
# Chapter 18 Electrochemistry

Shaun Williams, PhD

### Review of Terms

• Electrochemistry – the study of the interchange of chemical and electrical energy
• Oxidation–reduction (redox) reaction – involves a transfer of electrons from the reducing agent to the oxidizing agent
• Oxidation – loss of electrons
• Reduction – gain of electrons
• Reducing agent – electron donor
• Oxidizing agent – electron acceptor

### Half-Reactions

• The overall reaction is split into two half–reactions, one involving oxidation and one involving reduction. \begin{align} &\chem{8H^+ + MnO_4^- + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O} \\ &\text{Reduction: }\chem{8H^+ + MnO_4^- + 5e^- \rightarrow Mn^{2+} + 4H_2O} \\ &\text{Oxidation: }\chem{5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-} \end{align}

### The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution

1. Write separate equations for the oxidation and reduction half–reactions.
2. For each half–reaction:
• Balance all the elements except $$\chem{H}$$ and $$\chem{O}$$.
• Balance $$\chem{O}$$ using $$\chem{H_2O}$$.
• Balance $$\chem{H}$$ using $$\chem{H^+}$$.
• Balance the charge using electrons.
3. If necessary, multiply one or both balanced half–reactions by an integer to equalize the number of electrons transferred in the two half–reactions.
4. Add the half–reactions, and cancel identical species.
5. Check that the elements and charges are balanced.

### Example

$$\chem{Cr_2O_7^{2-}(aq) + SO_3^{2-}(aq) \rightarrow Cr^{3+}(aq) + SO_4^{2-}(aq)}$$

1. Seperate this into half-reactions \begin{align} \chem{Cr_2O_7^{2-}(aq)} &\rightarrow \chem{Cr^{3+}(aq)} \\ \chem{SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq)} \end{align}
2. Balance all the elements except $$\chem{H}$$ and $$\chem{O}$$ \begin{align} \chem{Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq)} \\ \chem{SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq)} \end{align}

### Example Continued

\begin{align} \chem{Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq)} \\ \chem{SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq)} \end{align}

1. Add electrons based on changes in oxidation states \begin{align} \chem{6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{Cr^{3+}(aq)} \\ \chem{SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq) + 2e^-} \end{align}
2. Balance the oxygen atoms using $$\chem{H_2O}$$ \begin{align} \chem{6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq) + 7H_2O} \\ \chem{H_2O + SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq) + 2e^-} \end{align}

### Example Continued Further

\begin{align} \chem{6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq) + 7H_2O} \\ \chem{H_2O + SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq) + 2e^-} \end{align}

1. Balance the hydrogen atoms using $$\chem{H^+}$$ \begin{align} \chem{14H^+ + 6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq) + 7H_2O} \\ \chem{H_2O + SO_3^{2-}(aq)} &\rightarrow \chem{SO_4^{2-}(aq) + 2e^- + 2H^+} \end{align}
2. Balance the electrons between the half-reactions \begin{align} \chem{14H^+ + 6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq) + 7H_2O} \\ 3\left[\chem{H_2O + SO_3^{2-}(aq)}\right. &\rightarrow \left.\chem{SO_4^{2-}(aq) + 2e^- + 2H^+}\right] \end{align}

### Example Concluded

\begin{align} \chem{14H^+ + 6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq) + 7H_2O} \\ 3\left[\chem{H_2O + SO_3^{2-}(aq)}\right. &\rightarrow \left.\chem{SO_4^{2-}(aq) + 2e^- + 2H^+}\right] \end{align}

1. Distributing \begin{align} \chem{14H^+ + 6e^- + Cr_2O_7^{2-}(aq)} &\rightarrow \chem{2Cr^{3+}(aq) + 7H_2O} \\ 3\chem{H_2O + 3SO_3^{2-}(aq)} &\rightarrow \chem{3SO_4^{2-}(aq) + 6e^- + 6H^+} \end{align}
2. Add the reactions together cancel out species on each side of the reaction to get the final balance equation $$\chem{Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O}$$

