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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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Electrochemistry

Electrochemistry

Chapter 20


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


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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