REVIEW:
Common Oxidation States (Periodic Trends)
 Common oxidation states are determined by the position
    of an element in the periodic table.

1- Alkali metal ions (GROUP I) ALL have oxidation number
       +1.
2- Alkaline earth metal ions (GROUP II) ALL have oxidation
      number +2.
   These numbers are the same as the number of valence
    shell electrons.

    In the first row, the maximum oxidation number initially
    increases regularly from left to right: +1, +2, +3, +4, +5.
    However, O and F are -2 and -1, respectively.

    For the second row, the most positive oxidation number
    increases regularly.
    It is possible for Si, P, S and Cl to have positive oxidation
    numbers (usually in oxides, or combined with more
    electronegative elements).

    The first transition metal series have many different
    oxidation numbers.  The maximum oxidation state
    increases regularly to +7 for Mn.

    Look to the name of the compound to know the oxidation
    number of the transition metal.
 

HINT:   IF THE OXIDATION NUMBER INCREASES, THEN THE
    ATOM OR ION HAS BEEN OXIDIZED.   IF IT DECREASES,
    (INCLUDING BECOMING MORE NEGATIVE), THE ATOM
    OR ION HAS BEEN REDUCED.

                    Fe(s) + Cu²⁺(aq) ------> Cu(s) + Fe²⁺(aq)

    are the following:

     Fe(s)  ------>  Fe²⁺(aq) + 2e^-  (oxidation 1/2 reaction)
     Cu²⁺(aq) + 2e^- ------> Cu(s)   (reduction 1/2 reaction)

Here, iron metal Fe(s) = Fe^o is oxidized to Fe²⁺ (loss of e^-)
and Cu²⁺ is reduced to copper metal Cu(s) = Cu^o (gain of e^-).
 

Balancing Oxidation-Reduction Equations:

 Consider the titration of an acidic solution of Na₂C₂O₄
    (sodium oxalate, colorless) with KMnO₄ (deep purple).
 MnO₄^-1 is reduced to Mn²⁺ (pale pink), while C₂O₄^2- is
    oxidized to CO₂.

What is the balanced chemical equation?  Follow these
    steps:

1. Write down the two half-reactions, INCLUDING the
     charges on ions!!
2. Balance each half-reaction:
    a.  First with all elements other than H and O.
    b.  Then balance OXYGEN by adding water to the side
          that needs O.
    c.  Then balance HYDROGEN by adding H⁺ to the side
          that needs H.
    d.  Finish by balancing the charges by adding
          electrons.
3.  Multiply each half reaction to make the number of
        electrons transferred in the 2 half-reactions equal.
4.  Add the reactions and simplify, by eliminating the same
        ions or molecules on each side.
5.  Check! Check! Check! Check! Check! Check! Check!

* For KMnO₄ + Na₂C₂O₄:  both are soluble salts!!!!!!!!WHY?

1.  The two incomplete half-reactions are:
    MnO₄^-(aq) ------> Mn²⁺(aq)     (reduction 1/2 reaction)
    C₂O₄^2-(aq) ------> 2CO₂(g)       (oxidation 1/2 reaction)

For the MnO₄^-(aq) ------> Mn²⁺(aq) reduction 1/2 reaction:

2. a,b:  Adding water yields:
         MnO₄^-(aq) ------> Mn²⁺(aq) + 4H₂O

2. c:  Adding H⁺ yields:
         MnO₄^-(aq) + 8H⁺  ------> Mn²⁺(aq) + 4H₂O

2. d:  There is a total charge of 7+ on the left of the arrow
        and 2+ on the right.  Therefore, 5 electrons (5
        negative charges) need to be added to the left of
        the arrow to have both sides equal at +2:
         5e^- + 8H⁺ + MnO₄^-(aq) ------> Mn²⁺(aq) + 4H₂O

    This half-reaction is BALANCED by mass and by charge.
 
 
For the C₂O₄^2-(aq) ------> 2CO₂(g) reaction:  
            oxidation 1/2 reaction

2. a.b.c:  No need to balance atoms, nor to add H₂O nor H⁺.
        They are equal on both sides.  CHECK THIS!!

2. d:  There is a 2- charge on the left of the arrow and a 0
        charge on the right; so we need to add 2 electrons to
        the right of the arrow to balance the charges:
         C₂O₄^2-(aq) ------> 2CO₂(g) + 2e^-

        Now, the charges are equal at 2- on each side; this
        half-reaction is BALANCED by mass and by charge.
 

