Principles of Reactivity: Electron Transfer Reactions Chapter 20

Electron Transfer Reactions In order to produce electricity, the electron transfer between substances must be carried out in an apparatus that allows the electrons to be transferred through an electrical circuit. Devices that use chemical reactions to produce an electric current are called voltaic cells or galvanic cells.

Electrochemistry refers to the interchange of electrical and chemical energy –Voltaic cells use product favored reactions to convert chemical energy to electrical energy. –Electrolysis is when electrical energy is used to effect a chemical change. Example: splitting water into its component elements

20.1 Oxidation-Reduction Reactions One reactant is oxidized and one is reduced. Oxidation and reduction must balance. Oxidizing agent is reduced. Reducing agent is oxidized. Oxidation numbers can be used to figure out what is oxidized (# increases) and what is reduced (# decreases).

Balancing Redox Equations ALL REDOX REACTIONS MUST BE BALANCED FOR BOTH MASS AND CHARGE! –The method most often used to balance redox equation is by writing half- reactions. –One half-reaction describes oxidation, one describes reduction. Balance electrons and add together! –The net ionic equation will therefore be balanced for both mass and charge.

Practice Problem Aluminum reacts with nonoxidizing acids to give Al 3+ (aq) and H 2 (g). The (unbalanced) equation is Al(s) + H + (aq)  Al 3+ (aq) + H 2 (g) Write balanced half-reactions and the balanced net ionic equation. Identify the oxidizing agent, the reducing agent, the substance oxidized, and the substance reduced.

Balancing Redox Sometimes water must be added and either hydrogen or hydroxide ions if the reaction occurs in acidic or basic solution. –equations with sulfate, nitrate, chromate, permanganate, etc…

Practice Problem The permanganate ion, MnO 4 -, is an oxidizing agent. A common laboratory analysis for iron is to titrate aqueous iron (II) ion with a solution of potassium permanganate of precisely known concentration. Use the half-reaction method to write the balanced net ionic equation for the reaction in acidic solution. MnO 4 - (aq) + Fe 2+ (aq)  Mn 2+ (aq) + Fe 3+ (aq)

Practice Problem Voltaic cells based on the oxidation of sulfur are under development. One such cell involves the reaction of sulfur with aluminum under basic conditions. Al(s) + S(s)  Al(OH) 3 (s) + HS - (aq) Balance this equation showing each balanced half-reaction. Identify the oxidizing and reducing agents, the substance oxidized, and the substance reduced.

20.2 Simple Voltaic Cells In a voltaic cell, the two half- reactions are separated so that electrons cannot be directly transferred between reactants. The two cells are connected with a salt bridge that allows cations and anions to move between them.

Simple Voltaic Cells The anode is the electrode at which oxidation occurs. The cathode is the electrode at which reduction occurs. In the salt bridge, cations move from anode to cathode, and anions move from cathode to anode in order to maintain electrical neutrality. Electrons flow spontaneously from anode to cathode.

Practice Problem Describe how to set up a voltaic cell using the following half-reactions: Reduction: Ag + (aq) + e -  Ag(s) Oxidation: Ni(s)  Ni 2+ (aq) + 2e - Which is the anode and which is the cathode? What is the overall cell reaction? What is the direction of electron flow in an external wire connecting the two electrodes? Describe the ion flow in a salt bridge connecting the cell compartments.

