Redox Lecture 2017

with

Dr. Nick What is RedOx Chemistry?

• The defining characteristic of a RedOx reaction is that electron(s) have completely moved from one atom / molecule to another. • The molecule receiving the electrons has been Reduced • The molecule giving the electrons has been Oxidized What is RedOx Chemistry?

OIL RIG LEO says GER Oxidation is Losing Losing Electron Oxidation Reduction is Gaining Gaining Electron Reduction (electrons)

Inorganic Chemistry Chapter 1: Table 5.2

• The standard potential (E˚) is a convenient metric for determining the relative strength of RedOx reactions at 1 M concentrations. • Think of E˚ as, “How easily a reaction will happen.” • Very positive means very easy, very negative means very difficult. • You can change the sign of E˚ by reversing the reaction.

Combining Half-Reactions

I. Write the half-reactions, including E˚ II. Check if the two are a RedOx pair. If not, reverse the more negative half reaction. III. Balance elements other than oxygen and hydrogen using coefficients.

IV. Balance oxygen by adding H2Os to one side. V. Balance hydrogen by adding H+s to one side. Combining Half-Reactions

VI. IF THE SOLUTION IS BASIC, neutralize all H+ by adding OH- to both sides. VII. Balance charges by adding e-. VIII. Multiply the half reactions to balance the electrons (e-). IX. Add half reactions and E˚, cancel terms. X. Final check: make sure all atoms and charges balance! Inorganic Chemistry Chapter 1: Figure 5.2

e- + - 2H + 2e H2 E˚= 0.00

Zn+2 + 2e- Zn E˚= -0.76 + - X- e- 2H + 2e H2 E˚= 0.00 Zn Zn+2 + 2e- E˚= +0.76

Galvanic Cathode

The source of the electrons is the Anode

∆G˚ = -nFE˚ • ∆G: Gibb’s free energy. Negative values are more thermodynamically favorable • n: Represented in the book as ν, is the coefficient in front of the electrons for the balanced RedOx reaction. • F: Faraday’s constant, 96,480 J/mol•V or C/mol. • E˚: Standard potential of the reaction. What we’ve been working with this whole time! Fun with Equations!

+ - 2H + 2e H2 E˚= 0.00 ∆G˚ = -nFE˚ Zn Zn+2 + 2e- E˚= +0.76

+ - +2 - Zn + 2H + 2e Zn + H2 + 2e E˚cell = + 0.76 V

∆G˚ = -(2)(96,480 J/mol•V)(0.76 V)

∆G˚ = -150 kJ/mol Fun with Equations! RT E = E˚ - lnQ cell cell nF NERNST! • The NERNST equation allows you to calculate the potential of a cell at non- standard conditions. • R: Ideal Gas Constant, use 8.314 J/mol•K for these calculations. • Q: Reaction Quotient, concentrations of products over reactants. Fun with Equations! RT E = E˚ - lnQ cell cell nF NERNST!

aA + bB cC + dD

[C]c[D]d Q = [A]a[B]b Fun with Equations! RT E = E˚ - lnQ cell cell nF NERNST!

+2 - +2 - Zn(s) + Sn (aq) + 2e Zn (aq) + Sn(s) + 2e E˚cell = + 0.62 V 0.1 M 0.5 M

0.5 M Q = 0.1 M

(8.314 J/mol•K)(298 K) 0.5 M Ecell = (0.62 V) - ln 2(96,480 J/mol•V) 0.1 M

Ecell = 0.60 V Fun with Equations! 0.05916 E = E˚ - logQ NERNST cell cell n LITE! • A shorthand of the NERNST equation that condenses the constants into a single value. • This assumes a 298 K (25 ˚C) temperature! • The logarithm is now base 10, not natural log! Fun with Equations! nFE˚ lnK = cell RT • If you know the standard potential of the cell, you can calculate the equilibrium constant by rearranging the NERNST equation. • Measuring the potential of a battery is one way of determining the ratio of products to reactants. Disproportionation!

