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Chapter 18.1-18.5

CHEM 001A Chapter 18.1-18.5: Lecture 30 prereading.docx

Course Code
Arnold Yuan

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Lecture 30: Got Electrons?
18.1-18.5 pg. 814-835
18.1 Pulling the Plug on the Power Grid
- connect to power plant
- small fuel-cell power plants, use heat to heat water, house
- 2H2 (g) + O2 (g)  2H 2O (l)
- (redox)
- hydrogen and oxygen are separated, so electrons must travel through wire to get
from hydrogen to oxygen
18.2 Balancing Oxidation-Reduction Equations
oxidation is loss, reduction is gain
balancing – use half-reaction method
- one for oxidation, one for reduction, then add
acidic solution and basic solution
(see examples)
1. oxidation states
2. separate half reactions
3. balance half reactions – elements other than H and O, O (H2O), then H (H+)
4. Balance charge (e-)
5. Multiply half reactions by appropriate coefficient
6. Add
- neutralize acid with OH-
18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical
Electrical current – flow of electric charge – electrons through wire or ions flowing
through solution
Redox reactions transfer electrons  electrical current
e.g. separate zinc atoms and copper ions, force electron transfer to happen through a wire
electric current
electrochemical cell
voltaic (galvanic) cell – electrochemical cell that produces electrical current from a
spontaneous chemical reaction
electrolytic cell – electrochemical cell that consumes electrical current to drive a
nonspontaneous reaction
half-cell – solid strip of zinc in Zn(NO3)2 solution; solid strip of copper in Cu(NO3)2
electrodes – conductive surfaces through which electrons can enter or leave the half-cells
Zn has greater tendency to ionize than copper – equilibrium of solubility reaction lies
further to right – more negatively charged
Electrons spontaneously flow from Zn electrode to Cu electrode (- to +)
Zn/Zn2+ shifts to right, Cu/Cu2+ shifts to the left
Rate of electrons flowing through wire = amperes (A) = 1C/s
- 1 electron = 1.602 x 10-19 C
- 1A = 6.242 x 1018 C/s
electrical current driven by a difference in potential energy
potential differencevolt (V) – difference in potential energy per unit charge (1 V = 1
large potential difference = large difference in “charge/potential energy between
electromotive force (emf) – force that moves electrons (potential difference)
cell potential (Ecell) cell emf – potential difference between two electrodes, depends on
tendencies of reactants to redox
strong tendency to oxidize + strong tendency to reduce = large Ecell
- also depends on concentrations and temperature
standard cell potential (E cell) standard emf - at 1 M (think Le Chatelier’s – if
concentration changes, then equilibrium shifts, so anode/cathode becomes more
negative cell potential means forward reaction is not spontaneous
anode – oxidation occurs (negative)
cathode – reduction occurs (positive)
buildup of positive charge in anode solution, negative charge in cathode solution  stops
electron flow
salt bridge – counterions flow between half cells to neutralize accumulation of positive
and negative charge
electrochemical cell notation
Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu(s)
- oxidation, then reduction, double line represents salt bridge
- single line separates different phases
if half reaction doesnt have a solid phase – use Pt(s)
Fe(s) | Fe2+ || MnO4-, H+, Mn2+| Pt(s)
18.4 Standard Electrode Potentials
Ecell depends on half reactions
Electrode in each half-cell has individual potential – standard electrode potential
Ecell is difference between two standard electrode potentials
Electrons flow from electrode with more negative charge to electrode with more positive
charge (more negative E to more positive E)
More negative E = higher potential energy (get rid of electrons more)
More positive E = lower potential energy (accept electrons more)
“0” potential is defined as standard hydrogen electrode (SHE) – inert platinum
electrode in 1M HCl with hydrogen gas bubbling through solution
2H+ + 2e- -> H2 (g) E cathode = 0.00 V
Ecathode - E anode = E cell
Connect SHE to electrode in another half-cell of interest
Determine half cell electrode potential using voltmeter
- More negative potential – then spontaneous flow from anode to cathode, H2
formed (E cell < 0)
- More positive potential than SHE – then nonspontaneous (E cell > 0)
o(assuming no hydrogen gas? Or else wouldn’t the cathode become the
anode, etc.?)
Convention – electrode potentials written for reduction half reactions
- electrode potential of SHE is 0
- electrode in half-cell with a greater tendency to undergo reduction is positively
charged relative to the SHE and has positive E
- electrode with lesser tendency to undergo reduction is negatively charged relative
to the SHE and has negative E
- cell potential of any electrochemical cell (E cell) is the difference between the
electrode potentials of the cathode and anode
- Ecell is positive for spontaneous reactions and negative for nonspontaneous
More negative half reaction loses electrons (oxidation), more positive half reaction gains
electrons (reduction)
Positive half reactions on reduction table accept electrons (forward direction); negative
half reactions reject electrons (reverse direction)
- half-reaction with more positive electrode potential attracts electrons more
strongly, undergoes reduction
- half-reaction with the more negative electrode potential repels electrons and
undergoes oxidation
- Any reduction reaction is spontaneous when paired with the reverse of the
reaction listed below it (more negative E)
Metals listed below reduction H+ to H2 will dissolve in acids
18.5 Cell Potential, Free Energy, and the Equilibrium Constant
Ecell and DG must be related, E cell and K must be related
- DG < 0, E cell > 0, K > 1
- DG > 0, E cell < 0, K < 1
potential difference – difference of potential energy per unit charge
E = potential energy difference (maximum work, wmax)/charge (q)
Wmax = -qE cell (negative because work done by system on surroundings is negative)
Faraday’s constant (F) – charge in coulombs of 1 mol of electrons (96,485 C/mol e)
Q = nF
Total charge = number of moles of electrons from balanced equation * F
Wmax = -nFE cell
DG = -nFE cell
-nFEcell = -RT ln K
Ecell = RT/nF (ln K)