### Exercise

Balance the following oxidation-reduction reaction that occurs in acidic solution $$\chem{Br^-(aq) + MnO_4^-(aq) \rightarrow Br_2(l) + Mn^{2+}(aq)}$$

\begin{align} \chem{10Br^-(aq)} + \chem{16H^+(aq)} + \chem{2MnO_4^-(aq)} \rightarrow \\ \chem{5Br_2(l)} + \chem{2Mn^{2+}(aq)} + \chem{8H_2O(l)} \end{align}

### The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution

1. Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if $$\chem{H^+}$$ ions were present.
2. To both sides of the equation obtained above, add a number of $$\chem{OH^–}$$ ions that is equal to the number of $$\chem{H^+}$$ ions. (We want to eliminate $$\chem{H^+}$$ by forming $$\chem{H_2O}$$.)
3. Form $$\chem{H_2O}$$ on the side containing both $$\chem{H^+}$$ and $$\chem{OH^–}$$ ions, and eliminate the number of H2O molecules that appear on both sides of the equation.
4. Check that elements and charges are balanced.

### Galvantic Cell

• Device in which chemical energy is changed to electrical energy.
• Uses a spontaneous redox reaction to produce a current that can be used to do work.
• Oxidation occurs at the anode.
• Reduction occurs at the cathode.
• Salt bridge or porous disk – devices that allow ions to flow without extensive mixing of the solutions.
• Salt bridge – contains a strong electrolyte held in a Jello–like matrix.
• Porous disk – contains tiny passages that allow hindered flow of ions.

### Cell Potential

• A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment.
• The “pull”, or driving force, on the electrons is called the cell potential ($$\mathcal{E}_{cell}$$), or the electromotive force (emf) of the cell.
• Unit of electrical potential is the volt (V).
• 1 joule of work per coulomb of charge transferred.

### More on Galvanic Cells

• All half-reactions are given as reduction processes in standard tables.
• Table 18.1
• $$1\,\mathrm{M}$$, $$1\,\mathrm{atm}$$, $$25^\circ\mathrm{C}$$
• When a half-reaction is reversed, the sign of $$E^\circ$$ is reversed.
• When a half-reaction is multiplied by an integer, $$E^\circ$$ remains the same.
• A galvanic cell runs spontaneously in the direction that gives a positive value for $$E^\circ_{cell}$$.

### Example: $$\chem{Fe^{3+}(aq) + Cu(s) → Cu^{2+}(aq) + Fe^{2+}(aq)}$$

• Half-Reactions: \begin{align} &\chem{Fe^{3+}+e^- \rightarrow Fe^{2+}} & E^\circ = 0.77\,\mathrm{V} \\ &\chem{Cu^{2+}+2e^- \rightarrow Cu} & E^\circ = 0.34\,\mathrm{V} \\ \end{align}
• To balance the cell reaction and calculate the cell potential, we must reverse reaction 2. $$\chem{Cu \rightarrow Cu^{2+} + 2e^-} \;\;\;\; E^\circ = -0.34\,\mathrm{V}$$
• Each $$\chem{Cu}$$ atom produces two electrons but each $$\chem{Fe^{3+}}$$ ion accepts only one electron, therefore reaction 1 must be multiplied by 2. $$\chem{2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}} \;\;\;\; E^\circ = 0.77\,\mathrm{V}$$

### Overall Balanced Cell Reaction

\begin{align} &\chem{2Fe^{3+}+2e^- \rightarrow 2Fe^{2+}} & E^\circ = 0.77\,\mathrm{V} \\ &\chem{Cu \rightarrow Cu^{2+}+2e^-} & E^\circ = -0.34\,\mathrm{V} \\ \end{align}

• Balanced Cell Reaction: $$\chem{Cu + 2Fe^{3+} \rightarrow Cu^{2+} + 2Fe^{2+}}$$
• Cell Potential: \begin{align} E^\circ_{cell} &= E^\circ (\mathrm{cathode}) - E^\circ (\mathrm{anode}) \\ E^\circ_{cell} &= 0.77\,\mathrm{V} - 0.34\,\mathrm{V} = 0.43\,\mathrm{V} \end{align}

