Electrochemistry

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Electrochemistry

• Chapter 20

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Half-reaction methodremember?

• Example
• Al(s) Cu2(aq) ? Al3(aq) Cu(s)
• Oxidation Al(s) ? Al3(aq) 3e-
• Reduction 2e- Cu2(aq) ? Cu(s)
• Use lowest common multiple to make both
  equivalent in number of electrons
• Oxidation ? multiply by 2
• 2Al(s) ? 2Al3(aq) 6e-
• Reduction ? multiply by 3
• 6e- 3Cu2(aq) ? 3Cu(s)
• Collate (electrons cross out)
• Net reaction
• 2Al(s) 3Cu2(aq) ? 2Al3(aq) 3Cu(s)
• DOES EVERYTHING BALANCE?
• (Make sure to balance after every step!)

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In acidic milieu

• Fe2(aq) MnO4-(aq) ? Fe3(aq) Mn2(aq)
• Oxidation
• Fe2(aq) ? Fe3(aq) e-
• Reduction
• 8H(aq) MnO4-(aq) ? 4H2O(l) Mn2(aq)
• What did I do in the above half-rxn?
• Is it fully balanced?
• 5e- 8H(aq) MnO4-(aq) ? 4H2O(l) Mn2(aq)
• Balance both half-reactions
• 5Fe2(aq) ? 5Fe3(aq) 5e- (multiply by 5 why?)
• 5e- 8H(aq) MnO4-(aq) ? 4H2O(l) Mn2(aq)
• Collate
• Net rxn
• 5Fe2(aq) 8H(aq) MnO4-(aq) ? 5Fe3(aq)
  4H2O(l) Mn2(aq)

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Solve

• VO2(aq) Zn(s) ? VO2(aq) Zn2(aq)

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Answer

• VO2(aq) Zn(s) ? VO2(aq) Zn2(aq)
• Oxidation
• Zn(s) ? Zn2(aq) 2e-
• Reduction
• e- 2H(aq) VO2(aq) ? VO2(aq) H2O(l)
• Balancing both half-reactions
• Zn(s) ? Zn2(aq) 2e-
• 2e- 4H(aq) 2VO2(aq) ? 2VO2(aq) 2H2O(l)
• Collate
• Net reaction
• Zn(s) 4H(aq) 2VO2(aq) ? 2VO2(aq) 2H2O(l)
  Zn2(aq)

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In basic milieu

• I-(aq) MnO4-(aq) ? I2(aq) MnO2(s)
• Oxidation
• I-(aq) ? I2(aq) e-
• 2I-(aq) ? I2(aq) 2e-
• Reduction
• MnO4-(aq) ? 2OH-(aq) MnO2(s)
• 2H(aq) 2OH-(aq) MnO4-(aq) ? 2OH-(aq)
  MnO2(s)
• 2H2O(l) MnO4-(aq) ? 2OH-(aq) MnO2(s)
• 2H2O(l) MnO4-(aq) ? 4OH-(aq) MnO2(s)
• 3e- 2H2O(l) MnO4-(aq) ? 4OH-(aq) MnO2(s)
• Balance both half-reactions
• 6I-(aq) ? 3I2(aq) 6e-
• 6e- 4H2O(l) 2MnO4-(aq) ? 8OH-(aq) 2MnO2(s)
• Collate
• Net rxn
• 6I-(aq) 4H2O(l) 2MnO4-(aq) ? 3I2(aq)
  8OH-(aq) 2MnO2(s)

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Solve

• Al(s) H2O(l) ? Al(OH)4-(aq) H2(g)

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Answer

• Al(s) H2O(l) ? Al(OH)4-(aq) H2(g)
• Oxidation
• Al(s) 4OH-(aq) ? Al(OH)4-(aq) 3e-
• Reduction
• 2e- 2H2O(l) ? 2OH-(aq) H2(g)
• Balance each half-reaction
• 2Al(s) 8OH-(aq) ? 2Al(OH)4-(aq) 6e-
• 6e- 6H2O(l) ? 6OH-(aq) 3H2(g)
• Collate
• Net-reaction
• 2Al(s) 2OH-(aq) 6H2O(l) ? 2Al(OH)4-(aq)
  3H2(g)

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Electricity

• Movt of electrons
• Movt of electrons through wire connecting 2
  half-reactions ? electrochemical cell
• Also called voltaic or galvanic cell
• Cell produces current from spontaneous rxn
• Example copper in solution of AgNO3 is
  spontaneous
• On the other hand, an electrolytic cell uses
  electrical current to drive a non-spontaneous
  chemical rxn