3.  To balance the full equation, there are 5 electrons
      transferred for the MnO₄^- equation and 2 electrons
      for the C₂O₄^2- equation; therefore, we need 10
      electrons for each half-reaction to make them equal.

      Multiplying each 1/2 reaction by the proper
            multiplier gives: 

       2x [MnO₄^-(aq) +5e^- + 8H⁺ ------> Mn²⁺(aq) + 4H₂O]
       5x [C₂O₄^2-(aq) ------> 2CO₂(g) + 2e^-]    

  to yield:
   2MnO₄^-(aq) + 10e^- + 16H⁺ ------> 2Mn²⁺(aq) + 8H₂O
                            5C₂O₄^2-(aq) ------> 10CO₂(g) + 10e^-

4.  Adding the 2 half-reactions gives:
    2MnO₄^-(aq) + 16H⁺(aq) + 5C₂O₄^2-(aq) ------>
                                    2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)

5.  All atoms (masses) and charges are now balanced
    (AND NO ELECTRONS ARE PRESENT in the balanced
    equation!!!!!)
 

Another example:  Calculate the number of grams of SO₂
in a sample of air if 7.37 mL of 0.00800 M KMnO₄ solution
are required for the titration:
        [HINT: MnO₄^-1 + SO₂ ----> Mn²⁺ + SO₄^2-]

Balance: MnO₄^-(aq) + 5e^- + 8H⁺ ------> Mn²⁺(aq) + 4H₂O
               SO₂  + 2H₂O  ------> SO₄^2-(aq) + 4H⁺ + 2e^-
These are 2 balanced 1/2 reactions, but with different
numbers of electrons in each.

To make the electrons cancel in each 1/2 reaction, we
must multiply the 1st 1/2 reaction by 2 and the 2nd
1/2 reaction by 5, to yield:

      2MnO₄^-(aq) + 5SO₂ + 2H₂O ------>
                            2Mn²⁺(aq) + 5SO₄^2-(aq) + 4H⁺

Now, we can solve the problem, because we know that 
    2 moles of MnO₄^- react with 5 moles of SO₂

_{V x M = moles} of KMnO_{4 = 7.37 mL x 0.00800 M
           =  0.05896 mmoles of KMnO4}
Therefore,
0.05896 mmoles _KMnO4 (5 mmoles SO₂/2 mmoles _KMnO4)
        = 0.1474 mmoles of SO₂
And, 0.1474 mmoles x 64.06 mg/mmole SO₂ = 9.44 mg
Answer:  0.00944 g SO₂
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

What about basic solutions?  We can't add protons to
balance!

Easiest way:  Balance AS IF you were doing it in acid
    medium!
AFTER you have it all balanced, then add OH^- ions to
each side to neutralize the H⁺ that are there, producing
H₂O.

Example:  Balance ClO^- + CrO₂^-  ---->   CrO₄^2-  + Cl^-
                 in basic medium
 

Do 2.a.b.c:  for CrO₂^- ----> CrO₄^2-  oxidation 1/2-reaction:
          to yield:  CrO₂^- + 2H₂O  ----> CrO₄^2- +  4H⁺

2. d:     CrO₂^-  + 2H₂O  ---->  CrO₄^2- +  4H⁺  + 3e^-

This is a balanced oxidation half-reaction in acidic
    medium!
 

Do 2. a.b.c:  For   ClO^-  ---->   Cl^-reduction half-reaction:
           to yield:   ClO^-  + 2H⁺  ---->   Cl^-  +  H₂O

2. d:              ClO^-  + 2H⁺ + 2e^- ---->   Cl^-  +  H₂O

This is a balanced reduction half-reaction in acidic
    medium!

3.  Balance the number of electrons (in this case,
                6 electrons total):
       2CrO₂^-  + 4H₂O  ---->  2CrO₄^2- +  8H⁺  + 6e^-
       3ClO^-  + 6H⁺ + 6e^- ---->   3Cl^-  +  3H₂O

4.  Balance the 2 half-reactions:

        2CrO₂^- + 4H₂O  + 3ClO^-  ----> 
                               3Cl^-  + 3H₂O + 2CrO₄^2-+ 2H⁺

  This is the balanced equation in acidic medium.

