Oxidation and Reduction Reactions

Description

Predict what might happen when a piece of copper wire in a solution of 2% AgNO₃.
If you try this experiment, you will initially see that the copper is a shiny copper color and the solution is clear. In less than one hour the solution is light blue and the wire is covered with shiny silver needles. What happened?
Copper metal became copper ions in solution and silver ions became silver metal.
Cu_(s) + Ag⁺_(aq) → Cu²⁺_(aq) + Ag_(s)       (unbalanced)
The Cu_(s) loses electrons to become Cu²⁺_(aq) ions and the Ag⁺_(aq) ions gain electrons to become Ag_(s).
Reactions that involve the exchange of electrons are called reduction and oxidation (redox) reactions. When a chemical species loses electrons we say that it is oxidized, and when a chemical species gains electrons we say that it is reduced.
The Cu_(s) loses electrons to be oxidized to Cu²⁺_(aq).
The Ag⁺_(aq) gain electrons to be reduced to Ag_(s).
We can break the reaction into the following two half reactions.
Cu_(s) →Cu²⁺_(aq) + 2e^-
Ag⁺_(aq) + e^- →Ag_(s)
What would you predict if you placed a piece of Ag metal in a solution of Cu²⁺?
Since we observed that the reaction of Ag⁺ and Cu is spontaneous, we would not expect the reverse reaction to be spontaneous. So no reaction occurs between Ag metal and Cu²⁺.

Balancing Redox Reactions

Balancing Half-Reactions

1. Balance the atoms being oxidized or reduced.
2. Balance oxidation numbers by adding electrons.
3. Balance charge by adding H⁺ (acidic solutions) or OH^- (basic solutions).
4. Balance H atoms by adding H₂O.

Balancing Redox Reactions

1. Write the half reactions.
2. Balance each half reaction.
3. Combine the half reactions to eliminate the electrons from the overall reaction.

Example

Fe²⁺_(aq) + MnO₄^-_(aq) →Fe³⁺_(aq) + Mn²⁺_(aq) (acidic solution)
1. Write the half reactions:
Fe²⁺_(aq)→ Fe³⁺_(aq) (oxidation)
MnO₄^-_(aq) →Mn²⁺_(aq) (reduction)
2. Balance each half reaction:
Fe²⁺_(aq) →Fe³⁺_(aq) + e^-
MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- →Mn²⁺_(aq) + 4H₂O
3. Combine the half reactions to give the overall reaction:

                    5Fe2+(aq) → 5Fe3+(aq) + 5e-

     MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O

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

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O

Try balancing the lead-acid battery reaction:
Pb_(s) + PbO_{2 (s)} + SO₄^2-_(aq) →PbSO_4(s)

Standard Reduction Potentials

Description

Consider again the spontaneous reaction of Zn metal in a solution of Cu²⁺.
    Zn_(s) + Cu²⁺_(aq)→  Zn²⁺_(aq) + Cu_(s)
Without doing the experiment, could we have predicted that this reaction would be spontaneous as written?
We can calculate ∆ G^o, or we can calculate a related quantity, E^o.
First write the half-reactions as reductions:
Cu²⁺_(aq) + 2e^- → Cu_(s)     E^o_red = +0.340 V
Zn²⁺_(aq) + 2e^- →Zn_(s)     E^o_red = -0.763 V
E^o_red are the standard reduction potentials for these two half-reactions, which are taken from tables of data.
The half-reaction with the more positive E^o_red will occur as written (as a reduction).
So for the Cu²⁺ and Zn²⁺ half-reactions, the Cu²⁺ E^o (+0.340 V) is more positive than the Zn²⁺ E^o (-0.763 V), so the Cu²⁺ will be reduced, and Zn metal will be oxidized.
To find the voltage produced by an electrochemical cell we simply sum all of the potentials in the circuit.
    E^o_cell = ∆ E^o_{half-reactions}
For simple cells:
E^o_cell = E^o_red + E^o_ox
The E^o of an oxidation half-reaction is the negative of the E^o of the reduction half-reaction.
For the Cu and Zn example: E^o_cell = 0.340 V + -(-0.763 V) = 1.103 V
When balancing reactions we sometimes multiply one or both half reactions by a factor so the electrons cancel. The E^o values do NOT get multiplied by these factors.

---

E^o_red Values

E^o_red is the standard reduction potential and is measured relative to:
    2H⁺_(aq) + 2e^- →H_2(g)
which is assigned the value E^o_red = 0.00 V
A positive E^o_red means that a half-reaction will go in the direction indicated (reduction) when paired with the hydrogen half-reaction:
Cu²⁺_(aq) + 2e^- →Cu_(s)     E^o_red = +0.340 V
H_2(g) →2H⁺_(aq) + 2e^-     E^o_red = 0.00 V
A negative E^o_red means that a half-reaction will go in the opposite direction indicated (oxidation) when paired with the hydrogen half-reaction:
2H⁺_(aq) + 2e^- →H_2(g)     E^o_red = 0.00 V
Zn_(s)→ Zn²⁺_(aq) + 2e^-     E^o_red = -0.763 V
In general for any two half-reactions, the half-reaction with the more positive E^o_red will proceed as a reduction, and the other half-reaction will proceed as an oxidation.
Note that for a reaction to occur the appropriate species must be present. One species must be reduced and another must be oxidized.
The species that is reduced is called an oxidizing agent and the species that is oxidized is called a reducing agent.

