Physical chem take home exam

G

Reaction coordinate

Gi

Gf ∆G < 0, spontaneous

Gf ∆G =0 equilibrium

Oxidation–Reduction • Reactions where electrons are transferred from one

atom to another are called oxidation–reduction reactions

• redox reactions for short • Atoms that lose electrons are being oxidized, atoms that

gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction16

Tro: Chemistry: A Molecular Approach, 2/e

Oxidation and Reduction • Oxidation occurs when an atom’s oxidation state

increases during a reaction • Reduction occurs when an atom’s oxidation state

decreases during a reaction

17 Tro: Chemistry: A Molecular Approach, 2/e

CH4 + 2 O2 → CO2 + 2 H2O −4 +1 0 +4 –2 +1 −2

oxidation

reduction

Oxidation–Reduction • Oxidation and reduction must occur simultaneously

• if an atom loses electrons another atom must take them • The reactant that reduces an element in another

reactant is called the reducing agent • the reducing agent contains the element that is oxidized

• The reactant that oxidizes an element in another reactant is called the oxidizing agent

• the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent

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Oxidation & Reduction • Oxidation is the process that occurs when

• the oxidation number of an element increases • an element loses electrons • a compound adds oxygen • a compound loses hydrogen • a half-reaction has electrons as products

• Reduction is the process that occurs when • the oxidation number of an element decreases • an element gains electrons • a compound loses oxygen • a compound gains hydrogen • a half-reaction has electrons as reactants

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Rules for Assigning Oxidation States • Rules are in order of priority 1. free elements have an oxidation state = 0

• Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 2. monatomic ions have an oxidation state equal to their

charge • Na+ = +1 and Cl- = −1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

• Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0

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Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in

a polyatomic ion equals the charge on the ion • N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

• Na = +1 in NaCl

4. (b) Group II metals have an oxidation state of +2 in all their compounds

• Mg = +2 in MgCl2

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Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states

according to the table below • nonmetals higher on the table take priority

Nonmetal Oxidation State Example F −1 CF4 H +1 CH4 O −2 CO2

Group 7A −1 CCl4 Group 6A −2 CS2 Group 5A −3 NH3

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Example: Determine the oxidation states of all the atoms in a propanoate ion, C3H5O2–

• There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b

(C3) + (H5) + (O2) = −1 • Because all the atoms are nonmetals, the next rule we

use is Rule 5, following the elements in order • H = +1 • O = −2

(C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔ Note: unlike charges, oxidation states can

be fractions! 23

Tro: Chemistry: A Molecular Approach, 2/e

Balancing Redox Reactions • Some redox reactions can be balanced by the method

we previously used, but many are hard to balance using that method

• many are written as net ionic equations • many have elements in multiple compounds

• The main principle is that electrons are transferred – so if we can find a method to keep track of the electrons it will allow us to balance the equation

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Half-Reactions • We generally split the redox reaction into two separate

half-reactions – a reaction just involving oxidation or reduction

• the oxidation half-reaction has electrons as products • the reduction half-reaction has electrons as reactants

3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+ 0 −1 +1 −2 −1 +5 −2 +1

oxidation: I− → IO3− + 6 e−

reduction: Cl2 + 2 e− → 2 Cl− 25

Tro: Chemistry: A Molecular Approach, 2/e

Balancing Redox Reactions 1. assign oxidation states

a) determine element oxidized and element reduced 2. write ox. & red. half-reactions, including electrons

a) ox. electrons on right, red. electrons on left of arrow

3. balance half-reactions by mass a) first balance elements other than H and O b) add H2O where need O c) add H+ where need H d) if reaction done in Base, neutralize H+ with OH−

4. balance half-reactions by charge a) balance charge by adjusting electrons

5. balance electrons between half-reactions 6. add half-reactions 7. check by counting atoms and total charge

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Practice – Balance the following equation in acidic solution I– + Cr2O72− → Cr3+ + I2

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Practice – Balance the following equation in acidic solution U3O9 → U2O8-2 + U Xe + Cr2O7-2 → Xe2O5 + Cr+2 C60 + Mo2O10-4 → CO3-2 + Mo

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Spontaneity, Free Energy of Redox Reactions

Oxidation: 6I−(aq) → 3I2(s) + 6e- ∆Goox

Reduction: 4H2O(l)+ 2MnO4−(aq) + 6e--→8OH- (aq)+2MnO2(s) ∆Gored

6I−(aq) +4H2O(l)+ 2MnO4−(aq) -→ 3I2(s) +8OH- (aq)+2MnO2(s)

∆Gorxn=∆Goox+∆Gored

How do we find ∆Goox and ∆Gored?

