Ch. 18 ElectrochemistryCh. 18 Electrochemistry

Dr. Namphol SinkasetChem 201: General Chemistry II

I. Chapter OutlineI. Chapter Outline

I. IntroductionII. Balancing Redox ReactionsIII. Galvanic CellsIV. Standard Reduction PotentialsV. E°cell, ΔG°, and K

VI. BatteriesVII. Electrolysis

I. IntroductionI. Introduction

• Reduction-oxidation (redox) reactions are a huge branch of chemistry.

• Basically, they are reactions in which electrons are in motion.

• Electrons in motion = electricity.• Concepts in this chapter can be used to

explain fuel cells, batteries, electroplating, etc.

I. Redox ReactionsI. Redox Reactions

• Recall that reduction and oxidation are coupled processes. Reduction is the gain of electrons. Oxidation is the loss of electrons.

• We identify what is oxidized and what is reduced by examining oxidation numbers.

I. Assigning Oxidation NumbersI. Assigning Oxidation Numbers

1) Atoms in elemental form have O.N. = 0.2) Charge on a monatomic ion equals its O.N.3) The sum of all O.N. must equal the total charge.4) For Group 1, O.N. = +1.5) For Group 2, O.N. = +2.6) For H, O.N. = +1 w/ nonmetals, -1 w/ metals and B.7) For F, O.N. = -1.8) For O, O.N. = -1 in peroxides and -2 in all others.9) For Group 17, typically O.N. = -1.

I. Sample Redox ReactionI. Sample Redox Reaction

• Ca(s) is oxidized; it is the reducing agent.

• H2O(l) is reduced; it is the oxidizing agent.

II. Balancing Redox ReactionsII. Balancing Redox Reactions

• Balancing a redox reaction is more involved because mass and charge must both be balanced.

• A common way to balance redox reactions is the “half-cell method.” In this method, we split the reaction into 2

half reactions: oxidation and reduction. Each half is balanced separately, and then

the two are recombined.

II. The Half-Cell MethodII. The Half-Cell Method1) Identify what is being oxidized and reduced.2) Write each half-cell reaction.3) Balance each half-cell w/ respect to mass:

Balance elements other than O and H. Balance O by adding H2O. Balance H by adding H+.

4) Balance each half-cell w/ respect to charge by adding e-’s.

5) Make the # of e-’s in each half-cell equal by multiplying by an appropriate factor.

6) Add the two half-cells together.7) Add required # of OH- to each side and simplify

if redox reaction takes place in basic solution.

II. Sample ProblemII. Sample Problem

• Balance the following reaction in acidic solution:

ClO-(aq) + Cr(OH)4

-(aq) CrO4

2-(aq) + Cl-(aq).

II. Sample ProblemII. Sample Problem

• Balance the following reaction in basic solution:

MnO4-(aq) + Br-

(aq) MnO2(s) + BrO3-(aq)

III. Generating ElectricityIII. Generating Electricity

• As stated earlier, electrical current is simply charge in motion.

• Since an e- is charged, movement of electrons generates electrical current.

• Some redox reaction occur spontaneously. e.g. Zn(s) + Cu2+

(aq) Zn2+(aq) + Cu(s)

III. Zn/CuIII. Zn/Cu2+2+ Redox Redox

• Zn(s) is oxidized to Zn2+ which goes into solution.

• Cu2+ is reduced to Cu(s) which “plates out.”

• 2 e-’s are transferred from Zn to Cu2+.

III. Making e-’s TravelIII. Making e-’s Travel

• In this arrangement, we don’t really get electrons flowing, more of a transfer.

• If we separate the components, we can force electrons to travel externally, and even do work for us.

• This kind of setup is called an electrochemical cell.

III. Electrochemical CellIII. Electrochemical Cell

III. Electrochemical CellsIII. Electrochemical Cells• Voltaic (or galvanic) cells produce

electrical current from a spontaneous redox reaction.

• Important aspects: Each part is called a half-cell. Oxidation occurs at the anode. Electrons

flow away from the anode. Reduction occurs at the cathode.

Electrons come to the cathode. A salt bridge is necessary to prevent

charge build up and complete the circuit.

III. Current and PotentialIII. Current and Potential

• Two important aspects about electrons in motion are current and potential.

• Electrical current is measured in amperes (A), which has units of coulombs (measure of charge) per second, C/s. 1 A = 1 C/s.

