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Electrolysis

  • Electrolysis is the process of using a direct current to cause the chemical decomposition of an electrolyte by means of two half-reactions which occur at the anode and cathode of an electrolytic cell

An Electrolytic Cell (Photo Credit: RSC)

  • An electrolyte is a compound which can conduct electricity in the molten form or in an aqueous solution by means of its mobile ions
  • Examples of electrolytes include acids, alkalis, and salts
  • Conductors are metals or poles of carbon graphite which conduct electricity by means of the mobile electrons in their solid or liquid states. For example, Na, Hg, etc.
  • Electrodes refer to anodes and cathodes of an electrolytic cell by which an electric current enters and leaves the electrolyte
  • Anode is the positive electrode in an electrolytic cell by which the conventional current enters the electrolyte or the pole by which the electrons leave an electrolyte
  • Cathode is the negative electrode in an electrolytic cell by which the conventional current leaves the electrolyte or the pole by which electrons enter the electrolyte
  • Inert electrodes are ones that do not change or get involved during the electrolytic processes. For example, Pt, and C
  • An active electrode is one which undergoes a chemical change during electrolysis. For example, Cu anode during the electrolysis of CuSO4 solution
  • A strong electrolyte is one which is completely ionized in solution. For example, solutions of mineral acids. They conduct a large amount of current because they are composed entirely of free moving ions
  • A weak electrolyte is one which is incompletely ionized in an aqueous solution. The few number of mobile ions in weak electrolytes accounts for their lower conductance or their relatively lower ability to conduct electric current
  • Before the electrodes are connected to a battery, the ions of the electrolyte move randomly about
  • The moment they are connected to a battery or a source of direct current, the cathode connected to the negative pole of the battery becomes negatively charged while the anode becomes positively charged as the positive terminal of the battery draws electrons from it
  • The ions of the electrolyte immediately begin directional movement. The cations (positive ions) move to the cathode while the anions (negative ions) move to the anode
  • At the cathode, the cations undergo reduction by gaining a number of electrons equal to the charge on the cation
  • At the anode, the anions undergo oxidation by losing a number of electrons equal to the charge on the anion
Factors Affecting the Discharge of Ions during Electrolysis
  • Position of the Ion in the Electrochemical Series
    • The electrochemical series (E.C.S.) gives an arrangement of the common elements in order of increasing ease of discharge of their ions

      Photo Credit: Science Zoneja

    • NOTE: The most reactive metals form the most stable ions which are therefore difficult to convert back to metals and the most reactive non-metals form the most stable anions which are therefor difficult  to convert to neutral atoms
    • When two cations get to the cathode, the cation which is lower in the E.C.S. is selectively discharged
    • When two anions get to the anode, the anion which is lower in the E.C.S. is preferably discharged
  • Effect of Concentration
    • When two ions at an electrode are two or three ions apart in the E.C.S., the ion higher up in the E.C.S. may be discharged preferentially if its concentration is much higher than that of the ion lower in the E.C.S.
    • In the electrolysis of brine (conc. NaCl), Cl ions are preferentially discharged and OH ions remain in the solution because the higher concentration of Cl causes its discharge potential to be lower at the surface of the electrode
    • In spit of the higher concentration of Na+, the large difference between the discharge potentials of Na+ and H+ causes hydrogen to be discharged
    • At the end of the electrolysis, hydrogen gas is liberated at the cathode, chlorine gas is liberated at the anode, and sodium hydroxide is left in the electrolytic cell
  • Nature of Electrode
    • An active electrode is one which takes part in the electrolytic process
    • In the electrolysis of CuSO4 solution with copper electrodes, the copper anode functions as an active electrode
    • Cu2+ and H+ ions in the electrolyte move to the copper cathode and Cu2+ ions are preferentially discharged. Cu2+ + 2e → Cu
    • The OH and SO42- move to the anode but neither OH nor SO42- ions are discharged at the anode. This is because the discharge potential of OH, which is lower in the E.C.S., is rather too high on the surface of the copper anode
    • The anodic reaction for this electrolysis is the dissolution of the active copper anode and this results in the introduction of Cu2+ ions back into the electrolyte
    • Anodic reaction: Cu – 2e → Cu2+
Faraday’s Laws of Electrolysis
  • The first law states that the mass of a substance produced during electrolysis is directly proportional to the quantity of electricity used up
  • Mathematically, M ∝ Q. Therefore, M1/Q1 = M2/Q2
  • Q is the quantity of electricity, it is expressed in Coulombs (C), and it is equal to the product of the current, I (in Amperes), and time, t (in seconds).
  • Q = It
  • Faraday, F, is defined as the quantity of charge on 1 mole of electrons or 6.02 × 10²³ electrons
  • Thus F = NAe
    • where NA is Avogadro number, and
    • e is electronic charge
  • Every mole of electrons is equivalent to the Faraday (F), which is equal to 96500 C
  • Faraday is linked to reactions occurring at the electrodes during electrolysis