Now, add the proper number of OH^- to counter the H⁺;
    therefore, add 2  OH^- to each side (to neutralize the
    2H⁺on the right), giving:

        2CrO₂^- + 4H₂O  + 3ClO^- + 2OH^- ---->
                               3Cl^-  + 3H₂O + 2CrO₄^2- + 2H₂O

Cancel out the redundancies, and get the balanced
    equation in basic medium:

    2CrO₂^-  + 3ClO^-  + 2OH^- ---->  3Cl^- + 2CrO₄^2- + H₂O

CHECK BOTH THE MASS BALANCE AND THE CHARGE
    BALANCE!
This is the balanced reaction in basic medium!!
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --

19.2 GALVANIC CELLS

VOLTAIC (GALVANIC) CELLS: spontaneous chemical
reactions produce electricity and supply it to an external
circuit.  HERE, THE ENERGY STORED IN THE CHEMICALS
THEMSELVES ARE GIVING US THE POWER!!(Ex.:  battery).

Voltaic cells consist of:
Anode:  Zn(s) ----> Zn²⁺(aq) + 2e^- (oxidation) = an ox
Cathode:  Cu²⁺(aq) + 2e^- ---> Cu(s)(reduction) = red cat

SALT BRIDGE (used to complete the electrical circuit): 
 cations in the SALT BRIDGE move from anode to cathode,
 anions in SALT BRIDGE move from cathode to anode.
The two solid metals are the electrodes (cathode and
        anode).

As oxidation occurs, Zn is converted to Zn²⁺ and 2e^-. The
    electrons flow from the anode to the cathode, where
    they are used in the reduction reaction.  We expect the
    Zn electrode to lose mass and the Cu electrode to gain
    mass as the reaction progresses.

Electrons flow from the anode to the cathode.
Therefore, the anode is negative and the cathode is
        positive.
Electrons cannot flow through the solution; they have to be
    transported through an external wire. (Rule 3.)
Counter anions and cations move through a porous barrier
    or SALT BRIDGE, usually filled with an aqueous solution
    of a salt like KCl.

Counter cations from bridge move into the cathodic
    compartment to neutralize the excess negatively
    charged ions (from Cu²⁺ being depleted).
Cathode:  Cu²⁺ + 2e^- ---->  Cu, so the counterion of Cu²⁺
        is in excess, i.e., Cl^- or NO₃^-; therefore, K⁺ ions move
        from SALT BRIDGE into the solution to maintain
        electroneutrality.

Counter anions move into the anodic compartment to
    neutralize the excess Zn²⁺ ions which are formed by the
    oxidation.
Anode:  Zn  + 2e^- ---->  Zn²⁺, so the Zn²⁺is in excess;
     therefore, Cl^- ions move from SALT BRIDGE into the
     solution to maintain electroneutrality.

The difference in electrical potential between the anode
and the cathode is measured by a voltmeter and is called
the cell voltage, E.
   

SHORTHAND NOTATION FOR THIS CELL:
            Zn/Zn²⁺(1M)//Cu²⁺(1M)/Cu

     (Anode = oxidation) //  (Cathode = reduction)

If you drop Zn metal into a solution of CuSO₄, then the
    Zn(s) will dissolve and the blue of the Cu²⁺ ions will
    disappear.  BUT, no electricity will flow because the 2
    half-reactions are not physically separated; the
    electrons have no place to flow.  You get a reaction,
    but no useful electrical work, like lighting a bulb or
    turning a motor.  When the two 1/2-reactions are
    separated from each other and joined by a SALT
    BRIDGE and a wire, then electromotive force (emf)
    or E can be harnessed and  used.

Can we make the following cell? 
                Zn/Zn²⁺(1M)//Cu²⁺(1M)/Pt
Yes, and the Cu(s) formed in the reaction will plate onto
        the Pt electrode.

Can we make the following cell? 
                Pt/Zn²⁺(1M)//Cu²⁺(1M)/Pt
No, because there is no source of Zn(s), which is necessary
        for oxidation to take place:  Zn  + 2e^- ---->  Zn²⁺.

Can we make the following cell? 
            Cu/Cu²⁺(1M)//Ag¹⁺(1M)/Ag
Yes, and the Ag(s) will plate onto the Ag electrode.

