Galvanic Nernst

Electrochemistry The study of the interchange of chemical and electrical energy. Sample electrochemical processes: 1) Corrosion • 4 Fe(s) + 3 O2(g) ⇌ 2 Fe2O3(s)

2) Biological processes • C6H12O6 + 6 O2 ⇌ 6 CO2 + 6 H2O

3) Batteries (Galvanic or Voltaic cells) • Electrochemical cells that produce a current (flow of electrons) as a result of a redox reaction

4) Electrolytic cells • Electrical energy is used to produce chemical change • Used to prepare or purify metals (such as sodium, aluminum,

Chemical Change → Electron Flow • Copper: Cu(s), Cu2+(aq) Cu(s) → Cu2+(aq) + 2eΔG°rxn = ΔG°f(Cu2+) = 65.6 kJ

• Silver: Ag(s), Ag+(aq)

Ag(s)

Ag(s) → Ag+(aq) + e-

Cu2+ in solution

ΔG°rxn = ΔG°f(Ag+) = 77.2 kJ

2(Ag+(aq) + e- → Ag(s) ) Cu(s) → Cu2+(aq) + 2e-

ΔG° = 2( -77.2 kJ) ΔG° =

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

+65.6 kJ ΔG° = -88.8 kJ Spontaneous wmax = -88.8 kJ

Harnessing the Energy • Separate the half-reactions – Creates a galvanic or voltaic cell

Luigi Galvani

Alessandro Volta

e-

salt bridge

−

“anode” and “oxidation” begin with vowels

+

NO3K+ KNO3(aq)

(produces electrons)

Anode

Red Cat (attracts electrons)

Cu

Ag Cu2+

Ag+

SO

NO3

24

-

1 M CuSO4

1 M AgNO3

Cu(s) → Cu2+(aq) + 2eOxidation

Ag+(aq) + e- → Ag(s) Reduction

Cathode “cathode” and “reduction” begin with consonants

Line Notation for Galvanic Cells e-

NO3-

Anode (−)

K+

Cu

Ag Cu2+

Ag+

SO42-

NO3-

1 M CuSO4

1 M AgNO3

Cu(s) → Cu2+(aq) + 2eOxidation

Ag+(aq) + e- → Ag(s) Reduction

Cathode (+)

Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) Anode always on the left

Cathode always on the right

Chemical Change → Electrical Work • Chemical change produces electrical energy • Electrical energy can be used to do work!

ΔG = wmax Electrical work: w = -nFℰ n = # of moles e- transferred F = charge on a mole of eℰ = electrical potential (electromotive force) Cell Potential (ℰ) or Electromotive Force (emf): The driving force pushing the electrons from the anode to the cathode. Units = Volts 1 Volt = 1 joule/coulomb

Standard Reduction Potentials • The cell potential ℰ°cell can be determined from the standard reduction potentials (ℰ°red) for the half-reactions: – Reduction potential = tendency for reduction to happen • Positive ℰ°red ⇒ spontaneous reduction reaction • Negative ℰ°red ⇒ non-spontaneous reduction or spontaneous oxidation (reverse reaction)

– Standard (o) = standard conditions (1 M solutions, 1 atm gases)

Standard Reduction Potentials Half-Reaction F2 + 2e- → 2FAu3+ + 3 e- → Au Ag+ + e- → Ag Cu2+ + 2e- → Cu 2H+ + 2e- → H2 Ni2+ + 2e- → Ni Zn2+ + 2e- → Zn Al3+ + 3e- → Al Li+ + e- → Li

ℰ° (V) 2.87 ℰ° > 0 1.50 Spontaneous reduction 0.80 0.34 0.00 ← ℰ° = 0 (SHE)

-0.23 -0.76 -1.66 -3.05

Standard Hydrogen Electrode ℰ° < 0 Non-Spontaneous reduction Spontaneous oxidation (reverse rxn)

•

Reduction potential = tendency for reduction to happen

•

Standard = standard conditions (1 M solutions, 1 atm gases)

Standard Reduction Potentials Half-Reaction F2 + 2e- → 2F-

ℰ° (V) 2.87

Au3+ + 3 e- → Au Ag+ + e- → Ag Cu2+ + 2e- → Cu 2H+ + 2e- → H2

1.50 0.80 0.34 0.00

Ni → Ni2+ + 2eZn → Zn2+ + 2eAl → Al3+ + 3eLi → Li+ + e-

+0.23 +0.76 +1.66 +3.05

ℰ° > 0 Spontaneous reduction

Spontaneous oxidation

But remember, an oxidation CANNOT happen without a reduction

Standard Reduction Potentials Half-Reaction F2 + 2e- → 2F-

ℰ° (V) 2.87 ←Strongest Oxidizing Agent

Au3+ + 3 e- → Au Ag+ + e- → Ag Cu2+ + 2e- → Cu 2H+ + 2e- → H2

1.50 0.80 0.34 0.00

Ni2+ + 2e- → Ni Zn2+ + 2e- → Zn Al3+ + 3e- → Al Li+ + e- → Li

-0.23 -0.76 -1.66 -3.05

(most easily reduced)