Examples

  1. Calculate the mass of copper deposited by passing 100 A current for 40 minutes through an electrolytic cell of copper (ii) tetraoxosulfate (vi) solution. (F = 96500 C, Cu = 64 g)
    • Cu2+ + 2e → Cu
    • Q1 = 2 × 96500 = 193000 C (because there are 2 electrons)
    • M1 = 64 g (because only 1 mole of Cu in the product)
    • Q2 = It = 100 × (40 × 60) = 240000 C
    • From M1/Q1 = M2/Q2
    • Therefore, M2 = (64 × 240000) ÷ 193000 = 79.59 g
  2. 0.325 g of a metal X was deposited when 0.5 A current flowed through a tetraoxosulfate (vi) solution of X for 32 minutes 10 seconds. Calculate the charge on the ion of metal X. (F = 96500 C, X = 65 g)
    • Let the ion X be represented Xn+
    • X is then deposited at the cathode according to the equation Xn+ + ne → X
    • Q1 = nF = n × 96500 = 96500n
    • M1 = 65 g
    • Q2 = It = 0.5 × ((32 × 60) + 10) = 965 C
    • M2 = 0.325 g
    • From M1/Q1 = M2/Q2
    • Therefore, Q1 = (65 × 965) ÷ .325 = 193000 C
    • But Q1 = 96500n
    • Therefore, n = 193000/96500 = 2
  • Faraday’s second law states the when the same quantity of electricity is passed through electrolytes in electrolytic cells connected in series, the number of moles of the element produced in each cell in inversely proportional to the charge on the ion of the element
  • Mathematically, n ∝ 1/c. Therefore, n1c1 = n2c2
  • NOTE: The magnitude of the charge on an ion represented by c must be equal to the number of moles of electrons or the number of Faradays required to produce 1 mole of the element

Examples

  1. A current is passed for 40 minutes through two electrolytic cells connected in series containing solutions of CuSO4 and brine respectively. If 12.8 g of copper are deposited in the CuSO4 cell, calculate the mass of chlorine gas liberated in the other cell. (Cu = 64g, Cl = 35.5 g)
    • nCu × cCu = nCl2 × cCl2
    • From Cu2+ + 2e → Cu, cCu = 2
    • From 2Cl + 2e → Cl2, cCl2 = 2
    • ∴ nCu × 2 = nCl2 × 2
    • ∴ nCu = nCl2 = 12.8/64 = 0.2 mole
    • But n = mass/MM
    • ∴ gCl2 = nCl2 × MMCl2 = 0.2 × 71 = 14.2 g
Uses of Electrolysis
  • For the extraction of highly electropositive metals such as K, Na, Ca, Mg, and Al from their ores
  • For the industrial production of some gases such as H2, F2, Cl2 from their fused salts or their aqueous solutions
  • For the purification of metals such as Cu, Ag
  • For electroplating one metal on another
  • For the preparation of some important compounds such as sodium hydroxide
ELECTRODE POTENTIALS
  • A piece of metal consists of metal cations and a cloud of delocalized valence electrons
  • If a metal is dipped into a solution of its ions, some of the cations in the metal may dissolve, leaving a build-up of electrons on the metal which will therefore become negatively charged.
    M(s) → Mn+ + ne
  • Relative to the solution, the metal is said to acquire a negative electrode potential
  • Alternatively, metal ions in the solution may take electrons from the metal and be discharged as metal atoms; thus leaving the metal positively charged. Mn+ + ne → M. The metal in this case acquires a positive electrode potential.

  • The sign of the electrode depends on which of these two processes predominates
  • For metals above hydrogen in the electrochemical series, the electrode potential is negative and this implies the dissolution process outweighs the deposition process
  • For metals below hydrogen in the electrochemical series, the electrode potential developed is positive because the deposition process outweighs the dissolution process
  • The positive or negative potential developed is known as absolute electrode potential but it cannot be measured experimentally because its value changes as soon as electrical wires are connected to the element
  • The potential that can be measured is known as relative electrode potential and this is carried out experimentally by joining the half=cell of the metal to a reference electrode known as Standard Hydrogen Electrode (S.H.E.)
Standard Hydrogen Electrode (S.H.E.)
  • S.H.E. is a reference electrode which is conveniently assigned an electrode or zero volts
  • It consists of hydrogen gas at 1 atm pressure bubbling over platinised platinum electrode or platinum foil coated with finely divided platinum known as platinum black which is immersed in a solution of 1 moldm-3 hydrogen ions at 25 °C

Photo Credit: Toppr

  • The platinum electrode conducts electricity into or out of the solution, provides a surface for electron transfer between H2(g) and H+ and catalyses the reaction
  • The potential which develops on the surface of the platinum according to the reaction
    2H(aq)+ + 2e → H2(g) or H2(g) → 2H(aq)+ + 2e is exactly zero volts
Measurement of Standard Electrode Potential
  • The electrode potential of an electrode or half cell can be found by combining it with a standard hydrogen electrode and measuring the e.m.f. of the cell formed
  • If the electrode is dipped into a 1 moldm-3 solution of its ions at 25 °C, then the e.m.f. obtained is equal to the standard electrode potential, Eθ, of the electrode or half cell because the standard electrode potential of S.H.E. is 0 volts
  • Electrode potential depends upon the nature of the electrode, the concentration of its ions in solution and temperature
  • Thus standard electrode potentials make room for meaningful comparison of the electrode potentials of different elements
  • By convention, standard electrode potential are also standard reduction potentials because the Eθ values simply indicate the inherent tendency of an ion to be reduced
  • A negative value indicates that the ion is unwilling to undergo reduction

