Electrochemistry Part II: Cell Potential, Free Energy, and Electrolysis

Relationship Between the Potential of a Galvanic Cell and Free Energy

In the study of electrochemistry, there is a fundamental relationship between the electrical potential of a galvanic cell and the Gibbs free energy ($\Delta G$). For standard conditions, this relationship is defined by the equation $\Delta G^\circ = -nFE^\circ$. In this formula, $\Delta G^\circ$ represents the change in standard free energy, $n$ represents the number of moles of electrons transferred in the balanced redox reaction, $F$ is the Faraday constant, and $E^\circ$ is the standard cell potential. The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell. This connection allows for the determination of whether a chemical reaction is spontaneous as written.

To calculate $\Delta G^\circ$ for a specific cell reaction, such as the reaction between copper and iron represented by the equation $Cu^{2+}(aq) + Fe(s) \rightarrow Cu(s) + Fe^{2+}(aq)$, one must utilize standard reduction potential data. By determining the $\Delta G^\circ$ value, one can answer whether the reaction is spontaneous. A negative $\Delta G^\circ$ value (or a positive $E^\circ$ value) indicates a spontaneous process.

Predicting spontaneity is critical in various chemical contexts. For instance, determining whether $1\,M$ $HNO_3$ will dissolve gold metal to form a $1\,M$ $Au^{3+}$ solution requires evaluating the cell potential for the proposed reaction. It is noted that under standard conditions, certain reactions involving $NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$ may not occur spontaneously if the calculated potential does not favor the oxidation of the metal in question.

Standard Reduction Potentials

The following standard reduction potentials ($E^\circ$) at $25^\circ C$ are used to determine cell behavior and calculate free energy changes:

$F_2 + 2e^- \rightarrow 2F^-$: $E^\circ = 2.87\,V$ $Ag^{2+} + e^- \rightarrow Ag^+$: $E^\circ = 1.99\,V$ $Co^{3+} + e^- \rightarrow Co^{2+}$: $E^\circ = 1.82\,V$ $H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$: $E^\circ = 1.78\,V$ $Ce^{4+} + e^- \rightarrow Ce^{3+}$: $E^\circ = 1.70\,V$ $PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4 + 2H_2O$: $E^\circ = 1.69\,V$ $MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$: $E^\circ = 1.68\,V$ $2e^- + 2H^+ + IO_4^- \rightarrow IO_3^- + H_2O$: $E^\circ = 1.60\,V$ $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$: $E^\circ = 1.51\,V$ $Au^{3+} + 3e^- \rightarrow Au$: $E^\circ = 1.50\,V$ $PbO_2 + 4H^+ + 2e^- \rightarrow Pb^{2+} + 2H_2O$: $E^\circ = 1.46\,V$ $Cl_2 + 2e^- \rightarrow 2Cl^-$: $E^\circ = 1.36\,V$ $Cr_{2}O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$: $E^\circ = 1.33\,V$ $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$: $E^\circ = 1.23\,V$ $MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$: $E^\circ = 1.21\,V$ $IO_3^- + 6H^+ + 5e^- \rightarrow \frac{1}{2}I_2 + 3H_2O$: $E^\circ = 1.20\,V$ $Br_2 + 2e^- \rightarrow 2Br^-$: $E^\circ = 1.09\,V$ $VO_2^+ + 2H^+ + e^- \rightarrow VO^{2+} + H_2O$: $E^\circ = 1.00\,V$ $AuCl_4^- + 3e^- \rightarrow Au + 4Cl^-$: $E^\circ = 0.99\,V$ $NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$: $E^\circ = 0.96\,V$ $ClO_2 + e^- \rightarrow ClO_2^-$: $E^\circ = 0.954\,V$ $2Hg^{2+} + 2e^- \rightarrow Hg_2^{2+}$: $E^\circ = 0.91\,V$ $Ag^+ + e^- \rightarrow Ag$: $E^\circ = 0.80\,V$ $Hg^{2+} + 2e^- \rightarrow 2Hg$: $E^\circ = 0.80\,V$ $Fe^{3+} + e^- \rightarrow Fe^{2+}$: $E^\circ = 0.77\,V$ $O_2 + 2H^+ + 2e^- \rightarrow H_2O_2$: $E^\circ = 0.68\,V$ $MnO_4^- + e^- \rightarrow MnO_4^{2-}$: $E^\circ = 0.56\,V$ $I_2 + 2e^- \rightarrow 2I^-$: $E^\circ = 0.54\,V$ $Cu^+ + e^- \rightarrow Cu$: $E^\circ = 0.52\,V$ $O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$: $E^\circ = 0.40\,V$ $Cu^{2+} + 2e^- \rightarrow Cu$: $E^\circ = 0.34\,V$ $Hg_2Cl_2 + 2e^- \rightarrow 2Hg + 2Cl^-$: $E^\circ = 0.27\,V$ $AgCl + e^- \rightarrow Ag + Cl^-$: $E^\circ = 0.22\,V$ $SO_4^{2-} + 4H^+ + 2e^- \rightarrow H_2SO_3 + H_2O$: $E^\circ = 0.20\,V$ $Cu^{2+} + e^- \rightarrow Cu^+$: $E^\circ = 0.16\,V$ $2H^+ + 2e^- \rightarrow H_2$: $E^\circ = 0.00\,V$ $Fe^{3+} + 3e^- \rightarrow Fe$: $E^\circ = -0.036\,V$ $Pb^{2+} + 2e^- \rightarrow Pb$: $E^\circ = -0.13\,V$ $Sn^{2+} + 2e^- \rightarrow Sn$: $E^\circ = -0.14\,V$ $Ni^{2+} + 2e^- \rightarrow Ni$: $E^\circ = -0.23\,V$ $PbSO_4 + 2e^- \rightarrow Pb + SO_4^{2-}$: $E^\circ = -0.35\,V$ $Cd^{2+} + 2e^- \rightarrow Cd$: $E^\circ = -0.40\,V$ $Fe^{2+} + 2e^- \rightarrow Fe$: $E^\circ = -0.44\,V$ $Cr^{3+} + e^- \rightarrow Cr^{2+}$: $E^\circ = -0.50\,V$ $Cr^{3+} + 3e^- \rightarrow Cr$: $E^\circ = -0.73\,V$ $Zn^{2+} + 2e^- \rightarrow Zn$: $E^\circ = -0.76\,V$ $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$: $E^\circ = -0.83\,V$ $Mn^{2+} + 2e^- \rightarrow Mn$: $E^\circ = -1.18\,V$ $Al^{3+} + 3e^- \rightarrow Al$: $E^\circ = -1.66\,V$ $H_2 + 2e^- \rightarrow 2H^-$: $E^\circ = -2.23\,V$ $Mg^{2+} + 2e^- \rightarrow Mg$: $E^\circ = -2.37\,V$ $La^{3+} + 3e^- \rightarrow La$: $E^\circ = -2.37\,V$ $Na^+ + e^- \rightarrow Na$: $E^\circ = -2.71\,V$ $Ca^{2+} + 2e^- \rightarrow Ca$: $E^\circ = -2.76\,V$ $Ba^{2+} + 2e^- \rightarrow Ba$: $E^\circ = -2.90\,V$ $K^+ + e^- \rightarrow K$: $E^\circ = -2.92\,V$ $Li^+ + e^- \rightarrow Li$: $E^\circ = -3.05\,V$