• The current is driven by a potential energy difference called the potential difference. Potential difference is a measure of the difference

in PE per unit of charge. The SI unit is the volt (V). 1 V = 1 J/C.

III. Cell PotentialsIII. Cell Potentials

• The potential difference is the force that drives the movement of e-’s, so it’s also called the electromotive force (emf).

• In a voltaic cell, the potential difference between the cathode and anode is called the cell potential (Ecell) or cell emf.

III. Standard Cell PotentialsIII. Standard Cell Potentials

• If the cell is under standard conditions, the cell potential is the standard cell potential (E°cell) or standard emf.

• For the Zn(s)/Cu2+(aq) cell, E°cell = 1.10 V.

• The cell potential is a measure of the tendency of the redox reaction to occur spontaneously.

III. Electrochemical Cell NotationIII. Electrochemical Cell Notation• Instead of drawing an electrochemical

cell, line notation can be used.• The Zn(s)/Cu2+

(aq) can be represented as Zn(s)|Zn2+

(aq)||Cu2+(aq)|Cu(s)

Oxidation is always written first, followed by the reduction.

Double line represents the salt bridge. Different phases are separated by single

lines. Multiple substances in solution are

separated by commas.

III. Sample Cell NotationIII. Sample Cell Notation

Fe(s)|Fe2+(aq)||MnO4

-(aq), H+

(aq), Mn2+(aq)|Pt(s)

III. Sample ProblemIII. Sample Problem

• Draw and completely label the electrochemical cell represented by the line notation shown below. Additionally, write the overall balanced equation for the redox reaction.

Sn(s)|Sn2+(aq)||NO(g)|NO3

-(aq), H+

(aq)|Pt(s)

IV. Calculating Cell PotentialsIV. Calculating Cell Potentials

• In an electrochemical cell, we can think of each half-cell as having its own potential.

• Thus, E°cell is a sum of both half-cell potentials.

• The half-cell with the higher potential will occur in the forward direction, forcing the other half-cell to occur in the reverse direction.

IV. Half-Cell PotentialsIV. Half-Cell Potentials

• We can’t create a cell that has only reduction or only oxidation; thus, we can’t measure an absolute value of a half-cell.

• We have to assign a particular half-cell a value of 0.00 V and measure everything else relative to that.

IV. Standard Hydrogen ElectrodeIV. Standard Hydrogen Electrode

• The standard hydrogen electrode (SHE) is normally assigned a potential of 0.00 V.

• We attach different half-cells to SHE, and whatever potential comes out is assigned to the half-cell.

IV. Measuring a Standard IV. Measuring a Standard Reduction PotentialReduction Potential

IV. Standard IV. Standard Reduction Reduction PotentialsPotentials

IV. Important PointsIV. Important Points• The reduction potential of SHE is 0.00 V.• Half-cells with greater tendency to

undergo reduction than SHE have a positive reduction potential.

• Half-cells with lesser tendency to undergo reduction than SHE have a negative reduction potential.

• Substances at the top have strong tendency to be reduced; they are strong oxidizing agents.

IV. Important PointsIV. Important Points

• Substances at the bottom have strong tendency to be oxidized; they are strong reducing agents.

• E°ox = -E°red

• For any cell, E°cell = E°ox + E°red.• Never multiply half-cell potentials by

coefficients used to balance redox reactions.

IV. Sample ProblemIV. Sample Problem

• Calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C.

3Pb2+(aq) + 2Cr(s) 3Pb(s) + 2Cr3+

(aq)

IV. Predicting Spontaneous IV. Predicting Spontaneous Redox ReactionsRedox Reactions

• Positive cell potentials are spontaneous.• We can identify the half-cells in a

reaction, look up their potentials, and sum to find overall cell potentials.

• Thus spontaneity can be determined for any redox reaction.

IV. Sample ProblemIV. Sample Problem

• Will the following redox reaction be spontaneous under standard conditions?

a) Zn(s) + Ni2+(aq) Zn2+

(aq) + Ni(s)

b) Zn(s) + Ca2+(aq) Zn2+

(aq) + Ca(s)

IV. Sample ProblemIV. Sample Problem

• Determine whether the following metals dissolve in HCl(aq), HNO3(aq), both, or neither.

a) Fe(s)

b) Au(s)

c) Ag(s)