Back to homeAP® Chemistry · AP Chemistry · Change subject
Start for free
AP Chem · AP® Chemistry
Chemical transformations · Oxidation and reduction

Electrolysis and Faraday's Law

Look at the shiny chrome trim on a classic car, the aluminum in a soda can, or the silver plating on a fork passed down at Thanksgiving. None of these metals formed by themselves. A battery-like power source pushed electrons into a solution and forced metal ions to become solid metal. How do we control how much metal we get, and which electrode does what?

Electrolysis: forcing a reaction to run

In a galvanic cell, a thermodynamically favorable redox reaction produces current. In an electrolytic cell, it is the other way around: an external direct-current source drives a reaction that would not happen on its own.

Try this: put two platinum wires in water with a little acid. Nothing happens. Connect them to a battery and bubbles of $\text{H}_2$ form at one wire and $\text{O}_2$ at the other. The battery supplies the energy that the reaction lacks.

Comparison of electrode polarity and reactions in a galvanic cell and an electrolytic cell. galvanic cell anode (−) oxidation cathode (+) reduction spontaneous: makes current electrolytic cell anode (+) oxidation cathode (−) reduction source forced by external source

Inside the solution, cations migrate toward the cathode and anions toward the anode, so the circuit is complete: electrons travel through the wire and ions carry charge through the electrolyte.

Practice★☆☆Not started
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 toward which electrode the cations and anions present in the electrolyte migrate.

Solve

Faraday's law

An electroplating bath deposits 0.67 g of silver with 1 A for 10 min. At 20 min it deposits 1.34 g. But with copper under the same conditions, only 0.20 g forms. What links the charge that flows to the mass deposited, and why does the element matter?

Flow chart for electrolysis calculations: current and time give charge, charge gives moles of electrons, the half-reaction gives moles of substance. q = I × t (charge, C) mol e⁻ (q ÷ F) mol substance then × M → g ÷ F half-reaction
Practice★☆☆Not started
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 $100\,\%$.

Solve

Electrolysis of molten salts

In a molten salt there is no water, only the ions of the salt. The cation is reduced at the cathode and the anion is oxidized at the anode. Nothing else competes. This is how very reactive metals such as sodium, magnesium, and aluminum are produced, because water would react before they could deposit.

Electrolytic cell with molten magnesium chloride: magnesium forms at the cathode and chlorine gas at the anode. molten MgCl₂ (750 °C) cathode (−) Mg²⁺ + 2e⁻ → Mg anode (+) 2Cl⁻ → Cl₂ + 2e⁻
Practice★☆☆Not started
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 $100\,\%$.

Solve

Electrolysis of aqueous solutions

Molten NaCl gives shiny sodium. Dissolve the same salt in water and run a current, and you get hydrogen gas and never any sodium. Why does the solvent get involved?

Comparison of reduction potentials at the cathode and anode to decide which species reacts instead of water. cathode: what is reduced? Na⁺/Na: −2.71 V H₂O/H₂: −0.41 V (wins) anode: what is oxidized? S₂O₈²⁻/SO₄²⁻: +2.01 V O₂/H₂O: +0.82 V (wins)
Practice★★☆Not started
To coat a metal object by electrodeposition, the experimental setup shown in the following diagram is used: . In the cell, a copper(II) sulfate solution,…

To coat a metal object by electrodeposition, the experimental setup shown in the following diagram is used:

I = 2.00 A t = 30.0 min + − e⁻ e⁻ Cu²⁺ Cu²⁺ SO₄²⁻ Aqueous CuSO₄ solution Cu anode (+) Oxidation: Cu → Cu²⁺ + 2e⁻ Cathode: object (−) Cu electrodeposition

. 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: $\mathrm{Cu^{2+}}(aq) + 2\,\mathrm{e^-} \rightarrow \mathrm{Cu}(s)$.

Solve
Practice★★☆Not started
An aqueous $1.0\text{ M}$ solution of sodium sulfate, $\mathrm{Na_2SO_4}$, is electrolyzed 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 electrolyzed 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 oxidized 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.

Solve

Cells in series and current efficiency

Two electrolytic cells connected in series carrying the same current and charge from one source. cell 1: ZnSO₄(aq) Zn²⁺ + 2e⁻ → Zn cell 2: Cr(NO₃)₃(aq) Cr³⁺ + 3e⁻ → Cr same I
Practice★★☆Not started
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.

Solve
Practice★★★Not started
$500\text{ mL}$ of an aqueous $0.200\text{ M}$ solution of copper(II) nitrate, $\mathrm{Cu(NO_3)2}$, is electrolyzed 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 electrolyzed 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 oxidized 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.

Solve

Purification and industrial uses

In electrolytic refining, impure copper is the anode and a thin sheet of pure copper is the cathode in a $\text{CuSO}_4$ solution. Each metal in the anode is judged by its reduction potential compared with $\text{Cu}^{2+}/\text{Cu}$ ($+0.34\ \text{V}$).

  • A metal with a lower potential than copper (zinc, $-0.76\ \text{V}$) is oxidized more easily, so it dissolves as $\text{Zn}^{2+}$. It is harder to reduce than $\text{Cu}^{2+}$, so it stays in solution.

  • A metal with a higher potential (silver, $+0.80\ \text{V}$; gold, $+1.50\ \text{V}$) is harder to oxidize at the voltage used. It does not dissolve and falls off as anode sludge, which is collected and sold.

  • Copper dissolves at the anode and deposits as pure copper at the cathode.

Practice★★★Not started
In the industrial production of aluminum 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 aluminum by the Hall-Héroult process, alumina, $\mathrm{Al_2O_3}$, dissolved in molten cryolite at $960\,^{\circ}\text{C}$ is electrolyzed. At the graphite cathode, aluminum ions are reduced to liquid aluminum: $\mathrm{Al^{3+}} + 3\,\mathrm{e^-} \rightarrow \mathrm{Al}(l)$. 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 aluminum 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 aluminum obtained; do not include parasitic oxidation of the anodes.

Solve
Practice★★★Not started
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 behavior 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 electrolyzed 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.

Solve

Back to the start: To silver-plate the fork, connect it as the cathode (negative terminal). The source feeds it electrons, and $\text{Ag}^+$ ions from the solution are reduced directly onto its surface as a shiny metal layer.

Summary

See all the exercises in this section