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°red. Note 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
*****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-).*****