Why does this happen, and can we predict it?
 

Cell EMF:  (electromotive force, or cell voltage)
Electrons flow from the anode to the cathode because
    the cathode has a lower electrical potential energy. 
    The potential difference is measured in Volts.
              1 Volt = 1 J/C = 1 Joule/Coulomb 

E°_cell is the standard cell potential:
            all 1M solutions and 25°C.
 

19.3 Standard Reduction Potentials (E°_red)

Convenient tabulation of electrochemical data.
Standard reduction potentials, E°_red , are measured
    relative to the Standard Hydrogen Electrode (SHE).

The SHE is the cathode.  It consists of a Pt electrode in a
    glass tube placed in 1 M H⁺ solution.  H₂(g) is bubbled
    through the tube to maintain 1 atm pressure.
 For the SHE, we assign:
  2H⁺(aq, 1M) + 2e^- -----> H₂(g)(1 atm) E_red = 0.00 Volts
                                 <-----                         E_ox = 0.00 Volts

The EMF of a cell can be calculated from standard
    reduction potentials:

            E°_cell =  E°(cathode) - E°(anode) 

    where both E°(cathode) and E°(anode) are standard
    reduction potentials of the electrodes.

 Consider Zn(s) ------> Zn²⁺(1M, aq) + 2e^- 
                       oxidation 1/2 reaction (anode)

Therefore, the other half-cell MUST be:
   2H⁺(aq, 1M) + 2e^- ----->  H₂(g, 1 atm)
                       reduction 1/2 reaction (cathode)

Overall:  
 Zn(s) + 2H⁺(aq, 1M) ----> Zn²⁺(1M, aq) +  H₂(g, 1 atm)

When we measure E_cell relative to the SHE (cathode),
we get:

        E°_cell = E°(cathode) - E°(anode)
        0.763 V = 0.00 - E°(anode) = 0.00 - E°(Zn²⁺/Zn)

        And, E°(cathode) = E°(Zn²⁺/Zn)  = -0.763 V

Standard reduction potentials must be written as
    reduction reactions:

    Zn²⁺(aq) + 2e^- ------> Zn(s)          E°_red = -0.763 V.
                                            <------

Since E°_red = -0.763 V, we conclude that the reduction of
    Zn²⁺ in the presence of the SHE is NOT spontaneous.

If this reaction is reversed, (e.g., the electrons are
products rather than reactants), then the sign of the
potential must be reversed!!

The oxidation of Zn(s) to Zn ²⁺(aq):

        Zn(s) -----> Zn²⁺(aq) + 2e^-      E° = +0.763 V.
                          <-----
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Changing the stoichiometric coefficient does not affect
        E°_red.
 Therefore,
            2Zn²⁺(aq) + 4e^- -----> 2Zn(s)  E° = -0.763 V
                                                            <-----

 Reactions with E° > 0 are spontaneous reductions
        relative to the SHE. 

 Reactions with E° < 0 are spontaneous oxidations
        relative to the SHE.

 The larger the difference between E° values, the
        larger E°_cell.

Consider  Cu²⁺(aq) + 2e^- ----> Cu(s) vs. SHE:

Anode(ox): H₂(g, 1 atm) ----> 2H⁺(aq, 1M) + 2e^- 
Cathode(red):  Cu²⁺(aq) + 2e^- ----> Cu(s)

Overall:
        H₂(g, 1 atm) + Cu²⁺(1M) ----> Cu(s) + 2H⁺(1M)

        E°_cell = E°(cathode) - E°(anode)
        0.34 V = E°(Cu²⁺/Cu) - E°(H⁺/H₂) = 0.34 - 0.00 

In a voltaic (galvanic) cell, when the E°_cell is positive,
then the reaction is spontaneous as written, and
vice-versa.
------------------------------------------------------------

For the Daniell cell:

Anode(ox):       Zn(s) ----> Zn²⁺(1M) + 2e^- 
Cathode(red):  Cu²⁺(aq) + 2e^- ----> Cu(s)

Overall: Zn(s) + Cu²⁺(1M) ----> Cu(s) + Zn²⁺(1M)

         E°_cell = E°(cathode) - E°(anode)
         E°_cell = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn)
         E°_cell = 0.34 V - (-0.763 V) = 1.103 V 

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