←Strongest Reducing Agent (most easily oxidized)

•

Reduction potential = tendency

for reduction to happen

•

Standard = standard conditions (1 M solutions, 1 atm gases)

Cell Potential ℰ°cell = ℰ°reduction + ℰ°oxidation Ag+(aq) + e- → Ag(s) ℰ° = 0.80 V Cu2+(aq) + 2 e- → Cu(s)

ℰ° = 0.34 V

Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) Reduction reaction: 2(Ag+(aq) + e- → Ag(s))

ℰ° = +0.80 V

Oxidation reaction: Cu(s) → Cu2+(aq) + 2 e-

ℰ° = - 0.34 V

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

ℰ°cell = +0.46 V

The ℰ°cell MUST be + and thus spontaneous for Galvanic cells

ℰ° is intensive, unlike ΔGo

Free Energy and Cell Potential

∆G° = wmax = −nFℰ° • n = number of moles of electrons transferred • F = Faraday’s constant = 96,485 coulombs per mole of electrons (C/mol e-) • ℰ° = standard cell potential (V or J/C)

Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) ℰ°cell = +0.46 V ΔG° = -nFℰ°cell ΔG° = -(2 mol e-)(96485 C/mol e-)(0.46 V) ΔG° = -88,800 J or -88.8 kJ

Michael Faraday

Practice Time Given the following information, draw a galvanic cell. Fe(s)Fe2+(1 M)Au3+(1 M)Au(s) Be sure to include the following: Anode/Cathode reactions Balanced overall reaction Complete circuit (external wire with e- flow direction, salt bridge) Label all parts of the cell (solution, electrode, etc.)

Fe(s)Fe2+(1 M)Au3+(1 M)Au(s) e-

anions cations

Anode

Fe

Au Fe2+

Cathode

Au3+

1 M Fe2+ Fe(s) → Fe2+(aq) + 2e-

1 M Au3+ Au3+(aq) + 3e- → Au(s)

Oxidation

Reduction

3Fe(s) + 2Au3+(aq) → 3Fe2+(aq) + 2Au(s) ℰ°cell = +0.440V (Fe rxn) + 1.50 V (Au rxn) = 1.94 V

Reaction Quotient • The reaction quotient (Q) sets up a ratio of products and reactants • For a reaction, A + 2B → 3C + 4D [C]3[D]4 Q= [A]1[B]2 Only concentrations (aq) or pressures (g) are used to solve for Q Solids (s) and liquids (l) are not included in the expression

Reaction Quotient practice • Write the Q expression for the following reaction CH4(g) + O2(g) → CO2(g) + H2O(g) Reaction must be balanced first CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) 2 (CO )(H O) 2 2 Q= (CH4)(O2)2

Reaction Quotient practice • Write the Q expression for the following reaction Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) (Cu2+)(Ag)2 Q= (Cu)(Ag+)2 Is this correct? NO: Solids aren’t included in the equation! (Cu2+) Q= (Ag+)2

Non-standard conditions: The Nernst Equation • We can calculate the potential of a cell in which some or all of the components are not in their standard states (not 1 M concentration or 1 atm pressure). ΔG = ΔG° + RT lnQ

ΔG = -nFℰ

ΔG° = -nFℰ°

-nFℰ = -nFℰ° + RT lnQ

RT ℰ = ℰ° lnQ nF Walther Nernst

R = 8.3415 J/mol K T = temperature n = moles of eF = Faraday’s constant 96,485 C/mol e-

Practice with the Nernst Equation • What will be the cell potential ℰ of a Cu/Ag cell using 0.10 M Cu2+ and 1.0 M Ag+ solutions at 25°C? Cu(s)Cu2+ (0.10 M)Ag+ (1.0 M)Ag(s) Cu(s) → Cu2+(aq) + 2e-

Ag+(aq) + e- → Ag(s)

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

[Cu 2 + ] Q= [Ag + ] 2

RT ℰ = ℰ° lnQ nF (8.314 J mol⋅K )(298 K) ln(0.10) ℰ = 0.46 V − C (2)(96485 mol )

ℰ = 0.46 V – (-0.03 V)

ℰ = 0.49 V

Cu

Ag Cu2+

Ag+

SO42-

NO3-

Brain Warmup Half-Reaction Ag+ + e- → Ag Cu2+ + 2e- → Cu Zn2+ + 2e- → Zn Al3+ + 3e- → Al

What is ℰ° for each of the following reactions? Which reaction(s) are spontaneous?

ℰ° (V) 0.80 0.34 -0.76 -1.66

ℰ°

Spontaneous?

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

2.46 V

Y

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

1.10 V

Y

-0.90 V

N

2 Al

3+

(aq)

+ 3 Zn(s) → 2 Al(s) + 3 Zn

2+

(aq)

Zn can reduce Cu2+, but not Al3+