 % faradaic efficiency for the formation of dry gases; the pH given corresponds to the initial medium.
Cells in series and current efficiency
Two electrolytic cells fitted with inert electrodes are connected in series. Cell 1 contains an aqueous solution of zinc sulfate, $\mathrm{ZnSO_4}$, and cell 2 contains…
Two electrolytic cells fitted with inert electrodes are connected in series. Cell 1 contains an aqueous solution of zinc sulfate, $\mathrm{ZnSO_4}$, and cell 2 contains an aqueous solution of chromium(III) nitrate, $\mathrm{Cr(NO_3)_3}$. A constant direct current is passed through the circuit for a certain time. a) Given that exactly $4.50\text{ g}$ of zinc metal has been deposited at the cathode of the first cell, calculate the total electric charge that has passed through the series of cells. b) Determine the mass of chromium metal deposited simultaneously at the cathode of the second electrolytic cell. Data: atomic molar masses: $M(\mathrm{Zn}) = 65.38\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{Cr}) = 52.00\text{ g}\times\text{mol}^{-1}$; $F = 96\,485\text{ C}\times\text{mol}^{-1}$. For this calculation model, assume that both depositions have a faradaic efficiency of 100 % and that no water is discharged; in a real aqueous electrodeposition, competing reactions would have to be considered and the bath conditions controlled.
In the industrial production of aluminium by the Hall-Héroult process, alumina, $\mathrm{Al_2O_3}$, dissolved in molten cryolite at $960\,^{\circ}\text{C}$ is…
In the industrial production of aluminium by the Hall-Héroult process, alumina, $\mathrm{Al_2O_3}$, dissolved in molten cryolite at $960\,^{\circ}\text{C}$ is electrolysed. At the graphite cathode, aluminium ions are reduced to liquid aluminium: $\mathrm{Al^{3+}} + 3\,\mathrm{e^-} \rightarrow \mathrm{Al}(l)$. At the graphite (carbon) anodes, oxide ions react, consuming the electrode according to the half-equation: $\mathrm{C}(s) + 2\,\mathrm{O^{2-}} \rightarrow \mathrm{CO_2}(g) + 4\,\mathrm{e^-}$. An industrial cell operates with a current of $120\,000\text{ A}$ and a current efficiency of $92.0\,\%$. a) Calculate the mass of aluminium produced by the cell in a continuous period of $24.0\text{ h}$. b) Determine the mass of carbon consumed at the anodes during the same period and the volume of $\mathrm{CO_2}$ emitted at $960\,^{\circ}\text{C}$ and $1.05\text{ atm}$. Data: $M(\mathrm{Al}) = 26.98\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $F = 96\,485\text{ C}\times\text{mol}^{-1}$; $R = 0.082\text{ atm}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$. In b), calculate the stoichiometric consumption of carbon and the production of carbon dioxide associated only with the net aluminium obtained; do not include parasitic oxidation of the anodes.
$500\text{ mL}$ of an aqueous $0.200\text{ M}$ solution of copper(II) nitrate, $\mathrm{Cu(NO_3)2}$, is electrolysed using inert platinum electrodes with a constant…
$500\text{ mL}$ of an aqueous $0.200\text{ M}$ solution of copper(II) nitrate, $\mathrm{Cu(NO_3)_2}$, is electrolysed using inert platinum electrodes with a constant current of $2.50\text{ A}$. Under these potential conditions, only $\mathrm{Cu^{2+}}$ is reduced at the cathode and water is oxidised at the anode according to: $2\,\mathrm{H_2O}(l) \rightarrow \mathrm{O_2}(g) + 4\,\mathrm{H^+}(aq) + 4\,\mathrm{e^-}$. Assuming that the faradaic efficiency is $100\,\%$ and that the volume of the solution remains constant at $500\text{ mL}$: a) Calculate the time required, in minutes, to completely reduce all the copper present in the solution. b) Determine the resulting $\mathrm{pH}$ of the solution after exactly $15.0\text{ min}$ of electrolysis from the start. Data: $F = 96\,485\text{ C}\times\text{mol}^{-1}$; $M(\mathrm{Cu}) = 63.55\text{ g}\times\text{mol}^{-1}$. Assume a homogeneous mixture, nitrate as a spectator ion, activities equal to concentrations and initial acidity negligible compared with the protons generated. Ignore the hydrolysis of copper.
In the electrolytic refining of impure copper (blister copper), an impure copper anode containing zinc, silver and gold impurities is used, immersed in a solution of…
In the electrolytic refining of impure copper (blister copper), an impure copper anode containing zinc, silver and gold impurities is used, immersed in a solution of $\mathrm{CuSO_4}$ and $\mathrm{H_2SO_4}$. A thin sheet of pure copper is placed at the cathode. The standard reduction potentials are: $E^{\circ}(\mathrm{Zn^{2+}}/\mathrm{Zn}) = -0.76\text{ V}$; $E^{\circ}(\mathrm{Cu^{2+}}/\mathrm{Cu}) = +0.34\text{ V}$; $E^{\circ}(\mathrm{Ag^+}/\mathrm{Ag}) = +0.80\text{ V}$; $E^{\circ}(\mathrm{Au^{3+}}/\mathrm{Au}) = +1.50\text{ V}$. a) Explain, with reasoning, the behaviour and final fate of each metal (Zn, Cu, Ag and Au) during the anodic and cathodic processes when the appropriate potential is applied. b) An anode sample of $500\text{ g}$ contains $96.0\,\%$ copper by mass and $4.0\,\%$ zinc (traces of noble metals are neglected in this calculation). If it is electrolysed with a constant current of $8.00\text{ A}$ until all the copper and zinc in the sample have completely dissolved, determine the time required in hours. Data: $M(\mathrm{Cu}) = 63.55\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{Zn}) = 65.38\text{ g}\times\text{mol}^{-1}$; $F = 96\,485\text{ C}\times\text{mol}^{-1}$. Assume that all the anodic current is used to dissolve copper and zinc as divalent ions.
Back to the start: to silver-plate the teaspoon, connect it as the cathode (negative terminal). The source supplies electrons and the $\mathrm{Ag^+}$ ions in the solution are reduced on its surface, forming a bright metallic layer.