Oxidation is the loss of electrons in a reaction, while reduction is the gain of electrons. In an oxidation-reduction or redox reaction, electrons are transferred from an atom that is oxidized to one that is reduced. The oxidation state or oxidation number is the hypothetical charge of an atom if all bonds to other atoms were fully ionic. It is important to recognize the theoretical nature of this state - the oxidation state does not reflect the actual charge on an atom, but is nonetheless a useful framework in which to describe certain types of chemical reactions. Consider the following redox reaction:
\(\ce{Zn + 2H+ -> Zn^2+ + H2}\)
Here, the oxidation states are as follows:
- Zn goes from an oxidation state of 0 (its elemental form) to an oxidation state of +2 in the products, indicating it has lost two electrons.
- Hydrogen goes from an oxidation state of +1 in the reactants to an oxidation state of 0 in the products (elemental form).
- Zinc is oxidized and hydrogen is reduced.
- Zinc in the reducing agent. Hydrogen is the oxidizing agent.
As you can see from the above example, the oxidizing agent causes another substance to be oxidized, and in doing so is itself reduced. The reducing agent causes another substance to be reduced by giving up electrons, and in doing so is itself oxidized.
As you have seen, chemical equations must obey the law of conservation of mass. Atoms are not created or destroyed during a chemical reaction. Instead, bonds between atoms are broken and formed. The same must hold for redox equations, with the added challenge of making sure electrons are balanced as well. While oxidation and reduction in a redox reaction occur at the same time, it is useful to split these processes into half-reactions in which they are addressed independently to aid in balancing the overall reaction. For reactions in acidic solutions, the procedure is as follows:
Steps to Balance the Reaction:
1. Divide into half-reactions:
Oxidation: \( \ce{Cl^- -> Cl2} \)
Reduction: \( \ce{Cr2O7^2- -> Cr^3+} \)
2. Balance atoms other than H and O. For reduction, chromium is balanced when considering the entire dichromate ion converting into two Cr\(^{3+}\) ions. For oxidation, chlorine is balanced upon forming \( \ce{Cl2} \).
3. Balance oxygen atoms by adding \( \ce{H2O} \): \( \ce{Cr2O7^2- -> 2Cr^3+ + 7H2O} \).
4. Balance hydrogen atoms by adding \( \ce{H+} \): \( \ce{Cr2O7^2- + 14H+ -> 2Cr^3+ + 7H2O} \).
5. Balance charges by adding electrons:
Oxidation: \( \ce{2Cl^- -> Cl2 + 2e^-} \)
Reduction: \( \ce{Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O} \).
6. Equalize and combine the half-reactions:
Oxidation (multiplied by 3): \( \ce{6Cl^- -> 3Cl2 + 6e^-} \)
Reduction: \( \ce{Cr2O7^2- + 14H+ + 6e^- -> 2Cr^3+ + 7H2O} \)
Notice that the electrons cancel out as they are equal on both sides.
7. Combine and simplify to get the final balanced equation: \( \ce{Cr2O7^2- + 14H+ + 6Cl^- -> 2Cr^3+ + 3Cl2 + 7H2O} \).
Consider the following spontaneous redox reaction between zinc and copper:
\( \ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}\)
The standard Gibb's free energy for this reaction is -212.27 kJ/mol, indicating it is spontaneous under standard conditions. At a constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings:
\(\Delta G = w_{\text{max}}\)
What if, instead of the reaction taking place in a single reaction vessel, it was possible to use the energy released during a spontaneous redox reaction to perform meaningful work? This is the idea behind voltaic cells. A voltaic cell, also known as a galvanic cell, converts chemical energy into electrical energy through a spontaneous redox reaction. In constructing a voltaic cell, the half-reactions must be separated. In doing so, the electrons move through an external circuit and in doing so generate a current that can be used to perform work.
How can we calculate the emf of a cell under nonstandard conditions? The Nernst equation involves the reaction quotient, Q, which represents reaction concentrations at any point during the reaction, not just equilibrium:
\(E = E^0 - \frac{RT}{nF} lnQ\)
For example, consider a voltaic cell in which one half-cell contains 0.5 M CuSO4 solution and the other half-cell contains 0.8 M Pb(NO3)2. The half-cells are maintained at 25 C. What is the cell potential, in volts?
Since the concentrations of the half cells are not at 1 M (non-standard conditions), we must use the Nernst equation to determine the cell potential. First, we must determine the standard cell potential of the reaction, E0. We can use a standard reduction potential table to determine this. Let's define the half-reactions:
\(\ce{Cu^2+ + 2e- -> Cu}\)
\(\ce{Pb -> Pb^2+ + 2e-}\)
Copper(II) has a standard reduction potential of 0.34 V, while Pb2+ has a standard reduction potential of -0.13 V. This means that copper is more likely to be reduced, and lead is more likely to be oxidized in the redox reaction. Since lead is oxidized, we switch the sign of the standard reduction potential to get the standard oxidation potential: 0.13 V.
\(\ce{E^\circ_{cell} = E^\circ_{oxidation} + E^\circ_{reduction} = 0.13 V + 0.34 V = 0.47 V}\)
Let's define the other terms in the Nernst equation. R is the ideal gas constant and F is the Faraday constant. n is the number of electrons transferred in the redox reaction, which in this case = 2. Next, let's determine Q for the reaction.
\(\ce{Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}}\)
The overall redox reaction is:
\(\ce{Pb + Cu^{2+} -> Pb^{2+} + Cu}\)
Thus, Q is:
\(Q = \frac{[\ce{Pb^{2+}}]}{[\ce{Cu^{2+}}]}\)
Substituting the given values we get Q = 0.8 M/0.5 M = 1.6. We can now solve for E:
\(E = 0.47\, \text{V} - \frac{0.0592}{2} \log(1.6)\)
E is 0.464 V.