The Nernst Equation

The Nernst Equation provides the mathematical relationship between the cell potential ($E$) and the concentrations of the cell components. It is derived from the thermodynamic relationship $\Delta G = \Delta G^\circ + RT \ln(Q)$. By substituting the expressions $\Delta G = -nFE$ and $\Delta G^\circ = -nFE^\circ$, the equation becomes $-nFE = -nFE^\circ + RT \ln(Q)$. Dividing each side of the equation by $-nF$ yields the general form of the Nernst Equation: $E = E^\circ - \frac{RT}{nF} \ln(Q)$.

In many applications, the Nernst Equation is used in a specialized form valid at $25^\circ C$ ($298.15\,K$), which utilizes log base 10 instead of natural logs: $E = E^\circ - \frac{0.0591}{n} \log(Q)$. This version is frequently used to calculate the potential of a cell in which some or all of the components are not in their standard states ($1\,M$ for solutions, $1\,atm$ for gases). The potential calculated represents the maximum potential before any current flow has occurred.

As an example of the Nernst Equation's application, consider a cell based on the following half-reactions at $25^\circ C$:

1. $VO_2^+ + 2H^+ + e^- \rightarrow VO^{2+} + H_2O$ with $E^\circ = 1.00\,V$

2. $Zn^{2+} + 2e^- \rightarrow Zn$ with $E^\circ = -0.76\,V$ To calculate the potential for this specific cell, one must use the provided concentrations: $[VO_2^+] = 2.0\,M$, $[H^+] = 0.50\,M$, $[VO^{2+}] = 1.0 \times 10^{-2}\,M$, and $[Zn^{2+}] = 1.0 \times 10^{-1}\,M$.

Corrosion: An Introduction

Corrosion is defined as the natural process of returning metals to their original, oxidized state. This chemical process involves the oxidation of the metal. When a metal's oxidation half-reaction is combined with the reduction half-reaction for oxygen, the resulting standard cell potential ($E^\circ$) is typically positive. This indicates that the oxidation of most metals by oxygen is a spontaneous process.

Conversely, noble metals are classified as those that are relatively difficult to oxidize. Familiar examples of noble metals include copper ($Cu$), gold ($Au$), silver ($Ag$), and platinum ($Pt$). Because of their high resistance to oxidation, these metals are often found in their metallic state in nature or resist environmental degradation longer than base metals.

