Chapter 19: Electrochemistry Renee Y. Becker Valencia Community College

Oxidation–Reduction Reactions Redox reactions are those involving the oxidation and reduction of species. Oxidation and reduction must occur together. They cannot exist alone. Fe2+ + Cu0  Fe0 + Cu2+ Reduced: Iron gained 2 electrons Fe2+ + 2 e  Fe0 Oxidized: Copper lost 2 electrons Cu0  Cu2+ + 2 e Remember that electrons are negative so if you gain electrons your oxidation # decreases and if you lose electrons your oxidation # increases

Oxidation–Reduction Reactions Fe2+ + Cu0  Fe0 + Cu2+ Fe2+ gains electrons, is reduced, and we call it an oxidizing agent Oxidizing agent is a species that can gain electrons and this facilitates in the oxidation of another species. (electron deficient) Cu0 loses electrons, is oxidized, and we call it a reducing agent Reducing agent is a species that can lose electrons and this facilitates in the reduction of another species. (electron rich)

Example 1: Oxidation–Reduction Reactions Which is a reduction half reaction? Fe  Fe2+ + 2e Fe2+  Fe3+ + 1e Fe  Fe3+ + 3e Fe3+ + 1e  Fe2+

Example 2: Oxidation–Reduction Reactions For each of the following, identify which species is the reducing agent and which is the oxidizing agent. Ca(s) + 2 H+(aq)  Ca2+(aq) + H2(g) 2 Fe2+(aq) + Cl2(aq)  2 Fe3+(aq) + 2 Cl–(aq) C) SnO2(s) + 2 C(s)  Sn(s) + 2 CO(g)

Oxidation–Reduction Reactions Assigning Oxidation Numbers: All atoms have an “oxidation number” regardless of whether it carries an ionic charge. 1. An atom in its elemental state has an oxidation number of zero. Elemental state as indicated by single elements with no charge. Exception: diatomics H2 N2 O2 F2 Cl2 Br2 and I2

Oxidation–Reduction Reactions 2. An atom in a monatomic ion has an oxidation number identical to its charge.

Oxidation–Reduction Reactions 3. An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. A. Hydrogen can be either +1 or –1. B. Oxygen usually has an oxidation number of –2. In peroxides, oxygen is –1. C. Halogens usually have an oxidation number of –1. When bonded to oxygen, chlorine, bromine, and iodine have positive oxidation numbers.

Oxidation–Reduction Reactions 4. The sum of the oxidation numbers must be zero for a neutral compound and must be equal to the net charge for a polyatomic ion. A. H2SO4 neutral atom, no net charge SO42- sulfate polyatomic ion [SO4]2- [Sx O42-] = -2 X + -8 = -2 X = 6 so sulfur has an oxidation # of +6

Oxidation–Reduction Reactions B. ClO4– , net charge of -1 [ClO4]-1 [Clx O42-] = -1 X + -8 = -1 X = 7 so the oxidation number of chloride is +7

Example 3: Oxidation–Reduction Reactions Assign oxidation numbers to each atom in the following substances: A. CdS G. V2O3 B. AlH3 H. HNO3 C. Na2Cr2O7 I. FeSO4 D. SnCl4 J. Fe2O3 E. MnO4– K. H2PO4- F. VOCl3

What is the oxidation number of arsenic in AsO43- ? 8 1 5 -1 Example 4: What is the oxidation number of arsenic in AsO43- ? 8 1 5 -1

Activity Series of Elements Lithium, Li can reduce anything under it. Lithium is a very strong reducing agent and this is in part why lithium batteries work so well. Light weight and produce a high voltage Anode: Li  Li+ + e Cathode: MnO2 + Li+ + e  LiMnO2

Activity Series of Elements Activity series looks at the relative reactivity of a free metal with an aqueous cation. Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Cu(s) + 2 Ag+(aq)  2 Ag(s) + Cu2+(aq) Mg(s) + 2 H+(aq)  Mg2+(aq) + H2(g)

Example 5: Activity Series of Elements Given the following three reactions, determine the activity series for Cu, Zn, & Fe. 1. Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s) 2. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) 3. Fe(s) + Zn2+ (aq)  NR

Balancing Redox Reactions Half-Reaction Method: Allows you to focus on the transfer of electrons. This is important when considering batteries and other aspects of electrochemistry. The key to this method is to realize that the overall reaction can be broken into two parts, or half-reactions. (oxidation half and reduction half)

