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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 EquationsALL 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 Al3+(aq) and H2(g). The (unbalanced) equation is
Al(s) + H+(aq) Al3+(aq) + H2(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, MnO4
-, 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.
MnO4-(aq) + Fe2+(aq) Mn2+(aq) + Fe3+(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) Ni2+(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)ICu2+(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; Eo
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• Eo
cell = Eocathode – Eo
anode
• When the Eocell 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 Eo
cell)• Best (strongest) reducing agents are
written at the bottom. (most negative Eo
cell)• Reversing a half-reaction reverses
the sign of Eo.• Northwest-southeast Rule.
Practice Problems• The net reaction that occurs in a
voltaic cell is
Zn(s) + 2Ag+(aq) Zn2+(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 Cl2.
• Decide which of the halogens is capable of oxidizing gold metal to Au3+(aq).
Practice Problem• Determine which of the following
redox equations are product-favored. Assume standard conditions.
• Ni2+(aq) + H2(g) Ni(s) + 2H+(aq)
• 2Fe3+(aq) + 2I-(aq) 2Fe2+(aq) + I2(s)
• Br2(l) + 2Cl-(aq) 2Br-(aq) + Cl2(g)
• Cr2O72-(aq) + 6Fe2+(aq) + 14H+(aq)
2Cr3+(aq) + 6Fe3+(aq) + 7H2O(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