Standard Reduction Potentials, Eor We can’t look up ∆Goox and ∆Gored , but we determine these from standard reduction potentials, Eor 2H2O(l)+ MnO4−(aq) + 3e--→4OH- (aq)+2MnO2(s) Eor= 1.68V

I2(s) + 2e-) → 2I−(aq) ) Eor= 0.54V

The Units of Eor is V (Volts). 1 V is equivalent to V = J/C where J = joule and C = Coulomb

(Coulomb unit of charge. Charge of an electron is -1.602 x 10-19 C)

Multiplying Voltage by Charge gives energy

∆Gor = - nFEor

1 electron volt, -eV = -e J/C = -1.6 x 10-19 J). The total charge of 1 mole of electrons is given by NA x (-1.6 x 10-19 C) =-96485.3 C = -1 F (F = Faraday).

To convert between Eor and ∆Gor, we simply multiply Eor by the total charge of electrons transferred, -nF (where n = moles of electrons).

2H2O(l)+ MnO4−(aq) + 3e--→4OH- (aq)+2MnO2(s) Eor= 1.68V

∆Gored= -3 mol e-(96485.3 C/mol e-) 1.68 J/C = -486.3 kJ

I2(s) + 2e-) → 2I−(aq) ) Eor= 0.54V

∆Gored= -2 mol e-(96485.3 C/mol e-) 0.54 J/C = -104.2 kJ

∆Goox = − ∆Gored

Spontaneity, Free Energy of Redox Reactions

6I−(aq) → 3I2(s) + 6e- ∆Goox = 3 x 104.2kJ= 312.6 kJ

4H2O(l)+ 2MnO4−(aq) + 6e--→8OH- (aq)+2MnO2(s) ∆Gored = 2*(-486.3 kJ) = -972.6 kJ

6I−(aq) +4H2O(l)+ 2MnO4−(aq) -→ 3I2(s) +8OH- (aq)+2MnO2(s)

∆Gorxn=∆Goox+∆Gored = -660 kJ

Spontaneity, Free Energy of Redox Reactions

∆Gorxn=∆Goox+∆Gored = - nF ( Eor (red step) – Eor(ox. step))

When the # of electrons in oxidation step = # of electrons in reduction step, then we have

∆Gorxn=-6F (Eor(red) – Eor(ox)) = -6F (1.68V-0.54V) = -6F (1.14 V) =-660 kJ

Eo = Eor(red) – Eor(ox)) is overall electrical potential (voltage) of Redox reaction at standard conditions (if Eo > 0, reaction spontaneous)

Tendencies from the Table of Standard Reduction Potentials

• A redox reaction will be spontaneous when there is a strong tendency for the oxidizing agent to be reduced and the reducing agent to be oxidized

• higher on the table of Standard Reduction Potentials = stronger tendency for the reactant to be reduced

• lower on the table of Standard Reduction Potentials = stronger tendency for the product to be oxidized

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Predicting Spontaneity of Redox Reactions

• A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table

• If paired the other way, the reverse reaction is spontaneous

Cu2+(aq) + 2 e− → Cu(s) E°red = +0.34 V Zn2+(aq) + 2 e− → Zn(s) E°red = −0.76 V

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) spontaneous E° = 1.1 V Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s) nonspontaneous E° = -1.1 V

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Practice – Decide whether each of the following will be spontaneous as written or in the reverse direction

Reduction Half-Reaction

F2(g) + 2e−→ 2 F−(aq)

IO3−(aq) + 6 H++ 5e−→ I2(s) + 3H2O(l)

Ag+(aq) + 1e−→ Ag(s)

I2(s) + 2e−→ 2 I−(aq)

Cu2+(aq) + 2e−→ Cu(s)

Cr3+(aq) + 1e−→ Cr2+(aq)

Mg2+(aq) + 2e−→ Mg(s)

F2(g) + 2 I−(aq) → I2(s) + 2 F−(aq)

Mg(s) + 2 Ag+(aq) → Mg2+(aq) + 2 Ag(s)

Cu2+(aq) + 2 I−(aq) → I2(s) + Cu(s)

Cu2+(aq) + 2 Cr2+(aq) → Cu(s) + 2 Cr3+(aq)

spontaneous as written

spontaneous as written

spontaneous as written

spontaneous in reverse

38 Tro: Chemistry: A Molecular Approach, 2/e

Eor

a) F−

b) I−

c) I2

d) Cr3+

Practice – Which of the following materials can be used to oxidize Cu without oxidizing Ag?