---

Example

Find a chemical species that will convert Ag⁺_(aq) to Ag_(s) without converting Cu⁺_(aq) to Cu_(s).
First determine what type of reaction occurs. The conversion of Ag⁺_(aq) to Ag_(s) is a reduction (the silver oxidation state goes from +1 to 0).
The half-reaction is:
    Ag⁺_(aq) + e^- →Ag_(s)     E^o_red = 0.799 V
First, do we need a half-reaction with a more positive or more negative E^o_red value than the silver half reaction? Since converting Ag⁺_(aq) to Ag_(s) is a reduction, we want a half-reaction that will go as an oxidation. That means we want a half reaction with a more negative E^o_red value than the E^o_red for silver.
Looking in a table of standard reduction potentials, all half reactions with E^o_red values more negative than 0.799 V will reduce Ag⁺_(aq) to Ag_(s).
We don't want to reduce Cu⁺_(aq) to Cu_(s). The half-reaction is:
    Cu⁺_(aq) + e^-→ Cu_(s)     E^o_red = 0.518 V
So we want a half reaction that has an E^o_red value less than 0.799 V but not less than or equal to 0.518 V.
Some choices are:
I_{2 (s)} + 2e^- →I^-_(aq)     E^o_red = 0.534 V
Fe³⁺_(aq) + e^- →Fe²⁺_(aq)     E^o_red = 0.769 V
Hg₂²⁺_(aq)(aq) + 2e^- →2Hg_(l)     E^o_red = 0.796 V
The Ag⁺_(aq) must react with something that can be oxidized. Species that can be oxidized are on the right side of the equation for reduction half reactions. Our choices are therefore I^-_(aq), Fe²⁺_(s), and Hg_(l).

Spontaneous Reactions

Redux Reaction Example

Demonstration: A piece of copper wire in a solution of 2% AgNO₃.
Initially the copper is a shiny copper color and the solution is clear. In less than one hour the solution is light blue and the wire is covered with shiny silver needles. What happened?
Copper metal became copper ions in solution and silver ions became silver metal.
Cu_(s) + Ag⁺_(aq)→ Cu²⁺_(aq) + Ag_(s)       (unbalanced)
The Cu_(s) loses electrons to become Cu²⁺_(aq) ions and the Ag⁺_(aq) ions gain electrons to become Ag_(s).
Reactions that involve the exchange of electrons are called reduction and oxidation (redox) reactions. When a chemical species loses electrons we say that it is oxidized, and when a chemical species gains electrons we say that it is reduced.
The Cu_(s) loses electrons to be oxidized to Cu²⁺_(aq).
The Ag⁺_(aq) gain electrons to be reduced to Ag_(s).
We observed that copper metal and silver ions undergo a spontaneous reaction to produce copper ions and silver metal.
Cu_(s) + 2 Ag⁺_(aq) →Cu²⁺_(aq) + 2 Ag_(s)
Will this reaction continue indefinitely?
At some point we reach equilibrium concentrations and the reaction stops. (Remember that microscopically reactants and products continue to interchange, but we say the reaction has stopped when the concentrations reach a steady state). When we reach equilibrium we write the reaction with arrows in both directions.
Cu_(s) + 2 Ag⁺_(aq) →Cu²⁺_(aq) + 2 Ag_(s)
Which of the following reactions are spontaneous?
H⁺_(aq) + OH^-_(aq) →H₂O
H₂O →H⁺_(aq) + OH^-_(aq)
The first reaction is spontaneous and the second is not. Water does not spontaneously decompose. It does reach the following equilibrium:
H₂O →H⁺_(aq) + OH^-_(aq)
We use the right arrow when we are talking about a reaction at non-equilibrium concentrations and the double arrows when a reaction has reached equilibrium. You will see the same reaction written both ways, the different notation indicates that we are considering either non-equilibrium or equilibrium concentrations of the reactants and products.
Reactants that are mixed in non-equilibrium concentrations have potential energy that drives the reaction towards equilibrium. This potential energy is the difference in free energy between the products and reactants, ∆ G. The reactants will react to achieve equilibrium and reach a lower energy condition.
∆ G = ∆ G^o + RTln(Q)
where
∆ G^o is the free energy of reaction at standard concentrations
R is the gas constant, 8.3145 J/mol·K
T is absolute temperature
Q is the reaction quotient
This relationship shows that reactants at concentrations far from equilibrium will have greater free energy than reactants near equilibrium concentrations.
At equilibrium ∆ G = 0. We cannot extract energy from a system at equilibrium.
Also note that ∆ G^o = -RTln(K)
That these two are related should make sense. K tells us relative concentrations of products to reactants at equilibrium, ∆ G^o tells us how much energy we could get out of a reaction when they are at standard concentrations.