Galvanic (or Voltaic) Cells Electrochemistry = the interchange of chemical and electrical energy = used constantly in batteries, chemical instruments, etc… I. Galvanic Cells A. Definitions 1) Redox Reaction = oxidation/reduction reaction = chemical reaction in which electrons are transferred from a reducing agent (which gets oxidized) to an oxidizing agent (which gets reduced) 2) Oxidation = loss of electron(s) to become more positively charged 3) Reduction = gain of electron(s) to become more negatively charged B. Using Redox Reactions to generate electric current (moving electrons) 1) 8H + (aq) + MnO 4 - (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O(l) a) Fe 2+ is oxidized and MnO 4 - is reduced b) Half Reaction = oxidation or reduction process only Reduction: 8H + + MnO 4 - + 5e - Mn 2+ + 4H 2 O Oxidation: 5(Fe 2+ Fe 3+ + e-) Sum = Redox Rxn
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Galvanic (or Voltaic) Cells Electrochemistry = the interchange of chemical and electrical energy = used constantly in batteries, chemical instruments,
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Galvanic (or Voltaic) Cells
Electrochemistry = the interchange of chemical and electrical energy
= used constantly in batteries, chemical instruments, etc…
I. Galvanic CellsA. Definitions
1) Redox Reaction = oxidation/reduction reaction = chemical reaction in which electrons are transferred from a reducing agent (which gets oxidized) to an oxidizing agent (which gets reduced)
2) Oxidation = loss of electron(s) to become more positively charged
3) Reduction = gain of electron(s) to become more negatively charged
B. Using Redox Reactions to generate electric current (moving electrons)
b) Half Reaction = oxidation or reduction process only
Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O
Oxidation: 5(Fe2+ Fe3+ + e-)
Sum = Redox Rxn
2) In solution:
a) Fe2+ and MnO4- collide and electrons are transferred
b) No work can be obtained; only heat is generated
3) In separate compartments, electrons must go through a wire = Galvanic Cell
a) Generates a current = moving electrons from Fe2+ side to MnO4- side
b) Current can produce work in a motor
c) Salt Bridge = allows ion flow without mixing solutions (Jello-like matrix)
d) Chemical reactions occur at the Electrodes = conducting solid dipped into the solution
i) Anode = electrode where oxidation occurs (production of e-)
ii) Cathode = electrode where reduction occurs (using up e-)
C. Cell Potential
1) Think of the Galvanic Cell as an oxidizing agent “pulling” electrons off of the reducing agent. The “pull” = Cell Potential
a) cell = Cell Potential = Electromotive Force = emf
b) Units for cell = Volt = V 1 V = 1 Joule/1 Coulomb
2) Voltmeter = instrument drawing current through a known resistance to find V
Potentiometer = voltmeter that doesn’t effect V by measuring it
II. Standard Reduction PotentialsA. Standard Hydrogen Electrode
1) When measuring a value, you must have a standard to compare it to
2) Cathode = Pt electrode in 1 M H+ and 1 atm of H2(g)
Half Reaction: 2H+ + 2e- H2(g) 1/2 = 0
3) We will use this cathode to find cell of other Half Reactions
4) Standard Reduction Potentials can be found in your text appendicesa) Always given as a reduction processb) All solutes are 1M, gases = 1 atm
5) Combining Half Reactions to find Cell Potentials
a) Reverse one of the half reactions to an oxidation; this reverses the sign of 1/2 b) Don’t need to multiply for coefficients = Intensive Property (color, flavor)c) Example: 2Fe3+(aq) + Cuo 2Fe2+(aq) + Cu2+(aq)
i. Fe3+ + e- Fe2+ 1/2 = +0.77 V
ii. Cu2+ + 2e- Cuo 1/2 = +0.34 V
iii. Reverse of (ii) added to (i) = -0.34 V + +0.77 V = +0.43 V = cell
B. Direction of electron flow in a cell
1) Cell always runs in a direction to produce a positive cell
2) Fe2+ + 2e- Feo 1/2 = -0.44 V
MnO4- + 5e- + 8H+ Mn2+ + 4H2O 1/2 = +1.51 V
3) We put the cell together to get a positive potential
e) As the reaction proceeds, cell 0 (Q K) = Dead Battery!
f) Calculating K:
lnQnF
RT - ε ε o
logQn
0.0592 - ε ε o
V 47.0log186
0.0592 - V 0.48 εcell
0.0592
nεlogKlogK
n
0.0592 - ε 0
oo
IV. Electrolysis = using electric energy to produce chemical change (opposite of cell)A. Example
1) Consider the Cu/Zn Galvanic Cell
a) Anode: Zn Zn2+ + 2e-
b) Cathode: Cu2+ + 2e- Cu ocell = +1.10 V
2) If we attach a power source of o > +1.10 V, we can force e- to go the other way
a) Anode: Cu Cu2+ + 2e-
b) Cathode: Zn2+ + 2e- Zn
c) Called an Electrolytic Cell
B. Calculations with Electrolytic Cells1) How much Chemical Change? Is usually the question.2) Find mass Cuo plated out passing 10 amps (10 C/s) through Cu2+ solution 30 min.
a) Cu2+ + 2e- Cuo(s) b) Steps: current/time, charge (C), moles e-, moles Cu, grams Cu
3) Example: How long must a current of 5.00 amps be applied to a Ag+ solution to produce 10.5 g of silver metal?