Balancing Redox Reactions Balance for an acidic solution: MnO4–(aq) + Br–(aq)  Mn2+(aq) + Br2(aq) 1. Determine oxidation and reduction half-reactions: Oxidation half-reaction: Br–(aq)  Br20(aq) Reduction half-reaction: MnO4–(aq)  Mn2+(aq) 2. Balance for atoms other than H and O: Oxidation: 2 Br–(aq)  Br2(aq) Reduction: MnO4–(aq)  Mn2+(aq)

Balancing Redox Reactions 3. Balance for oxygen by adding H2O to the side with less oxygen Oxidation: 2 Br–(aq)  Br2(aq) Reduction: MnO4–(aq)  Mn2+(aq) + 4 H2O(l) 4. Balance for hydrogen by adding H+ to the side with less hydrogens Oxidation: 2 Br–(aq)  Br2(aq) Reduction: MnO4–(aq) + 8 H+(aq)  Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions 5. Balance for charge by adding electrons (e–): Oxidation: 2 Br–(aq)  Br2(aq) + 2 e– Reduction: MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l) 6. Balance for numbers of electrons by multiplying: Oxidation: 5[2 Br–(aq)  Br2(aq) + 2 e–] Reduction: 2[MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l)]

Balancing Redox Reactions 7. Combine and cancel to form one equation: Oxidation: 10 Br–(aq)  5 Br2(aq) + 10 e– Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e– 2 Mn2+(aq) + 8 H2O(l) 2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l)

Example 6: Balancing Redox Reactions Balance the following in an acidic sol’n NO3–(aq) + Cu(s)  NO(g) + Cu2+ (aq)

Balancing Redox Reactions Balance for an basic solution: MnO4-(aq) + SO32-(aq)  MnO2(s) + SO42-(aq) 1. Determine oxidation and reduction half-reactions: Oxidation half-reaction: SO32–(aq)  SO42-(aq) Reduction half-reaction: MnO4–(aq)  MnO2(s) 2. Balance for atoms other than H and O: Oxidation: SO32– (aq)  SO42- (aq) Reduction: MnO4– (aq)  MnO2(aq)

Balancing Redox Reactions: Basic Sol’n 3. Balance for oxygen by adding H2O to the side with less oxygen Oxidation: H2O(l) + SO32–  SO42- Reduction: MnO4–  MnO2 + 2 H2O(l) 4. Balance for hydrogen by adding H+ to the side with less hydrogens Oxidation: H2O(l) + SO32–  SO42- + 2H+(aq) Reduction: 4H+(aq) + MnO4–  MnO2 + 2 H2O(l)

Balancing Redox Reactions: Basic Sol’n 5. Balance for charge by adding electrons (e–): Oxidation: H2O + SO32–  SO42- + 2H+ + 2 e- Reduction: 3 e- + 4H+ + MnO4–  MnO2 + 2 H2O 6. Balance for numbers of electrons by multiplying: Oxidation: 3[H2O + SO32–  SO42- + 2H++ 2 e-] Reduction: 2[3 e- + 4H+ + MnO4–  MnO2 + 2 H2O]

Balancing Redox Reactions: Basic Sol’n 7. Combine and cancel to form one equation: Ox: 3H2O(l) + 3SO32–(aq)  3SO42-(aq) + 6H+(aq) + 6 e- Red: 6 e- + 8H+(aq) + 2MnO4–(aq)  2MnO2(s) + 4 H2O(l) 2 1 2H+(aq) + 3SO32-(aq) + 2MnO4-(aq)  3SO42-(aq) + 2MnO2(s) + H2O(l) So far same as acidic sol’n, 2 more steps for basic sol’n

Balancing Redox Reactions: Basic Sol’n Note # of H+ ions in the equation. Add this # of OH- to both sides of equation 2OH- + 2H+ + 3SO32- + 2MnO4-  3SO42- + 2MnO2 + H2O + 2OH- Simplify equation by combining H+ and OH- to make H2O, reduce to lowest terms 2H2O + 3SO32- + 2MnO4-  3SO42- + 2MnO2 + H2O + 2OH- 1 Write balanced equation

