John Straub's lecture notes

Observing dramatic physical transformations mediated by reduction/oxidation reactions

A metallic copper plate is immersed in a solution of silver nitrate, leading to the formation of a "fur" of metallic silver.

Ingredients: silver nitrate, copper bar

Procedure: A complete recipe follows.

1. Prepare a solution of silver nitrate.

2. Place a copper bar in the silver nitrate solution.

3. Allow a few minutes to pass and observe silver "growth" on copper bar.

Understanding: When the copper bar is placed in contact with the silver nitrate solution, there is a driving force to transfer electrons from the copper metal to the silver ions, oxidizing the copper metal to cupric ion and reducing the silver ion to silver metal. The silver ions are mobile in solution, while the copper atoms are immobile on the copper bar, so the reaction occurs at the surface of the electrode

Cu(s) + 2 Ag⁺(aq) → Cu²⁺(aq) + 2 Ag(s)

Expanding the view of the deposition on the bar, one sees a "fur" of silver precipitate that has formed. As the silver is deposited, at first the reaction occurs near the copper surface. As the reaction proceeds, the reaction interface becomes the surface of the deposited silver which grows outward in solution, creating the furry silver precipitate and a blue solution of cupric ion.

Clearly, there is a driving force to reduce the silver metal and oxidize the copper atoms. Let's see how we can quantify the magnitude of the driving force by defining the reduction potential of each half-reaction.

Defining the half-reaction reduction potential

The reference electrode against which all other half-reactions are measured is the standard hydrogen electrode. Each electrochemical half-reaction has a corresponding standard reduction potential which is a measure of the tendency for that element to be reduced relative to the standard reference half-reaction

2 H⁺(aq) + 2 e^- → H₂(g)

where the concentration of acid is 1M and the pressure of hydrogen gas is 1 atmosphere. If an electrochemical cell is formed from two standard hydrogen electrodes, there will be no voltage across the cell and no flow of electrons between the half cells. We can say that relative to the standard hydrogen electrode, the reduction potential of the standard hydrogen electrode is by definition E^o_H⁺|H₂=0.0V which we write as

2 H⁺(aq) + 2 e^- → H₂(g) E^o_{H⁺|H₂=0.0V}

A galvanic cell can be formed from the reference standard hydrogen electrode and the standard copper electrode. The voltage across that electrochemical cell is measured to be 0.34V. We write

H₂(g) + Cu²⁺(aq) → Cu(s) + 2 H⁺(aq) E^o_{H₂|H⁺||Cu²⁺|Cu}=0.34V

The positive voltage indicates that there is a spontaneous flow of electrons from the hydrogen electrode (anode), where dihydrogen gas is oxidized to hydrogen ions in solution, to the copper electrode (cathode), where copper ions in solution are reduced to form copper metal. The notation used to label the voltage may seem cumbersome, but it is worth the effort!

We say that relative to the standard hydrogen electrode, the reduction potential of the standard copper half-cell is 0.34V or

Cu²⁺(aq) + 2 e^- → Cu(s) E^o_Cu²⁺|Cu=0.34V

We can make a similar measurement for the standard silver half-cell where the overall electrochemical cell is

H₂(g) + 2 Ag⁺(aq) → 2 Ag(s) + 2 H⁺(aq) E^o_{H₂|H⁺||Ag⁺|Ag}=0.80V

It follows that the standard reduction potential for the silver half-cell is

Ag⁺(aq) + e^- → Ag(s) E^o_Ag⁺|Ag=0.80V

The electrical work equals the change in Gibbs free energy

We now have a burning question to answer. What is the voltage that would be measured for the electrochemical cell composed of a standard silver electrode in contact with a standard copper electrode? If we imagine that the electrochemical reaction is carried out reversibly, the work that is done in moving electrons from one half cell to the other is the work that is done in moving electrons across the voltage E

w_elec = -Q E

where Q is the total charge that is moved. We can express the charge in terms of moles of electrons, n, where a mole of electrons has a total charge of

e x N₀ = 1.602 x 10^-19 C x 6.022 x 10²³ mol^-1 = 96,485 C/mol.

We call the mole of charge a Faraday of charge, after the famous chemist Michael Faraday, and Faraday's constant is F=96,485 C/mol. That leads to the final expression for the electrical work done in moving n moles of electrons across a voltage E

w_elec = -n F E

Now here is the kicker. We know that the change in Gibbs free energy for a chemical reaction at standard state will determine the position of equilibrium. The Gibbs free energy change is in general

ΔG = ΔH - Δ(TS)

and at constant temperature we find that

ΔG = ΔH - TΔS (constant T)

Suppose that we further constrain the reaction to occur at constant pressure. We can use the fact that

Δ H= ΔE + Δ (PV)

where Δ E = q + w = q + w_PV + w_elec, and the total work is the sum of the pressure-volume work and the electrical work. That means that

Δ H= ΔE + Δ (PV) = q - PΔV + PΔV + VΔP + w_elec = q + w_elec (constant T and P)

Combining these results we find that

ΔG = q - TΔS + w_elec (constant T and P)

What if the reaction is also reversible? We know that at constant temperature

ΔS = q_rev/T

which means that

ΔG = q_rev - TΔS + w_elec = q_rev - T(q_rev/T) + w_elec = w_elec (constant T and P for reversible process)

At constant temperature and pressure, for an electrochemical cell proceeding reversibly, the change in the Gibbs free energy is equal to the work done in the reversible process

Δ G = w_elec = -n F E (constant T and P for reversible process)

This is a fundamentally important result. It will allow us to prove something important for our electrochemical cell.

