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Concept 05 / 07Electrochemistry

Electrolysis: Faraday's Laws and Products of Electrolysis

Electrolytic cells, Faraday's first and second laws of electrolysis with the Faraday constant, quantitative calculations, and prediction of products of electrolysis of molten and aqueous electrolytes (NaCl, CuSO4, H2SO4, AgNO3) considering standard potentials, overpotential and the nature of electrodes.

Figure 1
In an electrolytic cell, an external source of electricity drives a non-spontaneous redox reaction. The cathode (connected to the negative terminal) is where reduction occurs; the anode (positive terminal) is where oxidation occurs. (Electrode signs are reversed compared with a galvanic cell, but "cathode = reduction" always holds.) Faraday's first law. The mass of substance deposited or liberated at an electrode is proportional to the quantity of electricity passed: m=ZQ=ZItm = ZQ = ZIt, where ZZ is the electrochemical equivalent (mass deposited by 1 C1\,\mathrm{C}). Faraday's second law. When the same quantity of electricity is passed through different electrolytes in series, the masses liberated are proportional to their equivalent weights: m1m2=E1E2\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}. Faraday constant: one mole of electrons carries F=NAe=96487 CF = N_{A}e = 96487\,\mathrm{C} (about 96500 C96500\,\mathrm{C}). For Mn++ne−→M\mathrm{M^{n+} + ne^{-} \rightarrow M}, nn faradays deposit one mole of M, so m=M I tnFm = \frac{M\,I\,t}{nF}. Products of electrolysis. At the cathode, the species with the most positive reduction potential is reduced; at the anode, the species with the most negative reduction potential (most easily oxidised) is oxidised, taking overpotential into account. Molten NaCl: cathode Na, anode Cl2\mathrm{Cl_{2}}. Aqueous NaCl (see figure): cathode 2H2O+2e−→H2+2OH−\mathrm{2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}} (E∘=−0.83 VE^{\circ} = -0.83\,\mathrm{V}, far easier than Na+\mathrm{Na^{+}} at −2.71 V-2.71\,\mathrm{V}); anode Cl2\mathrm{Cl_{2}}, although oxidation of water (E∘=1.23 VE^{\circ} = 1.23\,\mathrm{V}) is thermodynamically easier than of Cl−\mathrm{Cl^{-}} (1.36 V1.36\,\mathrm{V}), because oxygen evolution has a large overpotential. NaOH is left in solution (chlor-alkali industry). Very dilute NaCl gives O2\mathrm{O_{2}} at the anode. Aqueous CuSO4\mathrm{CuSO_{4}} with inert Pt electrodes: cathode Cu; anode O2\mathrm{O_{2}} (water is oxidised; sulfate is not); the solution becomes acidic. Aqueous CuSO4\mathrm{CuSO_{4}} with copper electrodes: cathode Cu deposits; anode Cu dissolves (Cu→Cu2++2e−\mathrm{Cu \rightarrow Cu^{2+} + 2e^{-}}); concentration is unchanged. This is the principle of electrorefining and electroplating. Dilute H2SO4\mathrm{H_{2}SO_{4}}: H2\mathrm{H_{2}} at the cathode, O2\mathrm{O_{2}} at the anode (electrolysis of water). Concentrated H2SO4\mathrm{H_{2}SO_{4}}: peroxodisulfate forms at the anode (2HSO4−→S2O82−+2H++2e−\mathrm{2HSO_{4}^{-} \rightarrow S_{2}O_{8}^{2-} + 2H^{+} + 2e^{-}}). Aqueous AgNO3\mathrm{AgNO_{3}} with silver electrodes: Ag deposits at the cathode and dissolves at the anode. Key Takeaways: - m=MItnFm = \frac{MIt}{nF}; F=96487 C mol−1F = 96487\,\mathrm{C\,mol^{-1}} of electrons. - Second law: masses proportional to equivalent weights for the same charge. - Aqueous NaCl: H2\mathrm{H_{2}} at the cathode, Cl2\mathrm{Cl_{2}} at the anode (overpotential), NaOH in solution. - Active electrodes (Cu, Ag) dissolve at the anode instead of liberating O2\mathrm{O_{2}}. - Pitfall: in electrolysis the cathode is negative; in a galvanic cell it is positive; reduction always occurs at the cathode. Brain-Teaser 1: A current of 1.5 A1.5\,\mathrm{A} is passed through CuSO4\mathrm{CuSO_{4}} solution for 10 min10\,\mathrm{min}. What mass of copper is deposited? Solution: Q=1.5×600=900 CQ = 1.5 \times 600 = 900\,\mathrm{C}. m=63.5×9002×96487=0.296 gm = \frac{63.5 \times 900}{2 \times 96487} = 0.296\,\mathrm{g}. Brain-Teaser 2: Three cells containing AgNO3\mathrm{AgNO_{3}}, CuSO4\mathrm{CuSO_{4}} and ZnSO4\mathrm{ZnSO_{4}} are connected in series. If 1.45 g1.45\,\mathrm{g} of silver is deposited, find the masses of Cu and Zn deposited. Solution: Moles of electrons =1.45108=0.01343= \frac{1.45}{108} = 0.01343. Cu: 0.013432×63.5=0.426 g\frac{0.01343}{2} \times 63.5 = 0.426\,\mathrm{g}. Zn: 0.013432×65.4=0.439 g\frac{0.01343}{2} \times 65.4 = 0.439\,\mathrm{g}. Brain-Teaser 3: How many coulombs are needed to reduce 1 mol1\,\mathrm{mol} of MnO4−\mathrm{MnO_{4}^{-}} to Mn2+\mathrm{Mn^{2+}}, and to oxidise 1 mol1\,\mathrm{mol} of H2O\mathrm{H_{2}O} to O2\mathrm{O_{2}}? Solution: MnO4−→Mn2+\mathrm{MnO_{4}^{-} \rightarrow Mn^{2+}} needs 55 electrons: 5F=4.82×105 C5F = 4.82 \times 10^{5}\,\mathrm{C}. H2O→12O2+2H++2e−\mathrm{H_{2}O \rightarrow \frac{1}{2}O_{2} + 2H^{+} + 2e^{-}} releases 22 electrons: 2F=1.93×105 C2F = 1.93 \times 10^{5}\,\mathrm{C}.

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