Monday, June 22, 2020

SCH 4U - Ecell

Standard Reduction Potentials, E°red
There are tables of standard reduction potentials, E°red, created by chemists.  All the reactions listed in the table are reversible.  As written, you have reduction half-reactions.  Flip the equation and you have oxidation half-reactions, which require the E° value to have a change of sign.  

Some extra half-reactions that aren't on the table, but might prove useful:

Cr2+(aq) + 2e →  Cr(s)    Eored = -0.91 V

Au3+(aq) + 3e → Au(s)    Eored = +1.50 V

SO42-(aq) + H2O(l) + 2e    SO32-(aq) + 2OH-(aq)    Eored = -0.93 V




Oxidizing & Reducing Agents
The more positive the E° for a ½-reaction, the more likely the reaction will occur as written.  A negative E° indicates that the species is more difficult to reduce than H+(aq).

From the table, 
F2(g)  +  2e-    2F-(aq)          E°red = 2.87 V
Li+(aq)  +  e-    Li(s)              E°red = -3.05 V

So F2(g) is the most easily reduced (strongest oxidizing agent) and Li+(aq) is the most difficult to reduce (poorest oxidizing agent).

Example  Which is the strongest oxidizing agent, MnO4- (in acid solution), I2(s) or Zn2+(aq)?

Answer  Using the table, find and write out the half reaction for each substance and its corresponding E° value.  I have dropped the state symbols here, so that the info would fit on the line - normally, we include the states for each entity.
MnO4-  +  8H+  +  5e-    Mn2+  +  4H2O        E°red = +1.51 V
I2  +  2e-    2I-                                                E°red = +0.536 V
Zn2+  +  2e-    Zn                                           E°red = -0.763 V

Thus, since the MnO4-  has the most positive E°red, it is the most easily reduced so it is the strongest oxidizing agent.



Spontaneity of Redox Reactions
We have seen that voltaic cells use redox reactions that proceed spontaneously.  Any reaction that occurs in a voltaic cell that produces a positive E°cell will be spontaneous.

Example  Are the following reactions spontaneous under standard conditions?
(a) Cu(s)  +  2K+(aq)    Cu2+(aq)  +  2K(s)
(b) Cl2(g)  +  2I-(aq)    2Cl-(aq)  +  I2(s)              

Answer
(a)
  • Split the reaction into two half-reactions – one for oxidation and one for reduction.  Do this by grouping like substances together, maintaining their positions as reactants or products.
Cu(s)    Cu2+(aq) 
K+(aq)    K(s)

  • Add electrons to each half-reaction to balance out the charges.
Cu(s)    Cu2+(aq)  +  2e-
K+(aq)  +  e-    K(s)

  • If the number of electrons in each half reaction does not match, multiply each half reaction by the lowest factor needed to balance.
Cu(s)    Cu2+(aq)  +  2e-
2K+(aq)  +  2e-    2K(s)

  • Look up the half reactions in the table.  If you find the half reaction as written, it is the reduction.  Write down the E°redNote that we DO NOT double the E°red even though the half reaction is doubled.
2K+(aq)  +  2e-    2K(s)        E°red = -2.93 V

  • If the reaction in the table but it is backward, flip it.  Write down the E°ox.  
Cu(s)    Cu2+(aq)  +  2e-       E°ox = -0.34 V

  • Now add together the two half reactions and their corresponding values.  The electrons should cancel out completely.
                  Cu(s)    Cu2+(aq)  +  2e-       E°ox = -0.337 V
    2K+(aq)  +  2e-    2K(s)                       E°red = -2.925 V
______________________________________________
Cu(s)  +  2K+(aq)    Cu2+(aq)  +  2K(s)   E°cell = -3.262 V

If the E°cell is positive, the reaction is spontaneous; if it is negative, the reaction is not spontaneous.
∴ not spontaneous

(b)
     Cl2(g)  +  2e-    2Cl-(aq)                E°red = +1.359 V
               2I-(aq)    I2(s)  +  2e-           E°ox = -0.536 V
__________________________________________
Cl2(g)  +  2I-(aq)    2Cl-(aq)  +  I2(s)   E°cell = +0.823 V
spontaneous


Homework: #10, 11a, 18a

*****TIP: When faced with multiple half-reactions containing the same substance (for instance copper(II)), use the half-reaction that has the solid metal in combination with an ion (for instance, Cu(s)    Cu2+(aq)  +  2e-).*****