Balancing Redox Reactions: Basic Sol’n H2O(l) + 3SO32-(aq) + 2MnO4-(aq)  3SO42-(aq) + 2MnO2(s) + 2OH-(aq)

2 types of electrochemical cells 1. Galvanic cells (voltaic cells) Electrochemistry 2 types of electrochemical cells 1.  Galvanic cells (voltaic cells) Spontaneous chemical rxn generates an electric current 2.  Electrolytic cells An electric current drives a non-spontaneous rxn

Cu2+ is reduced and it is the oxidizing agent Review of Redox Overall rxn Zn + Cu2+  Zn2+ + Cu ½ ox. rxn Zn  Zn2+ + 2e- ½ red. rxn Cu2+ + 2e-  Cu   Cu2+ is reduced and it is the oxidizing agent Zn is oxidized and it is the reducing agent Electrons are transferred directly from Zn to Cu2+ 4. Enthalpy of rxn is lost to the surroundings as heat

Same rxn with electrochemical cell (galvanic) Same rxn with electrochemical cell (galvanic) Some of the chemical energy released by the rxn is converted to electrical energy which can be used to light a light bulb Apparatus

Anode electrode is negative, oxidation takes place Zn strip Apparatus Electrodes- strips of zinc and copper connected by an electrically conductive wire (electrons transferred through wire make current) Anode electrode is negative, oxidation takes place Zn strip Cathode electrode reduction takes place Cu strip Salt bridge- U-shaped tube that contains a gel permeated with a sol’n of inert electrolyte (will not react)   Anode: Ox ½ Zn  Zn2+ + 2e- Cathode: Red ½ Cu2+ + 2e-  Cu Overall Rxn Zn + Cu2+  Zn2+ + Cu

Why is the salt bridge necessary? 1. Completes the electrical circuit Apparatus Why is the salt bridge necessary? 1. Completes the electrical circuit Without it the anode cell would become positively charged as Zn2+ ions appear the sol’n in the cathode beaker would become negatively charged as Cu2+ ions are removed 2. Because of the charge imbalance the elctrode rxns would quickly stop and electron flow through the wire stops 3. With the salt bridge the electrical neutrality is maintained in both beakers by a flow of ions Anions- SO42- flow through salt bridge from cathode to anode Cations- from anode to cathode

Remember: An ox ate a red cat Overview Anode: Oxidation occurs, electrons are produced, anions move toward, neg. sign Cathode: Reduction occurs, Electrons are consumed, cations move towards, positive sign Remember: An ox ate a red cat

Shorthand Notation for Galvanic Cells Example: Zn(s) + Cu2+  Zn2+ + Cu(s) Shorthand notation: Zn(s)  Zn2+(aq)  Cu2+(aq)  Cu(s) 1. , single vertical line represents a phase boundary (between a solid electrode and an aqueous solution 2. , double vertical line denotes a salt bridge 3. The anode half-cell is always on the left of the salt bridge, with the solid electrode to the far left 4. The cathode half-cell is always on the right of the salt bridge, with the solid electrode on the far right 5. The reactants in each half cell are always written first, followed by the products 6. Electrons move through the external circuit from left to right (from anode to cathode)

Fe(s) + Sn2+(aq)  Fe2+(aq) + Sns) Example 7: Write the shorthand notation for a galvanic cell that uses the reaction   Fe(s) + Sn2+(aq)  Fe2+(aq) + Sns)

Cell Potentials and Free-Energy Changes for Cell Reactions Electrons move through the external circuit from the zinc anode to the copper cathode because they have lower energy when on copper than on zinc. The driving force that pushes the negatively charges electrons away from the anode (- electrode) and pulls them toward the cathode (+ electrode) is an electrical potential called the electromotive force (emf) also known as cell potential (E) or the cell voltage. SI units in volts (V)

Cell Potentials and Free-Energy Changes for Cell Reactions 1 J = 1C x 1V G = -nFE n = number of moles of electrons transferred in the reaction F = faraday constant 9.648534 x 104 C/mol G = gibbs free energy   G = -nFE Standard free-energy change and standard cell potential Because both are directly proportional a voltmeter can be regarded as a “free- energy meter” When a voltmeter measures E, it is also indirectly measuring G

Al(s) + Cr3+(aq)  Al3+(aq) + Cr(s) Example 8: The standard cell potential at 25C is 0.92 V for the reaction   Al(s) + Cr3+(aq)  Al3+(aq) + Cr(s) What is the standard free-energy change for this reaction at 25C in kJ?

