Predicting the Direction of Reactions:
At standard conditions, we can use E^o to predict in which
direction a reaction will go:

Will Fe²⁺ oxidize metallic Hg(l) to Hg₂Cl₂
or will Hg₂Cl₂ oxidize Fe(s) to Fe²⁺ spontaneously? 
We will write the necessary reactions, and then
calculate E^o for the two possible cells:

Given:
Hg₂Cl₂ + 2e^- -----> 2Hg(l) + 2 Cl^-     E^o = +0.27 V
Fe²⁺ + 2e^- -----> Fe(s)                      E^o = -0.44 V
 

There are 2 possibilities:

1)   2Hg(l) + 2 Cl^------>  Hg₂Cl₂  + 2e^-     E^o = -0.27 V (ox)
       Fe²⁺ + 2e^-    ----->  Fe(s)                  E^o = -0.44 V(red)
   Therefore, E^o for this cell = -0.71V (NOT spontaneous!)
^{- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -}

2)  The other possible reaction is:

         Hg₂Cl₂ + 2e^-   ----->   2Hg(l) + 2 Cl^-   E^o = +0.27 V(red)
      Fe(s)  ----->  Fe²⁺ + 2e^-                    E^o = +0.44 V(ox)
   Therefore, E^o for the cell = +0.71V for the spontaneous
   reaction.
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19.4 SPONTANEITY OF REDOX REACTIONS

Relationship of E^o(cell) to DG^o and K:
   DG^o = -RTlnK = -nFE^o(cell)

Therefore, if we know the standard cell potential for a
reaction, we can calculate the equilibrium constant K
for the reaction if we know the T and the number
of electrons transferred.

Calculate the value of K for a reaction in which 4
electrons are transferred, the E^o is measured to be
+0.28V, and the reaction is carried out at 25^oC.
                           RTlnK = nFE^o(cell)
(8.314J/mol-K)(298K)lnK =
                            (4mol e^-)(96500J/V mol e^-)(0.28V)
              2477.6 J(lnK) = 108080 J
                              lnK = 108080/2477.6 = 43.6
                                 K = 8.81 x 10¹⁸   (VERY LARGE!)
                 Therefore, forward reaction is favored!!!

We can also calculate the standard Gibbs free energy:
     DG^o = -RTlnK = -(8.314J/mol K)(298K)(43.6)
     DG^o = -108022 J/mol of reaction = -108 kJ/mol
- - - - - - -  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

ALSO:    Since    DG = -nFE
                           DG = DG^o + RTlnQ
                       -nFE = -nFE^o + RTlnQ
                       E = E^o - [(RT/nF)(lnQ)]   Nernst equation

At equilibrium, E = 0, and Q = K, therefore:  

19.5 THE EFFECT OF CONCENTRATION ON CELL EMF

The Nernst Equation

For the following reaction, predict whether the reaction
is spontaneous or non-spontaneous at the given
conditions:
    Au(s) + Cl₂(g) ----> Au³⁺ + Cl^-1

The Cl₂(g) is 1 atm. and the concentration of AuCl₃ is
1.0 x 10^-3 M; therefore, the concentration of
Cl^-1 = 3.0  x 10^-3 M , and the temperature is 298K.

    2x   [Au(s)  ----> Au³⁺ + 3e^-           E^o= -1.42V
    3x   [Cl₂(g) + 2e^-----> 2 Cl^-1           E^o = +1.36V

Overall reaction: 2Au(s) + 3Cl₂(g) ----> 2Au³⁺ + 6Cl^-1
           E^ototal =  -0.06V   with a 6 electron transfer

E = E^o - [(0.0592/n)(log Q)]
E = -0.06V - [(0.0592/6)(log [Au³⁺]²[Cl^-1]⁶/P_Cl2³)]
E=-0.06V-[(0.0592/6)log(1.0 x 10^-3)²(3.0 x 10^-3)⁶/(1 atm)³]
E = -0.06V - (-0.21V) = +0.15V  Therefore, spontaneous!!
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Concentration cells

The E^ototal =  0.00V 

    E = E^o - [(0.0592 V/n)(log Q)] 
    E = E^o - [(0.0592 V/n)(log [Prods]/[Reacts])]
  Therefore, E = 0.00V - [(0.0592/2)(log [Ni²⁺]/Ni²⁺)]
      E = 0.00V - [(0.0592/2)log(0.10M/0.50M)]
      E = 0.00V - [(0.0592/2)(-0.70)]
      E = 0.00V -(-0.021V) = +0.021V 
     Therefore, spontaneous!!

We can also use the Nernst equation to determine the
concentration of ions in these concentration cells.........
if we know both E and E^o
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19.6 BATTERIES

The Dry Cell Battery

Leclanche (1839-1882):
Cathode: 2NH₄⁺(aq) + 2MnO₂(s) + 2e^- ---->
                                      Mn₂O₃(s) + 2NH₃(aq) + H₂O(l)

Anode (ox):    Zn(s) ----> Zn²⁺(aq) + 2e^-
E(cell) = 1.6 V

The NH₃ produced combines with the Zn²⁺ ions and
forms the soluble complex ion [Zn(NH₃)₄]²⁺.

For an alkaline dry cell: NH₄Cl is replaced by KOH;
approximately 50% more total energy than a common
dry cell of the same size.
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Anode: Zn(Hg) + 2OH^-(aq) ---->  ZnO(s) + H₂O(l) + 2e^-

Cathode: HgO(s) + H₂O(l)+ 2e^- ---->Hg(l) +  2OH^-(aq)
                                                
Overall:  Zn(Hg) + HgO(s) ----> ZnO(s)  + Hg(l)
        E(cell) = 1.35 V

- - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The Lead Storage Battery

Anode:  Pb(s) + SO₄^2-(aq) ---->  PbSO₄(s) + 2e^-
    E(anode) = 0.356 V

Cathode: PbO₂(s) + 4H⁺ (aq) + SO₄^2-(aq) 2e^-   ---->
                                                PbSO₄(s) + 2H₂O(l)
    E(cathode) = 1.70 V

Overall:
 Pb(s) + PbO₂ + 4H⁺(aq)+ 2SO₄^2-(aq)  ---->
                        2PbSO₄(s)+ 2H₂O(l) 
         E(cell) = 2.041 V

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The Nickel-Cadmium (NICAD) cell

    Cd(s) + 2OH^-(aq) ----->  Cd(OH)₂(s) + 2e^-    (an ox)
NiO₂(s) + 2 H₂O(l) + 2e^- ---->Ni(OH)₂(s) + 2OH^-(aq) (red)

Overall:
 Cd(s) + NiO₂(s) + 2 H₂O(l)  ----> Ni(OH)₂(s) + Cd(OH)₂(s)
             E^o(cell) = +1.4V

19.7 CORROSION

Corrosion:  metals are oxidized by oxygen in the presence
of water.
     Fe(s)  ----->  Fe²⁺ + 2e^-               (an ox)
     O₂ + 2H₂O + 4e^-  ----->  4 OH^-      (red cat)