2A A+ + A-

• Disproportionation reactions occur when two identical molecules react to donate / accept electron(s). • These reactions, like all RedOx reactions, can be broken down into half reactions and examined for spontaneity. Disproportionation!

+ +2 2Cu Cu + Cu(s)

+ - Cu + e Cu(s) E˚= +0.52 V Cu+ Cu+2 + e- E˚= -0.16 V

E˚= +0.36 V

SPONTANEOUS!!! Comproportionation!

A+ + A- 2A • Simply the reverse reaction of disproportionation, these occur when two different molecules donate / accept electron(s) to form two identical molecules. • These reactions, like all RedOx reactions, can be broken down into half reactions and examined for spontaneity. Comproportionation!

+2 + Ag + Ag(s) 2 Ag

Ag+2 + e- Ag+ E˚= +1.99 V + - Ag(s) Ag + e E˚= -0.80 V

E˚= +1.19 V

SPONTANEOUS!!! Latimer Diagrams

• Latimer Diagrams provide a quick way to assess the successive reductions of a compound. • Organized with the most oxidized species on the left, each reduction, represented by an arrow, denotes the standard potential associated with the transformation. • Disproportionation reactions are easy to spot in a Latimer diagram. Latimer Diagrams

+0.16 +0.52 +2 +1 Cu Cu Cu(s)

+2 +1.99 + +0.80 Ag Ag Ag(s)

+ 1.40

n1E˚1 + n2E˚2 E˚1 + E˚2 = n1+n2 Inorganic Chemistry Chapter 1: Figure 5.17

Redox Reactions

Redox

Inorganic Chemistry Chapter 1: Box 5.1

Anode

+ - 2H2 4H + 4e

Cathode

+ - 4H + 4e + O2 2H2O Overall

2H2 + O2 2H2O Alternate Sources of Energy

Anode

+ - 2H2 4H + 4e Specific Energy Cathode (MJ/Kg) + - 2H O 4H + 4e + O2 Uranium-2352 83,140,000 Overall Hydrogen 123

2H2 + O2 Gasoline2H2O 46 Propane 46 Fat 37 Platinum is currently the best catalyst for both the productionCoal and 24 oxidation of hydrogen.Carbohydrates 17 TOF = 6,000 s-1 Platinum: $51,076.75 per kg Iron: $0.20 per kg Nickel: $2.94 per kg

http://www.keepbanderabeautiful.org 30 Lead Acid Battery

Overall Reaction: a “comproportionation reaction” Lithium Batteries

• Disposable

• Produce voltages from 1.5 V to about 3.7 V. • (comparable to a zinc–carbon or alkaline battery) Lithium Ion Batteries Lithium Ion Battery Lithium Ion Battery: The Electrochemistry Lithium Ion Batteries: Configurations A zinc/acid battery Daniell Cell. Copper Deposition/Zinc Dissolution

Electrolytic Cell Electroplating

• amp = C/s

• F = C/mol e-

• n = mol e-/mol product

• How many grams of silver will be plated if the cell is run at 5 amps for 1 minute? Electroplating • How many grams of silver will be plated if the cell is run at 5 amps for 1 minute? 5 amps x 60 sec = 300 C 300 C ÷ 96,500 C/mol e- = 0.00311 mol e- - - 0.00311 mol e ÷ 1 mol e /mol Ag(s) = 0.00311 mol Ag(s)

0.00311 mol Ag(s) x 107.87 g/mol =

0.335 grams Water Reduction

- 2 H2O(l) + 2 e- H2(g) + 2OH (aq)

• If the electron source is a metal, the reaction will yield H2 gas and the metal hydroxide. This is why the first group metals are called “alkali metals.”

• https://youtu.be/ofNN1b7xzFw Water Oxidation

+ - 2 H2O(l) O2(g) + 4 H (aq) + 4 e Water Splitting

2 H2O(l) O2(g) + 4 H2(g) Passivation

• A metal can react with O2 to form a metal-oxide layer that is impervious to further oxidation.