The relationship between cell voltage and Gibbs free energy

We have measured the standard reduction potentials of our silver half-cell and copper half-cells, relative to the standard hydrogen electrode. We now see that those reduction potentials correspond to changes in Gibbs free energy!

H₂(g) + Cu²⁺(aq) → Cu(s) + 2 H⁺(aq) Δ G^o = -n F E^o_{H₂|H⁺||Cu²⁺|Cu}

and

H₂(g) + 2 Ag⁺(aq) → 2 Ag(s) + 2 H⁺(aq) Δ G^o = -n F E^o_{H₂|H⁺||Ag⁺|Ag}

For the overall reaction

Cu(s) + 2 Ag⁺ → Cu²⁺(aq) + Ag(s)

we see that

Δ G^o = -n F E^o _{H₂|H⁺||Ag⁺|Ag} - (-n F E^o_{H₂|H⁺||Cu²⁺|Cu}) = -n F E^o_{Cu|Cu²⁺||Ag⁺|Ag}

using the basic properties of adding the Gibbs free energies of reaction. This leads to the fundamental result that

E^o_cell = E^o_Ag⁺|Ag - E^o_Cu²⁺|Cu

In this cell, oxidation occurs at the copper cell, making it the anode, and reduction occurs at the silver cell, making it the cathode. So in general we can say that

E^o_cell = E^o_cathode - E^o_anode

but most fundamentally

Δ G^o_cell = Δ G^o_cathode - Δ G^o_anode

Returning to our reaction of the copper bar in the silver nitrate solution, we find that at standard state the electrochemical cell voltage would be

E^o_{Cu|Cu²⁺||Ag⁺|Ag} = E^o_Ag⁺|Ag - E^o_Cu²⁺|Cu = 0.80V - 0.34V = 0.46V

with a change in Gibbs free energy of

Δ G^o_{Cu|Cu²⁺||Ag⁺|Ag} = -n F E^o_{Cu|Cu²⁺||Ag⁺|Ag} = -(2 mol) (96,485 C/mol) (0.46V) = -88,766 CV = -88.8 kJ

making the overall reaction favor the product state by 88.8 kJ. That is the driving force that favors the transfer of electrons from the copper metal to the silver ions, reducing the silver, and forming the metallic silver fur over the copper electrode.

We can now see how the more cumbersome notation for the reduction and cell potentials pays off. The notation makes it clear that the standard reduction potentials are always for the reduction reactions. The addition and subtraction of half-cell reduction potentials, when labeled in this manner, will always lead to the correct result.

A zinc bar is placed in a solution of copper sulfate, leading to the deposition of copper metal.

Ingredients: copper sulfate, zinc bar

Procedure: A complete recipe follows.

1. Prepare a solution of copper sulfate.

2. Place a zinc bar in the copper sulfate solution.

3. Allow a few minutes to pass and observe copper deposition on the zinc bar.

Understanding: The spontaneous formation of copper metal deposited on the zinc electrode is rapid and dramatic. The zinc metal is oxidized to divalent zinc ions, while divalent copper ions are reduced to copper metal. The copper metal deposits on the zinc surface.

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

At first, the deposited copper appears black on the bar, and when it is rubbed off of the zinc bar on a white paper towel. However, if the deposition is allowed to proceed, the characteristic lustrous metallic copper surface is formed The initial deposit of copper lacks the metallic "shine" that we associate with the metal. However, after a time the deposit thickens to form ordered metallic copper that has the characteristic copper color.

What is the measure of the driving force of this reaction? We know that

Δ G^o_cell = Δ G^o_cathode - Δ G^o_anode = Δ G^o_Cu²⁺|Cu - Δ G^o_Zn²⁺|Zn = -n F E^o_{Zn|Zn²⁺||Cu²⁺|Cu}

which we measure to be E^o_cell=1.10V. Given our prior result that E^o_Cu²⁺|Cu=0.34V we find that

E^o_cell = 1.10V = E^o_Cu²⁺|Cu - E^o_Zn²⁺|Zn = 0.34V -E^o_Zn²⁺|Zn

with the result that E^o_Zn²⁺|Zn=-0.76V. Note that we have not measured the standard zinc half-cell against the standard hydrogen electron. Yet we were able to compute the standard reduction potential for the zinc half-cell using thermodynamics!

What does it mean that E^o_Zn²⁺|Zn < E^o_Cu²⁺|Cu? Since ΔG = -n F E we know that if E is positive then ΔG is negative and the reaction is favorable. So we can say that the more positive reduction potential of copper relative to zinc implies that copper will be reduced while the zinc will be oxidized.

In general we can organize the standard reduction potentials of a variety of metals

The elements platinum and gold have large and positive reduction potentials. They are referred to as noble metals as they are reluctant to be oxidized and so remain in their metallic state. In contrast, aluminum and zinc are metals that tend to be easily oxidized, making both metals very reactive.

This result informs the use of zinc as a sacrificial electrode. Suppose that there is a metal such as iron that one would like to prevent from oxidative damage and corrosion. We can place the iron in electrical contact with a bar of zinc. As long as there is zinc metal, the electrons will always prefer to oxidize zinc while allowing the iron to remain in its reduced metallic state.

Galvanized steel is steel treated with zinc. The zinc is sacrificially oxidized, protecting the steel from oxidative damage. Anodized aluminum is aluminum metal that has been coated with aluminum oxide, subsequently forming aluminum hydrate. In that process, the aluminum surface is oxidized, preventing it from further corrosion.

Direct measures of relative reduction potentials

Question: When the copper bar was placed in a solution of silver nitrate, silver metal precipitated out of solution, and the solution became blue with copper ion. What would happen if you placed a silver bar in a solution of copper sulfate?

You can check your answers here.