Electrochemical Corrosion of Iron

The corrosion of iron, particularly in the form of steel, is an electrochemical process. Steel has a nonuniform surface because its chemical composition is not completely homogeneous. Furthermore, physical strains during manufacturing or use leave stress points in the metal. these irregularities cause specific areas to become anodic regions (where the iron is more easily oxidized) and other areas to become cathodic regions.

In the anodic region, iron atoms lose electrons to form ions: $2Fe \rightarrow 2Fe^{2+} + 4e^-$. The electrons released during this oxidation flow through the conductive steel to the cathodic region. In the cathodic region, these electrons react with oxygen and moisture: $O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$. The $Fe^{2+}$ ions formed in the anodic regions migrate to the cathodic regions through the moisture (electrolyte) present on the surface of the steel.

In the cathodic region, the following reaction ultimately occurs: $4Fe^{2+}(aq) + O_2(g) + (4 + 2n)H_2O(l) \rightarrow 2Fe_2O_3 \cdot nH_2O(s) + 8H^+(aq)$. The product is hydrated iron(III) oxide, commonly known as rust. The degree of hydration (represented by $n$) of the iron oxide affects the specific color of the rust, which can vary from black to yellow to the characteristic reddish-brown.

Prevention of Corrosion

There are several primary methods utilized to prevent or slow the corrosion of metals:

1. Application of a Coating: This creates a physical barrier between the metal and the environment. One specific method is galvanizing, which involves coating steel with a layer of zinc. Zinc reacts to form a mixed oxide–carbonate coating that protects the underlying metal. Based on potential half-reactions, any oxidation that occurs will dissolve the zinc rather than the iron, as Zinc's potential ($E^\circ = 0.76\,V$) is more active than Iron's ($E^\circ = 0.44\,V$) when viewed as oxidation potentials.

2. Alloying: Metals can be made corrosion-resistant by mixing them with other elements. Stainless steel, for example, contains chromium ($Cr$) and nickel ($Ni$). Both of these elements form oxide coatings that change the reduction potential of the steel to one characteristic of noble metals. An alternative, more cost-effective method is treating carbon steel via ion bombardment. In this process, a plasma or ion gas of alloying ions is formed at high temperatures and directed onto the surface of the metal to produce a thin layer of stainless steel or another desirable alloy.

3. Cathodic Protection: This method is extensively used to protect steel in buried fuel tanks and pipelines. An active metal, such as magnesium ($Mg$), is connected by a wire to the pipeline or tank. Because magnesium is a better reducing agent than iron, the magnesium acts as the anode and furnishes electrons to the iron system. This keeps the iron from being oxidized. As the process continues, the magnesium anode dissolves and must be replaced periodically.

Electrolysis and Electrolytic Cells

An electrolytic cell utilizes electrical energy to produce a chemical change that would not occur spontaneously. The process, known as electrolysis, involves forcing a direct current through a cell to drive a chemical change for which the cell potential is negative. This is the opposite of a galvanic cell, which generates electricity from a spontaneous chemical reaction.

Stoichiometry of Electrolysis

The stoichiometry of electrolysis refers to the quantitative determination of how much chemical change occurs based on the flow of a given current for a specified period of time. This process is generally broken down into four steps. To illustrate, consider determining the mass of copper ($Cu$) plated out when a current of $10.0\,amps$ is passed for $30.0\,minutes$ through a solution containing $Cu^{2+}$. Each $Cu^{2+}$ ion requires two electrons to become a copper atom ($Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$). This reduction process occurs at the cathode.

Step 1: Multiply the current by the time in seconds to obtain the total coulombs of charge passed. For the example: $10.0\,A \times (30.0\,min \times 60\,s/min) = 1.80 \times 10^4\,C$.

Step 2: Convert the total coulombs to moles of electrons. Since $1\,mole$ of electrons carries a charge of $1\,Faraday$ ($96,485\,coulombs$), we find: $\frac{1.80 \times 10^4\,C}{96,485\,C/mol\,e^-} \approx 0.187\,mol$ of electrons.

Step 3: Use the stoichiometry of the half-reaction to find the moles of metal. In this case, each $Cu^{2+}$ requires two electrons. Thus, each mole of electrons produces $1/2\,mole$ of copper metal.

Step 4: Calculate the final mass of the plated metal using its molar mass and the number of moles calculated in Step 3.

Questions & Discussion

Question 1: Identification of an Unknown Metal An unknown metal ($M$) is electrolyzed in a solution containing $M(NO_3)_3$. It took $52.8\,sec$ for a current of $2.00\,amp$ to plate $0.0719\,g$ of the metal. What is the identity of the metal?

Question 2: Chromium Deposition Time In a industrial application, a $0.010\text{-mm}$ layer of chromium must be deposited on a part with a total surface area of $3.3\,m^2$ from a solution containing chromium(III) ions. Given that the density of chromium metal is $7.19\,g/cm^3$ and the current is $33.46\,A$, how long would it take to deposit this layer?