Reduction Half-Reaction F2(g) + 2e−→ 2 F−(aq) IO3−(aq) + 6 H++ 5e−→ I2(s) + 3H2O(l) Ag+(aq) + 1e−→ Ag(s) I2(s) + 2e−→ 2 I−(aq) Cu2+(aq) + 2e−→ Cu(s) Cr3+(aq) + 1e−→ Cr2+(aq) Mg2+(aq) + 2e−→ Mg(s)

a) F−

b) I−

c) I2

d) Cr3+

39 Tro: Chemistry: A Molecular Approach, 2/e

Eor

Practice – Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction.

a) Ag

b) Cu

c) Fe

d) Cr

Reduction Half-Reaction Au3+(aq) + 3e−→ Au(s) Ag+(aq) + 1e−→ Ag(s) Cu2+(aq) + 2e−→ Cu(s) 2H+(aq) + 2e−→ H2(g) Fe3+(aq) + 3e−→ Fe(s) Cr3+(aq) + 3e−→ Cr(s) Mg2+(aq) + 2e−→ Mg(s)

40 Tro: Chemistry: A Molecular Approach, 2/e

Eor

Practice – Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction.

Reduction Half-Reaction Au3+(aq) + 3e−→ Au(s) Ag+(aq) + 1e−→ Ag(s) Cu2+(aq) + 2e−→ Cu(s) 2H+(aq) + 2e−→ H2(g) Fe3+(aq) + 3e−→ Fe(s) Cr3+(aq) + 3e−→ Cr(s) Mg2+(aq) + 2e−→ Mg(s)

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From the following: Cu2+(aq) + 2 e− → Cu(s) E°red = +0.34 V Cu1+(aq) + e− → Cu(s) E°red = +0.52 V

Determine Cu2+(aq) + e− → Cu+1(aq) E°red = ?

Use Hess’s Law

From the following: Cu2+(aq) + 2 e− → Cu(s) ∆G° = -2F x 0.34 V Cu(s) → Cu1+(aq) + e− ∆Go = + F x 0.52 V

Cu2+(aq) + e− → Cu+1(aq) ∆Go = - F x 0.16 V = -F E°red

E°red = 0.16 V

Free Energy of Redox Reactions at Q ≠ 1

6I−(aq) +4H2O(l)+ 2MnO4−(aq) -→ 3I2(s) +8OH- (aq)+2MnO2(s)

∆Gorxn=-660 kJ, Eo = 1.14 V at standard conditions (meaning Q =1 when referring to Free energy)

At non-standard conditions, ∆G =∆Go + RT ln(Q)

Nernst Equation for E for redox reaction

At non-standard conditions, ∆G =∆Go + RT ln(Q).

We have that ∆Go = -nF Eo. Take the electrical potential at arbitrary Q, E, to be related to ∆G by ∆G = -nF E.

∆G =∆Go + RT ln(Q).

-nFE =-nFEo + RT ln(Q). Divide both sides by - nF

E =Eo – RT/nF ln(Q). (The Nernst Equation)

Equilibrium and Electrochemical potential

At equilibrium ∆G =∆Go + RT ln(Keq) = 0

E =Eo – RT/nF ln(Keq) = 0

Keq = exp (-∆Go /RT) From Nernst Equation, we have

Keq = exp(Eo nF/ RT)

Equilibrium and Electrochemical potential

Practice – Calculate K for the reaction at 25°C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)

Reduction Half-Reaction E°red, V F2(g) + 2e−→ 2 F−(aq) +2.87 IO3−(aq) + 6 H++ 5e−→ ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e−→ Ag(s) +0.80 I2(s) + 2e−→ 2 I−(aq) +0.54 Cu2+(aq) + 2e−→ Cu(s) +0.34 Cr3+(aq) + 1e−→ Cr2+(aq) −0.50 Mg2+(aq) + 2e−→ Mg(s) −2.37

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Eo = 1.20 V – 0.54 V = 0.66 V Keq = exp(Eo nF/ RT) = e257 = 4.2 x 10111.

∆G and maximum work from a redox reaction

Remember, ∆G =wmax, maximum amount of work that we can get out of a process (∆G <0) or minimum amount of work required to make a process happen (∆G > 0)

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) spontaneous

Zn Cu2+

Zn Cu2+

Zn2+

Cu

No useful work can be obtained.

Electric Current Flowing Directly Between Atoms

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Electric Current Flowing Indirectly Between Atoms

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When the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)

When the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow

Concentration Cell

Cu(s)| Cu2+(aq) (0.010 M) || Cu2+(aq) (2.0 M) | Cu(s)

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E = 0 – 8.314 J/K *298 K/2(96485 C) ln(0.01/2)= 0.068 V