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Voltaic cell

• Solid Zn in zinc ion solution half-cell
• Likewise, Cu/Cu-ion solution
• Wire attached to each solid
• Salt bridge
• 1. contains electrolytes,
• 2. connects 2 half-cells,
• 3. anions flow to neutralize accumulated cations
  at anode and cations flow to neutralize
  accumulated anions at cathode (completes circuit)

• An Ox anode oxidation
• Has negative charge because releases electrons
• Red Cat reduction cathode
• Has positive charge because takes up electrons

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Electrical current

• Measured in amperes (A)
• 1 A 1 C/s
• Coulomb unit of electric charge
• e- 1.602 x 10-19 C
• 1 A 6.242 x 1018 e-/s
• Electric current driven by difference in
  potential energy per unit of charge J/C
• Potential difference (electromotive force or emf)
  volt (V)
• Where 1 V 1 J/C

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More

• In the voltaic cell, potential difference (emf)
  between cathode and anode is referred to as
• Cell potential (Ecell)
• Under standard conditions (1 M, 1 atm, 25C),
  cell potential is
• Standard cell potential Ecell
• Cell potential measure of overall tendency of
  redox rxn to occur spontaneously
• Thus, the higher the Ecell, the greater the
  spontaneity

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Electrochemical notation

• Cu(s)Cu2(aq)Zn2(aq)Zn(s)
• Notation describes voltaic cell
• An ox on left
• Red cat on right
• Separated by double vertical line (salt bridge)
• Single vertical line separates diff phases

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Electrochemical notation

• Some redox rxns reactants products in same
  phase
• Mn doesnt precipitate out ? uses Pt at cathode
• Pt is inert, but provides area for electron
  gain/loss
• Fe(s)Fe2(aq)MnO4-(aq), H(aq), Mn2(aq)Pt(s)
• Write out net reaction

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Answer

• Fe(s)Fe2(aq)MnO4-(aq), H(aq), Mn2(aq)Pt(s)
• Oxidation
• Fe(s) ? Fe2(aq) 2e-
• Reduction
• 5e- MnO4-(aq) 8H(aq) ? Mn2(aq) 4H2O(l)
• Net-reaction
• 5Fe(s) 2MnO4-(aq) 16H(aq) ? 5Fe2(aq)
  2Mn2(aq) 8H2O(l)

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Standard reduction potentials

• One half-cell must have a potential of zero to
  serve as reference
• Standard hydrogen electrode (SHE) half-cell
• Comprises Pt electrode in 1 M HCl w/ H2 bubbling
  at 1 atm
• 2H(aq) 2e- ? H2(g) Ered 0.00 V

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Example

• Throw zinc into 1M HCl
• Zn(s)Zn2(aq)2H(aq)H2(g)
• Ecell Eox Ered 0.76 V
• If Ered 0.00 V (as the reference)
• Then Eox 0.76 V ( oxid of Zn half-rxn)
• Reduction of Zn-ion
• Is -0.76 V (non-spontaneous)

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Problem

• Cr(s)Cr3(aq)Cl-(aq)Cl2(g)
• What is the std cell pot (Ecell) given oxid of
  Cr 0.73 V and Cl red 1.36V?
• Hint standard electrode potentials are intensive
  properties e.g., like density
• Stoichiometry irrelevant!

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Solution

• Ecell Eox Ered 0.73V 1.36V
• 2.09V

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Appendix M, pages A-33-35

• Standard reduction potentials in aqueous solution
  _at_ 25C
• Also, pg. 967, Table 20.1 (gives increasing
  strengths of ox/red agents)
• Lets take a look at it
• Does increasing strengths of ox/red agents make
  sense?
• What happens to oxidizing agent, reducing agent?

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Problem

• Calculate the standard cell potential for the
  following
• Al(s) NO3-(aq) 4H(aq) ? Al3(aq) NO(g)
  2H2O(l)

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Answer

• Oxidation
• Al(s) ? Al3(aq) 3e- Eox 1.66V
• Reduction
• NO3-(aq) 4H(aq) 3e- ? NO(g) 2H2O(l)
• Ered 0.96V
• Ecell Eox Ered 1.66V 0.96V 2.62V

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Predicting the spontaneous direction of a redox
rxn

• Generally, any reduction half-rxn is spontaneous
  when paired w/reverse of half-rxn below it in
  table of standard reduction potentials
• Lets look at table
• Predict the exact value and spontaneity for the
  following
• Fe(s) Mg2(aq) ? Fe2(aq) Mg(s)

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Answers

• Fe(s) Mg2(aq) ? Fe2(aq) Mg(s)
• Oxidation
• Fe(s) ? Fe2(aq) 2e- Eox 0.45V
• Reduction
• Mg2(aq) 2e- ? Mg(s) Ered -2.37V
• Ecell Eox Ered 0.45V -2.37V
• -1.92V
• nonspontaneous

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Will metal X dissolve in acid?