Standard Reduction Potentials The standard potential of any galvanic cell is the sum of the standard half-cell potentials for oxidation at the anode and reduction at the cathode: Ecell = Eox + Ered H2 gas is oxidized to H+ ions at the anode and Cu2+ ions are reduced to copper metal at the cathode

Standard Reduction Potentials Anode (oxidation): H2(g)  2 H+(aq) + 2 e Cathode (reduction): Cu2+(aq) + 2 e  Cu(s)   Overall: H2(g) + Cu2+(aq)  2 H+(aq) + Cu(s) Ecell = Eox + Ered = EH2H+ + ECu2+  Cu = .34 V To come up with a table of standard potentials we must first chose a reference half-cell to measure all others with.

Standard Reduction Potentials Standard Hydrogen Electrode (SHE) (last cell we considered!!) E = 0 V   .34 V = 0 V + .34 V Because the Cu2+/Cu half-reaction is a reduction the half-cell potential is called a standard reduction potential Cu2+ + 2 e  Cu E = .34 V Cu  Cu2+ + 2 e E = -.34 V

Standard Reduction Potential Table 1. The half-reactions are all written as reductions. Oxidizing agents and electrons are on the left side of each half-reaction and reducing agents are on the right side 2. The listed half-cell potentials are standard reduction potentials, also known as standard electrode potentials 3. The half-reactions are listed in order of decreasing standard reduction potential(decreasing tendency to occur in the forward direction; increasing tendency to occur in the reverse direction) The strongest oxidizing agents are located in the upper left of the table (F2, H2O2, MNO4-) the strongest reducing agents are found in the lower right of the table (Li, Na, Mg

Using Standard Reduction Potentials Table arranges oxidizing or reducing agents in order of increasing strength, this allows us to predict the spontaneity or nonspontaneity of thousands of redox rxns. Let’s calculate E for the oxidation of Zn(s) by Ag+(aq): 2 Ag+(aq) + Zn(s)  2 Ag(s) + Zn2+(aq)

Using Standard Reduction Potentials Step 1: Find half-rxns in Table and write them in the appropriate direction Step 2: Multiply the Ag+/Ag half-reaction by a factor of 2 so that the electrons cancel. Do not multiply the E by 2 because electric potential is an intensive property, which does not depend on the amount of substance. E = -G / nF G will double and n will double so that E will remain constant Step 3: Change E for oxidations since the table is based on reductions Step 4: Add the half-reactions to get the overall reaction.

Using Standard Reduction Potentials Reduction: 2 x [Ag+ + e-  Ag] E = 0.80 V Oxidation: Zn  Zn2+ + 2e- E = -(-0.76 V) Overall reaction: 2 Ag+(aq) + Zn(s)  2 Ag(s) + Zn2+(aq) E= 1.56 V E = -G / nF G =   -E(nF) = -3.01 x 105 Because E is positive and G is negative, oxidation of zinc by Ag+ is a spontaneous reaction under SS conditions. Note: Ag+ can oxidize any reducing agent that lies below it in the table. Can’t oxidize a reducing agent that appears above it on the table. Because E will have a negative value.

Br2(l) , Fe3+(aq) , Cr2O72-(aq) Example 9: Arrange the following oxidizing agents in order of increasing strength under SS conditions: Br2(l) , Fe3+(aq) , Cr2O72-(aq)

Example 10: Arrange the following reducing agents in order of increasing strength under SS conditions: Al(s) , Na(s) , Zn(s)

a) 2 Fe3+(aq) + 2 I-(aq)  2 Fe2+(aq) + I2(s) Example 11: Predict from the Reduction table whether each of the following reactions can occur spontaneously under SS conditions: a)      2 Fe3+(aq) + 2 I-(aq)  2 Fe2+(aq) + I2(s) b)      3 Ni(s) + 2 Al3+(aq)  3 ni2+(aq) + 2 Al(s)

-Concentration of solutes -partial pressures of gases Cell potentials and composition of the rxn mixture: The Nernst Equation Cell pot. and gibbs free energy depend on temp. and composition of rxn mixture -Concentration of solutes -partial pressures of gases   G = G + RT ln Q Q = Rxn quotient nFE = -nFE + RT ln Q

