chapter 20. calculating oxidation numbers each oxide ion has a charge of -2 7 oxide ions have a...

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Page 1: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Chapter 20

Page 2: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Oxidation Numbers

1. Elements always have oxidation number = 0.2. Column I alkalai metals in compounds always

have oxidation number = +1.3. Column II alkaline earth metals in compounds

always have oxidation number = +2.4. Aluminum and gallium oxidation numbers are +3,

zinc and cadmium are +2, and silver is +1.5. Hydrogen normally has oxidation number = +1 in

compounds except when combined with Colunm Ior Column II elements; then rules 2 and 3 apply.

6. Oxygen normally has oxidation number = -2except in H2O2 (rule 4 has priority), or Column Iand Column II oxides, where rules 2 and 3 apply.

7. To calculate other oxidation numbers pretendoxidation numbers are per atom charges andmake all “charges” on all atoms total up to overallcharge on ion or molecule containing atoms.

Page 3: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Calculating Oxidation Numbers

Each oxide ion has a charge of -2

7 oxide ions have a subtotal charge of -2 x 7 = -14

Since the formula has to be uncharged the 2 manganese ions have to have a +14 subtotal

The +14 subtotal divided evenly over 2 manganese ions gives each manganese +14 / 2 = +7

This compound is manganese(VII) oxide

Work oxidation numbers of Cr and S in Cr2(SO4)3 (Hint: treat SO42- as a single particle)

Page 4: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Oxidation and Reduction

1. In simple chemical reduction-oxidation (redox)reactions one reactant substance contains anatom whose oxidation number increases whenproduct is created. This substance becomesoxidized in the reaction. This substance is calleda reducing agent or reductant because it causesanother substance to become reduced.

2. The other reactant substance contains an atomwhose oxidation number decreases when productis created. This substance becomes reduced inthe reaction. This substance is called anoxidizing agent or oxidant because it causesanother substance to become oxidized.

Page 8: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Balancing Redox Reactions

An alternative to textbook method called the REDOHXmethod is outlined below.

1. Balance Reducing and oxidizing atoms first.2. Update total “charges” on reducing and oxidizing

atoms by multiplying oxidation numbers by allappropriate subscripts and coefficients.

3. Balance Electrons by adjusting coefficients infront of oxidizing and reducing agents.

4. Balance atoms which Don’t fit other categories.5. Balance Oxygens by adding H2O.6. Balance H by adding H+

7. Xtra work only when balancing in base solution.Add OH- to both sides to destroy H+ and theneliminate redundant H2O.

Page 9: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 14: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 15: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 16: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 17: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 18: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Standard Cell Potentials

A standard reduction potential (previous slide) is astoichiometry-independent potential energy differencemeasured in voltage units associated with a reductionreaction done under standard conditions.

An oxidation reaction is simply the reverse of areduction reaction and therefore has the same voltagewith the sign reversed as the voltage of the (opposite)reduction reaction found in a reduction potential table.

The total cell potential for a redox reaction done understandard conditions is sum of the voltages (energies)for the oxidation part plus the reduction part.

Remember to change the sign of voltage for oxidationreaction before adding voltage for reduction reaction.

Page 19: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Relationship Between E° and ΔG°

To convert an an energy in voltage units into a Gibbsfree energy three things need to be done:

1. Voltages are stoichiometry-independent (ΔG° isnot). Need to balance redox reaction to figure outhow many electrons involved in redox reaction(“e”) and multiply by this number to fix this.

2. Sign convention of voltages are opposite that ofΔG°. Need to change sign of energy.

3. Voltages based on coulomb quantities and ΔG°based on mole quantities. Need to multiply byFaraday’s constant (F = 96,500 C/mol) to fix this.

In Summary: ΔG° = -eFE°

Page 20: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Nonstandard E ValuesNernst Equation: Analogous to nonstandard ∆G equation

∆G = ∆Go + RTlnQ

E = Eo - (RT/eF)lnQ

• Notice effect of opposite sign convention on direction of deviation from standard value

• Notice RT (kJ/mol) becomes RT/eF (J/coul)

• R = 0.008314 kJ/mol-K (∆G) vs. R = 8.314 J/mol-K (E)

Page 21: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Nonstandard E Values

Batteries in real world seldom have standard conc’s ofall redox reactants and products. To figure out ifvoltage higher or lower than calculated for standardconditions use Le Châtelier’s principle to decide ifreaction more or less spontaneous than standard andinterpret this in terms of more or less positive E value.

Problem 20.58:

Al(s) + 3 Ag+(aq) → Al3+(aq) + 3 Ag(s)

(a) Dilute anode cell (b) Increase Al(s)(c) Increase AgNO3 vol. (d) Add HCl to AgNO3 cell

Page 22: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 23: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 24: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 25: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 26: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has
Page 27: Chapter 20. Calculating Oxidation Numbers Each oxide ion has a charge of -2 7 oxide ions have a subtotal charge of -2 x 7 = -14 Since the formula has

Faraday’s Law

To convert between number of moles of electrolysisproduct (n) and amps of current (I), or t (time inseconds), or coulombs of charge (C = It), or number ofelectrons involved in redox reaction (“e”) Faraday’slaw is used:

n = It/(eF)

Problem 20.86:

(a) What mass of Mg formed by passsing 5.25 A thrumolten MgCl2 for 2.50 days?

(b) How many minutes to make 10.00 g of Mg frommolten MgCl2 using I = 3.50 A?