Photo Credit: Chegg

  • When electrons flow from an electrode through the external circuit to the hydrogen electrode, the standard electrode potential is negative
  • If electrons flow from the standard hydrogen electrode to an electrode, then the standard electrode potential of that electrode is positive
Deductions from Eθ Values
  • Any element with a negative electrode potential is a good reducing agent
  • The higher the negative electrode potential, the better the reducing ability of the element
  • An element with a higher electrode potential will displace or kick out the ion of another element with a lower negative electrode potential form an aqueous solution. For example, Zn will kick out Fe ions because Zn has a higher negative electrode potential
  • An element or ion with a positive electrode potential may serve as an oxidizing agent
  • The greater the positive electrode potential of the element or ion, the greater its oxidizing ability or its ability to accept electrons from a rich source of electrons
  • An element with a higher positive electrode potential will displace the ion of another element with a lower positive electrode potential from an aqueous solution. For example, Cl ions would kick outI ions because Cl has a higher positive electrode potential
ELECTROCHEMICAL CELLS
  • Any two suitable half cells joined together make up an electrochemical cell known as a Galvanic (Voltaic) Cell
  • In such a cell, the half cell with the higher negative electrode potential acts as the anode while the other half cell acts as the cathode
Cell Notation
  • By convention, the standard notation for representing galvanic cells is a cell diagram which puts the anode on the left, salt bridge in the middle, and the cathode on the right
  • The diagram takes the general form Anode / Anode electrolyte // Cathode electrolyte/ Cathode
  • A single vertical line indicates physical contact between species in different phases or physical states
  • A double line indicates a salt bridge or porous divider which permits ion flow without allowing the electrolytes to mix
  • The cell diagram for the electrochemical cell formed using zinc and copper electrodes may be represented as follows:
    • Zn / ZnSO4 // CuSO4 / Cu OR Zn / Zn2+ // Cu2+ /Cu
  • The anode half cell is always written to depict oxidation while the cathode half cell is always written to depict reduction
  • When a voltaic cell has a gas electrode such as hydrogen electrode or chlorine electrode as one of its half cells, an inert electrode such as platinum that helps in effecting electrical connections is also indicated in the cell notation
  • Thus, an electrochemical cell made up of a magnesium half cell and a hydrogen half cell is represented as Mg / Mg2+ // H+ / H2 / (Pt)
Cell Reaction and Cell E.M.F.
  • A galvanic cell is a device for converting chemical energy into electrical energy by means of a redox reaction which occurs in two separate half cells connected by an electrical conductor
  • If a shiny piece of metallic copper is placed into a solution of AgNO3, a spontaneous reaction occurs.
  • No usable energy can be tapped from this reaction because the energy change that accompanies the reaction is lost as heat
  • This same chemical reaction will produce usable energy if the two half reactions involved in the net reaction are made to occur in separate containers.
  • When the switched is closed, copper is oxidized to Cu2+ in the anode compartment
  • The electrons released travel through the external circuit as an electric current to effect the reduction of Ag+ to Ag in the cathode compartment
  • A salt bridge is an inverted U-tube filled with porous plugs at both ends to hold in place a saturated solution of KCl or KNO3 filled into it.
  • The salt bridge maintains electrical neutrality in both the anode and cathode compartments
  • During the operation of the cell, negative ions diffuse from the salt bridge into the copper half cell to keep the electrolyte neutral
  • At the other half cell, positive ions form the salt bridge move into the cathode compartment to maintain neutrality
  • The overall cell reaction of a galvanic cell is the sum of the anodic and the cathodic reaction
    • Anodic reaction: Cu → Cu2+ + 2e
    • Cathodic reaciton: 2Ag+ + 2e → 2Ag
    • Overall reaction: Cu + 2Ag+ → Cu2+ + 2Ag
  • The cell notation for the copper – silver cell is given by Cu / Cu2+ // Ag+ / Ag
  • The e.m.f. or voltage produced by a cell is equal to the standard cell potential (Eθ cell) when all ion concentrations are 1 moldm-3 and the partial pressure of any gas involved is equal to 1 atmosphere
  • The e.m.f. or the standard cell potential of a cell is the difference between the standard electrode potential of a cell is the difference between the standard electrode potential of the cathode and the standard electrode potential of the anode
    • E.M.F. = Eθcell = Eθcathode – Eθanode
  • For the Cu – Ag cell given that Eθ(Cu2+/Cu) and Eθ(Ag+/Ag) are +0.34 and +0.80 V respectively
    • Eθcell = +0.80 – (+0.34) = 0.46V
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