Nernst Equation: E = E - RT ln Q nF or E = E - 2.303 RT ln Q The Nernst Equation Nernst Equation: E = E - RT ln Q nF or E = E - 2.303 RT ln Q   or E = E - 0.0592 V log Q in Volts @ 25C n This allows us to calculate cell potentials under non-standard state conditions

Zn(s) + 2 H+(aq)  Zn2+(aq) + H2(g) Example 12: Consider a galvanic cell that uses the reaction   Zn(s) + 2 H+(aq)  Zn2+(aq) + H2(g) Calculate the cell potential @ 25C when [H+] = 1.0 M, [Zn2+] = 0.0010 M, and PH2 = 0.10 atm Going to use: E = Eox + Ered Going to use: E = E - 0.0592 V log Q n

Cu(s) + 2 Fe3+(aq)  Cu2+(aq) + 2 Fe2+(aq) Example 13: Consider a galvanic cell that uses the reaction Cu(s) + 2 Fe3+(aq)  Cu2+(aq) + 2 Fe2+(aq)   What is the potential of a cell at 25C that has the following ion concentrations [Fe3+] = 1.0 x10-4 M [Cu2+] = 0.25 M [Fe2+] = 0.20 M Going to use: E = Eox + Ered Going to use: E = E - 0.0592 V log Q n

Standard Cell potentials and equilibrium constants E = - 0.0592 V log K in Volts @ 25C n + E = + log K therefore K>1   - E = - log K therefore K<1 Remember: when K is large rxn essentially to 100% completion When K is very small rxn doesn’t proceed at all

Standard Cell potentials and equilibrium constants Now we have 3 ways to find an equilibrium constant 1.    aA + bB  cC + dD from concentration or partial pressures K = [C]c [D]d [A]a [B]b 2. G = -RT ln K, ln K = -G RT From thermochemical data   3. E = RT ln K or ln K = nFE or log K = nE nF RT .0592 V From electrochemical data

6 Br-(aq) + Cr2O72-(aq) + 14 H+(aq)  3 Br2(l) + 2 Cr3+(aq) + 7 H2O(l) Example 14: Use the standard reduction potentials in Table to calculate the equilibrium constant at 25C for the reaction 6 Br-(aq) + Cr2O72-(aq) + 14 H+(aq)  3 Br2(l) + 2 Cr3+(aq) + 7 H2O(l)

4 Fe2+(aq) + O2(g) + 4 H+(aq)  4 Fe3+(aq) + 2 H2O(l) Example 15: Use the data in the standard reduction table to calculate the equilibrium constant @ 25C for the reaction   4 Fe2+(aq) + O2(g) + 4 H+(aq)  4 Fe3+(aq) + 2 H2O(l)

Single-cell batteries consist of one galvanic cell. Batteries are the most important practical application of galvanic cells. Single-cell batteries consist of one galvanic cell. Multicell batteries consist of several galvanic cells linked in series to obtain the desired voltage. (car batteries)

Anode: Lead grid packed with spongy lead. Batteries Lead Storage Battery: A typical 12 volt battery consists of six individual cells connected in series. Anode: Lead grid packed with spongy lead. Cathode: Lead grid packed with lead oxide. Electrolyte: 38% by mass sulfuric acid. Cell Potential: 1.924 V

Anode: Zinc metal can on outside of cell. Batteries Zinc Dry-Cell: Also called a Leclanché cell, uses a viscous paste rather than a liquid solution. Anode: Zinc metal can on outside of cell. Cathode: MnO2 and carbon black paste on graphite. Electrolyte: NH4Cl and ZnCl2 paste. Cell Potential: 1.5 V but deteriorates to 0.8 V with use.

Anode: Zinc metal can on outside of cell. Batteries Alkaline Dry-Cell: Modified Leclanché cell which replaces NH4Cl with NaOH or KOH. Anode: Zinc metal can on outside of cell. Cathode: MnO2 and carbon black paste on graphite. Electrolyte: NaOH or KOH, and Zn(OH)2 paste. Cell Potential: 1.5 V but longer lasting, higher power, and more stable current and voltage.

Anode: Zinc metal can on outside of cell. Batteries Mercury Dry-Cell: Modified Leclanché cell which replaces MnO2 with HgO and uses a steel cathode. Anode: Zinc metal can on outside of cell. Cathode: HgO in contact with steel. Electrolyte: KOH, and Zn(OH)2 paste. Cell Potential: 1.3 V with small size, longer lasting, and stable current and voltage.