Inert Electrodes Sometimes there has to be an inert electrode because there aren’t two conductive metals as reactants. Shorthand is often used to symbolize electrochemical cells. –The anode is written on the left. A single vertical line indicates a phase boundary, and double vertical lines indicate a salt bridge. Cu(s)ICu 2+ (aq, 1.0M)IIAg + (aq, 1.0M)IAg(s)

20.4 Standard Electrochemical Potentials Electrons move from anode toward the cathode due to the difference in potential energy of electrons at the two electrodes. –electromotive force (emf) = difference in potential energy –units of volts (V) 1 Joule = 1 volt x 1 coulomb One coulomb is the quantity of charge that passes a point in an electric circuit when a current of one ampere flows for one second (1 coulomb = 1 amp x 1 sec)

Standard Conditions Half-cell potentials assume the following: –Reactants and products are present as pure liquids or solids. –Solutes in aqueous solution have a concentration of 1.0M. –Gaseous reactants or products have a pressure of 1.0 atm or 1.0 bar. Standard Potentials are measured under these conditions; E o cell

Deviations from Standard Conditions The farther the reaction is from equilibrium, the greater the magnitude of the cell potential –As the system approaches equilibrium, the magnitude of cell potential decreases, reaching zero at equilibrium –Deviations that take the cell further from equilibrium increase cell potential –Deviations that take the cell closer to equilibrium decreases the cell potential

Standard Cell Potentials Predict that the reaction occurring is the one in which the reactants are stronger reducing and oxidizing agents than the products. Electrons move from the electrode of higher potential energy to the one of lower potential energy. Cell potential can be calculated to determine the relative oxidizing or reducing ability.

Standard Reduction Potentials E o cell = E o cathode – E o anode When the E o cell has a positive value, the reaction is predicted to be product-favored as written.

Tables of Standard Reduction Potentials Table on page 920 All potentials are for reduction reactions. Best (strongest) oxidizing agent is written at the top. (most positive E o cell ) Best (strongest) reducing agents are written at the bottom. (most negative E o cell ) Reversing a half-reaction reverses the sign of E o. Northwest-southeast Rule.

Practice Problems The net reaction that occurs in a voltaic cell is Zn(s) + 2Ag + (aq)  Zn 2+ (aq) + 2Ag(s) Assuming standard conditions, identify the half-reactions that occur at the anode and the cathode and calculate a potential for the cell.

Practice Problem Rank the halogens in order of their strength as oxidizing agents. Decide if hydrogen peroxide in acidic solution is a stronger oxidizing agent than Cl 2. Decide which of the halogens is capable of oxidizing gold metal to Au 3+ (aq).

Practice Problem Determine which of the following redox equations are product- favored. Assume standard conditions. Ni 2+ (aq) + H 2 (g)  Ni(s) + 2H + (aq) 2Fe 3+ (aq) + 2I - (aq)  2Fe 2+ (aq) + I 2 (s) Br 2 (l) + 2Cl - (aq)  2Br - (aq) + Cl 2 (g) Cr 2 O 7 2- (aq) + 6Fe 2+ (aq) + 14H + (aq)  2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2 O(l)

Recognizing Redox Reactions One reactant is a metal and the other is an aqueous metal ion. Metal will be oxidized to form an aqueous ion, usually by not always having a charge of 2+ and the aqueous ion will be reduced to the corresponding metal. –A piece of solid zinc is placed in an aqueous solution of copper (II) sulfate.

Recognizing Redox Reactions One reactant is a polatomic anion with a metallic element displaying its highest oxidation number and the other is an anion displaying an oxidation number lower than its max. The polyatomic ion reduces to an ion displaying the metal in a lower oxidation state. Anion is oxidized to a higher oxidation state. –Acidic aqueous sodium dichromate is mixed with a solution of potassium bromide.

Recognizing Redox Reactions An organic compound burned in air (or oxygen) produces carbon dioxide and water. –Ethanol is burned in air.

Recognizing Redox Reactions A metal reacts with a non-metal to produce a binary salt. –Solid sodium is mixed with chlorine gas.

Recognizing Redox Reactions An active metal reacts with water to produce hydrogen gas and an hydroxide base. Solid lithium is placed in water.

Faraday’s Laws Used to determine stoichiometry of the redox reactions in cells with respect to: –Number of electrons transferred –Mass of material deposited or removed from an electrode –Current –Time elapsed –Charge of ionic species

Homework After reading Chapter 20, you should be able to do the following problems… P. 944 (50-59)