• Metals whose reduction half-rxns lie below
  reduction of proton to hydrogen gas will dissolve
  in acids
• Why?
• Just look at the table!
• Nitric acid is exception
• Lets take a look

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Ecell, ?G, K

• What must the values for Ecell, ?G, K be in
  order to have a spontaneous rxn?
• ?Glt0
• Ecellgt0
• Kgt1
• Product-favored

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Relationship between ?G Ecell

• Faradays Constant (F) 96,485 C/mol e-
• ?G -ne-FEcell
• Problem
• Calculate ?G for
• I2(s) 2Br-(aq) ? 2I-(aq) Br2(l)
• Is it spontaneous?

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Solution its nonspontaneous!

• I2(s) 2Br-(aq) ? 2I-(aq) Br2(l)
• Oxidation
• 2Br-(aq) ? Br2(l) 2e- Eox -1.09V
• Reduction
• I2(s) 2e- ? 2I-(aq) Ered 0.54V
• Ecell -1.09V 0.54V -0.55V

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Problem

• 2Na(s) 2H2O(l) ? H2(g) 2OH-(aq) 2Na(aq)
• Is it spontaneous?

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Solution its spontaneous!

• Oxidation
• 2Na(s) ? 2Na(aq) 2e- Eox 2.71V
• Reduction
• 2H2O(l) 2e- ? H2(g) 2OH-(aq) Ered -0.83V
• Ecell 2.71V -0.83V 1.88V

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Relationship between Ecell K
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Problem

• Calculate K for
• 2Cu(s) 2H(aq) ? Cu2(aq) H2(g)

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Solution is it product-favored?

• Oxidation
• 2Cu(s) ? Cu2(aq) 2e- Eox -0.34V
• Reduction
• 2H(aq) 2e- ? H2(g) Ered 0.00V
• Ecell -0.34V

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Cell potential concentration Nernst Equation

• Concentration ? 1M
• Non-standard conditions
• Under standard conditions, Q 1
• ? Ecell Ecell

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Problem

• Compute the cell potential, given
• Cu(s) ? Cu2(aq, 0.010 M) 2e-
• MnO4-(aq, 2.0 M) 4H(aq, 1.0M) 3e- ? MnO2(s)
  2H2O(l)

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Solution

• Balance the equation!
• Oxidation
• 3Cu(s) ? 3Cu2(aq) 6e- Eox -0.34V
• Reduction
• 2MnO4-(aq) 8H(aq) 6e- ? 2MnO2(s) 4H2O(l)
  Ered 1.68V

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To summarize

• If Qlt1, rxn goes to products
• Ecell gt Ecell
• If Qgt1, rxn goes to reactants
• Ecell lt Ecell
• If Q K, _at_ eq.,
• Ecell 0 ( Ecell 0)
• Explains why all batteries die

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Concentration cells

• Voltaic cells can be constructed from two similar
  half-rxns where difference in concentration
  drives current flow
• Cu(s) Cu2(aq, 2.0M) ? Cu2(aq, 0.010M) Cu(s)
• Ecell 0 since both half-rxns are the same
• However, using Nernst equation, different
  concentrations yield 0.068V
• Lets take a look
• Flow is from lower Cu-ion concentration half-cell
  to higher one
• Down the concentration gradient
• The electrons will flow to the concentrated cell
  where they dilute the Cu-ion concentration
• Results in ? Cu-ion concentration in dilute cell
  ? Cu-ion concentration in concentrated cell

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Batteries

• Dry-cell batteries
• Dont contain large amounts of water
• Anode
• Zn(s) ? Zn2(aq) 2e-
• Cathode
• 2MnO2(s) 2NH4(aq) 2e- ? Mn2O3(s) 2NH3(g)
  H2O(l)
• Cathode is carbon-rod immersed in moist (acidic)
  paste of MnO2 that houses NH4Cl
• 1.5 V

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Batteries

• More common dry-cell type
• Alkaline battery
• Anode
• Zn(s) 2OH-(aq) ? Zn(OH)2(s) 2e-
• Cathode
• 2MnO2(s) 2H2O(l) 2e- ? 2MnO(OH)(s)
  2OH-(aq)
• Longer shelf-life, live longer
• Cathode in basic paste

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Car Batteries

• Lead-acid storage batteries
• 6 electrochemical cells (2V) in series
• Anode
• Pb(s) HSO4-(aq) ? PbSO4(s) H(aq) 2e-
• Cathode
• PbO2(s) HSO4-(aq) 3H(aq) 2e- ? PbSO4(s)
  2H2O(l)
• In 30 soln of sulfuric acid
• If dead due to excess PbSO4 covering electrode
  surfaces
• Re-charge (reverse rxn) ? converts PbSO4 to Pb
  and PbO2

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Rechargeable batteries

• Ni-Cd
• Anode
• Cd(s) 2OH-(aq) ? Cd(OH)2(s) 2e-
• Cathode
• 2NiO(OH)(s) 2H2O(l) 2e- ? 2Ni(OH)2(s)
  2OH-(aq)
• KOH, usually, used
• 1.30 V
• Reverse rxn recharges battery
• Excess recharging ? electrolysis of water
• EXPLOSION!!!
• Muhahahaha!

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Rechargeable batteries

• Since Cd is toxic
• Developed safer alternative
• Ni-MH
• Hybrid car batteries high energy density
• Same cathode rxn as previous
• Anode
• MH(s) OH-(aq) ? M(s) H2O(l) e-
• Commonly, M AB5, where A is rare earth mixture
  of La, Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mn
• Very few use AB2, where A Ti and/or V

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Rechargeable batteries

• Anode made of graphite w/incorporated Li-ions
  between carbon layers
• Ions spontaneously migrate to cathode
• Cathode LiCoO2 or LiMn2O4
• Transition metal reduced
• Used in laptop computers, cell phones, digital
  cameras
• Light weight and high E density

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Fuel cell

• Reactants flow through battery
• Undergo redox rxn
• Generate electricity
• Hydrogen-oxygen fuel cell
• Anode
• 2H2(g) 4OH-(aq) ? 4H2O(l) 4e-
• Cathode
• O2(g) 2H2O(l) 4e- ? 4OH-(aq)
• Used in space-shuttle program
• And Arnolds Hummah

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Electrolysis

• Electrical current used to drive nonspontaneous
  redox rxn
• In electrolytic cells
• Used in
• Electrolysis of water
• Metal plating silver coated on metal, jewelry,
  etc.

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Electrolytic cells using electricity to run a rxn

• Anode is ? gives electrons, connected to
  positive terminal of power source
• Cathode is - ? takes electrons, connected to
  negative terminal of power source
• Opposite scheme of voltaic cell!

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Predicting the products of electrolysis

• Pure molten salts
• Anion oxidized/cation reduced
• Obtain 2Na(s) and Cl2(g) from electrolysis of
  NaCl
• Mixture of cations or anions
• K/Na and Cl-/Br- present
• Look at page 967 compare half-cell potentials
• Cation/anion preferably reduced that has least
  negative, or most positive, half-cell potential

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Example

• Predict the half-rxn occurring at the anode and
  the cathode for electrolysis of
• AlBr3 MgBr2
• Oxidation
• Br-(l) ? Br2(g) 2e- Eox -1.09V
• Bromide will be oxidized at the anode
• Reduction
• Al3(l) 3e- ? Al(s) Ered -1.66V
• Mg2(l) 2e- ? Mg(s) Ered -2.37V
• Reduction of Al will occur at the cathode since
  its potential is greater than Mgs

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Predicting the products of electrolysis

• aqueous solns same as 2
• Water redox might occur simultaneously
• 2H2O(l) ? O2(g) 4H(aq) 4e-
• 2H2O(l) 2e- ? H2(g) 2OH-(aq)
• Eox -0.82 V Ered -0.41 V
• ?Ecell -1.23 V

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Problem

• Given oxidation of I- -0.54 V reduction of
  Li -3.04 V, which, if any, gases would be
  formed and where i.e., at cathode/anode?

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Solution

• Oxidation
• 2I-(aq) ? 2I2(aq) 2e- Eox -0.54V
• 2H2O(l) ? O2(g) 4H(aq) 4e- Eox -0.82V
• I- will be oxidized at the anode
• Reduction
• 2Li(aq) 2e- ? 2Li(s) Ered -3.04 V
• 2H2O(l) 2e- ? H2(g) 2OH-(aq) Ered -0.41 V
  
• Water will be reduced at the cathode

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Stoichiometry of electryolysis

• Can use e- stoichiometric relations to predict
  moles and/or grams of substances
• Remember, unit of current ampere A 1 C
  (magnitude of current)/s (time of current flow)
• Also, F 96,485 C/mole e-

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Problem

• Gold can be plated out of a soln containing the
  Au3 according to
• Au3(aq) 3e- ? Au(s)
• What mass of gold (in grams) will be plated by
  the flow of 5.5 A of current for 25 mins?

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Solution