Nickel–Cadmium Battery: Modified Leclanché cell which is rechargeable. Batteries Nickel–Cadmium Battery: Modified Leclanché cell which is rechargeable. Anode: Cadmium metal. Cd(s) + 2 OH–(aq)  Cd(OH)2(s) + 2 e– Cathode: Nickel(III) compound on nickel metal. NiO(OH) (s) + H2O(l) + e–  Ni(OH)2(s) + OH–(aq) Electrolyte: Nickel oxyhydroxide, NiO(OH). Cell Potential: 1.30 V

Nickel–Metal–Hydride (NiMH): Batteries Nickel–Metal–Hydride (NiMH): Replaces toxic Cd anode with a hydrogen atom impregnated ZrNi2 metal alloy. During oxidation at the anode, hydrogen atoms are released as H2O. Recharging reverses this reaction.

Anode: Li metal, or Li atom impregnated graphite. Li(s)  Li+ + e– Batteries Lithium Ion (Li–ion): The newest rechargeable battery is based on the migration of Li+ ions. Anode: Li metal, or Li atom impregnated graphite. Li(s)  Li+ + e– Cathode: Metal oxide or sulfide that can accept Li+. MnO2(s) + Li+(aq) + e–  LiMnO2(s) Electrolyte: Lithium-containing salt such as LiClO4, in organic solvent. Solid state polymers can also be used. Cell Potential: 3.0 V

Anode: Porous carbon containing metallic catalysts. Fuel Cell Fuel Cell: Uses externally fed CH4 or H2, which react to form water. Most common is H2. Anode: Porous carbon containing metallic catalysts. 2 H2(s) + 4 OH–(aq)  4 H2O(l) + 4 e– Cathode: Porous carbon containing metallic catalysts. O2(s) + 2 H2O(l) + 4 e–  4 OH–(aq) Electrolyte: Hot aqueous KOH solution. Cell Potential: 1.23 V, but is only 40% of cell capacity.

Fuel cells are not batteries because they are not self-contained. Fuel cells typically have about 40% conversion to electricity; the remainder is lost as heat. Excess heat can be used to drive turbine generators.

b) Rusting results from tiny galvanic cells formed by water droplets Corrosion Corrosion is the oxidative deterioration of a metal, such as iron to rust. To prevent corrosion first we have to understand how it occurs a) Rusting requires both oxygen and water b) Rusting results from tiny galvanic cells formed by water droplets

Oxidation: Fe(s)  Fe2+(aq) + 2 e– Corrosion Oxidation: Fe(s)  Fe2+(aq) + 2 e– Reduction: O2(g) + 4 H+(aq) + 4 e–  2 H2O(l) Overall: 2 Fe(s) + O2(g) + 4 H+(aq)  2 Fe2+(aq) + 2 H2O(l) This electrochemical mechanism for corrosion also explains why automobiles rust more rapidly in parts of the country where road sat is used to melt snow and ice. Dissolved salts in the water droplet greatly increase the conductivity of the electrolyte, thus accelerating the pace of corrosion.

Prevention of corrosion Galvanizing: is the coating of iron with zinc. Zinc is more easily oxidized than iron, which protects and reverses oxidation of the iron. Cathodic Protection: is the protection of a metal from corrosion by connecting it to a metal (a sacrificial anode) that is more easily oxidized. All that is required is an electrical connection to the sacrificial anode (usually magnesium or zinc).

Prevention of corrosion Cathodic Protection with Zinc Layer

Prevention of corrosion Cathodic Protection with Magnesium Anode

An electrolytic cell is an apparatus for carrying out electrolysis. Electrolysis: is the process in which electrical energy is used to drive a nonspontaneous chemical reaction. An electrolytic cell is an apparatus for carrying out electrolysis. Processes in an electrolytic cell are the reverse of those in a galvanic cell.

Applications of Electrolysis Manufacture of Sodium (Downs Cell):

Applications of Electrolysis Manufacture of Cl2 and NaOH (Chlor–Alkali):

Applications of Electrolysis Manufacture of Aluminum (Hall–Heroult):

Applications of Electrolysis Electrorefining and Electroplating: