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Electrochemistry. Chapter 20. Half-reaction method…remember?. Example Al (s) + Cu 2+ (aq)  Al 3+ (aq) + Cu (s) Oxidation: Al (s)  Al 3+ (aq) + 3e - Reduction: 2e - + Cu 2+ (aq)  Cu (s) Use lowest common multiple to make both equivalent in number of electrons

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half reaction method remember
Half-reaction method…remember?
  • 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!)

in acidic milieu
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)

solve
Solve

VO2+(aq) + Zn(s) VO2+(aq) + Zn2+(aq)

answer
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)

in basic milieu
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)

solve1
Solve
  • Al(s) + H2O(l) Al(OH)4-(aq) + H2(g)
answer1
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)

electricity
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 celluses electrical current to drive a non-spontaneous chemical rxn
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

Voltaic cell
electrical current
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
slide12
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, 25°C), cell potential is
  • Standard cell potential = E°cell
  • Cell potential = measure of overall tendency of redox rxn to occur spontaneously
  • Thus, the higher the E°cell, the greater the spontaneity
electrochemical notation
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
electrochemical notation1
Electrochemical notation
  • Some redox rxns reactants & products in same phase
  • Mn doesn’t 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
answer2
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)

standard reduction potentials
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); E°red = 0.00 V
example
Example
  • Throw zinc into 1M HCl
  • Zn(s)|Zn2+(aq)||2H+(aq)|H2(g)
  • E°cell = E°ox + E°red= 0.76 V
  • If E°red = 0.00 V (as the reference)
  • Then E°ox = 0.76 V (= oxid of Zn half-rxn)
  • Reduction of Zn-ion
    • Is = -0.76 V (non-spontaneous)
problem
Problem
  • Cr(s)|Cr3+(aq)||Cl-(aq)|Cl2(g)
  • What is the std cell pot (E°cell) 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!
solution
Solution
  • E°cell = E°ox + E°red = 0.73V + 1.36V

= 2.09V

appendix m pages a 33 35
Appendix M, pages A-33-35
  • Standard reduction potentials in aqueous solution @ 25°C
  • Also, pg. 967, Table 20.1 (gives increasing strengths of ox/red agents)
    • Let’s take a look at it
      • Does increasing strengths of ox/red agents make sense?
      • What happens to oxidizing agent, reducing agent?
problem1
Problem
  • Calculate the standard cell potential for the following:

Al(s) + NO3-(aq) + 4H+(aq) Al3+(aq) + NO(g) + 2H2O(l)

answer3
Answer
  • Oxidation

Al(s) Al3+(aq) + 3e-; E°ox = 1.66V

  • Reduction

NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l);

E°red = 0.96V

  • E°cell = E°ox + E°red =1.66V + 0.96V = 2.62V
predicting the spontaneous direction of a redox rxn
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
  • Let’s look at table
  • Predict the exact value and spontaneity for the following:

Fe(s) + Mg2+(aq) Fe2+(aq) + Mg(s)

answers
Answers

Fe(s) + Mg2+(aq) Fe2+(aq) + Mg(s)

  • Oxidation

Fe(s) Fe2+(aq) + 2e-; E°ox = 0.45V

  • Reduction

Mg2+(aq) + 2e- Mg(s); E°red = -2.37V

  • E°cell = E°ox + E°red =0.45V + -2.37V =

-1.92V

nonspontaneous

will metal x dissolve in acid
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
    • Let’s take a look
e cell g k
E°cell, G°, K
  • What must the values for E°cell, G°, & K be in order to have a spontaneous rxn?
  • G°<0
  • E°cell>0
  • K>1
    • Product-favored
relationship between g e cell
Relationship between G° & E°cell
  • Faraday’s Constant (F) = 96,485 C/mol e-
  • G° =-ne-FE°cell
  • Problem:
  • CalculateG° for

I2(s) + 2Br-(aq) 2I-(aq) + Br2(l)

Is it spontaneous?

solution it s nonspontaneous
Solution: it’s nonspontaneous!

I2(s) + 2Br-(aq) 2I-(aq) + Br2(l)

  • Oxidation

2Br-(aq) Br2(l) + 2e-; E°ox = -1.09V

  • Reduction

I2(s) + 2e- 2I-(aq); E°red = 0.54V

  • E°cell = -1.09V + 0.54V = -0.55V
problem2
Problem

2Na(s) + 2H2O(l) H2(g) + 2OH-(aq) + 2Na+(aq)

Is it spontaneous?

solution it s spontaneous
Solution: it’s spontaneous!
  • Oxidation

2Na(s) 2Na+(aq) + 2e-; E°ox = 2.71V

  • Reduction

2H2O(l) + 2e-  H2(g) + 2OH-(aq) E°red = -0.83V

  • E°cell = 2.71V + -0.83V = 1.88V
problem3
Problem
  • Calculate K for

2Cu(s) + 2H+(aq) Cu2+(aq) + H2(g)

solution is it product favored
Solution: is it product-favored?
  • Oxidation

2Cu(s) Cu2+(aq) + 2e-; E°ox = -0.34V

  • Reduction

2H+(aq)+ 2e- H2(g); E°red = 0.00V

E°cell = -0.34V

cell potential concentration nernst equation
Cell potential & concentration: Nernst Equation
  • Concentration ≠ 1M
    • Non-standard conditions
  • Under standard conditions, Q = 1
  •  Ecell = E°cell
problem4
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)

solution1
Solution
  • Balance the equation!
  • Oxidation

3Cu(s) 3Cu2+(aq)+ 6e-; E°ox = -0.34V

  • Reduction

2MnO4-(aq)+ 8H+(aq) + 6e-  2MnO2(s) + 4H2O(l) ; E°red = 1.68V

to summarize
To summarize
  • If Q<1, rxn goes to products
    • Ecell > E°cell
  • If Q>1, rxn goes to reactants
    • Ecell < E°cell
  • If Q = K, @ eq.,
    • E°cell = 0 (& Ecell = 0)
      • Explains why all batteries die
concentration cells
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)

    • E°cell = 0 since both half-rxns are the same
  • However, using Nernst equation, different concentrations yield 0.068V
    • Let’s 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
batteries
Batteries
  • Dry-cell batteries
    • Don’t 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
batteries1
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
car batteries
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
rechargeable batteries
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!
rechargeable batteries1
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
rechargeable batteries2
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
fuel cell
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 Arnold’s Hummah
electrolysis
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.
electrolytic cells using electricity to run a rxn
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!
predicting the products of electrolysis
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
example1
Example
  • Predict the half-rxn occurring at the anode and the cathode for electrolysis of

AlBr3 & MgBr2

  • Oxidation

Br-(l) Br2(g) + 2e-;E°ox = -1.09V

Bromide will be oxidized at the anode

  • Reduction

Al3+(l) + 3e-  Al(s);E°red = -1.66V

Mg2+(l) + 2e-  Mg(s) ;E°red = -2.37V

Reduction of Al will occur at the cathode since its potential is greater than Mg’s

predicting the products of electrolysis1
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)

E°ox = -0.82 V & E°red = -0.41 V

E°cell = -1.23 V

problem5
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?
solution2
Solution
  • Oxidation

2I-(aq) 2I2(aq) + 2e-; E°ox = -0.54V

2H2O(l) O2(g) + 4H+(aq) + 4e-; E°ox = -0.82V

I- will be oxidized at the anode

  • Reduction

2Li+(aq) + 2e-  2Li(s); E°red = -3.04 V

2H2O(l) + 2e-  H2(g) + 2OH-(aq); E°red = -0.41 V

Water will be reduced at the cathode

stoichiometry of electryolysis
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-
problem6
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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