### Line Notation

• Used to describe electrochemical cells.
• Anode components are listed on the left.
• Cathode components are listed on the right.
• Separated by double vertical lines which indicated salt bridge or porous disk.
• The concentration of aqueous solutions should be specified in the notation when known.
• Example: $$\chem{Mg(s) | Mg^{2+}(aq) || Al^{3+}(aq) | Al(s)}$$
• $$\chem{Mg \rightarrow Mg^{2+} + 2e^-}\;\;\;\mathrm{(anode)}$$
• $$\chem{Al^{3+} + 3e^- \rightarrow Al}\;\;\;\;\;\mathrm{(cathode)}$$

### Description of a Galvanic Cell

• The cell potential (always positive for a galvanic cell where $$E^\circ_{cell} = E^\circ (\mathrm{cathode}) – E^\circ (\mathrm{anode})$$ and the balanced cell reaction.
• The direction of electron flow, obtained by inspecting the half–reactions and using the direction that gives a positive $$E^\circ_{cell}$$.
• Designation of the anode and cathode.
• The nature of each electrode and the ions present in each compartment. A chemically inert conductor is required if none of the substances participating in the half–reaction is a conducting solid.

### Work

• Work is never the maximum possible if any current is flowing.
• In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.

### Maximum Cell Potential

• Directly related to the free energy difference between the reactants and the products in the cell.
• $$\Delta G^\circ = –nFE^\circ$$
• $$F = 96485\,\bfrac{\mathrm{C}}{\mathrm{mol\cdot e^-}}$$

### Nernst Equation

• The relationship between cell potential and concentrations of cell components
• At $$25^\circ\mathrm{C}$$: \begin{align} \mathcal{E} &= \mathcal{E}^\circ - \frac{0.0591}{n} \log (Q) \\ &\mathrm{or} \\ \mathcal{E}^\circ &= \frac{0.0591}{n}\log (K) \;\;\; \text{(at equilibrium)} \end{align}

### Exercise

A concentration cell is constructed using two nickel electrodes with $$\chem{Ni^{2+}}$$ concentrations of $$1.0\,\mathrm{M}$$ and $$1.00 \times 10^{-4}\,\mathrm{M}$$ in the two half-cells.

Calculate the potential of this cell at $$25^\circ\mathrm{C}$$.

$$0.118\,\mathrm{V}$$

### Concept Check

You make a galvanic cell at $$25^\circ\mathrm{C}$$ containing:

• A nickel electrode in $$1.0\,\mathrm{M}\,\chem{Ni^{2+}(aq)}$$
• A silver electrode in $$1.0\,\mathrm{M}\,\chem{Ag^+(aq)}$$

Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential.

$$1.03\,\mathrm{V}$$

### Corrosion

• Process of returning metals to their natural state – the ores from which they were originally obtained.
• Involves oxidation of the metal.

### Corrosion Prevention

• Application of a coating (like paint or metal plating)
• Galvanizing
• Alloying
• Cathodic Protection
• Protects steel in buried fuel tanks and pipelines.

### Electrolysis

• Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
• Steps of stoichiometry of electrolysis
1. current and time $$\rightarrow$$ quantity of charge $$\text{Coulombs of charge} = \text{current (C/s)} \times \text{time (s)}$$
2. quantity of charge $$\rightarrow$$ moles of electrons $$\text{mole }e^- = \text{Coulombs of charge} \times \frac{1\,\mathrm{mol}\,e^-}{96485\,\mathrm{C}}$$
3. moles of electrons $$\rightarrow$$ moles of analyte
4. moles of analyte $$\rightarrow$$ grams of analyte

### Concept Check

An unknown metal ($$\chem{M}$$) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing $$\chem{M\left(NO_3\right)_3}$$.

What is the metal?

gold (Au)

### Commercial Electrolytic Processes

• Production of aluminum
• Purification of metals
• Metal plating
• Electrolysis of sodium chloride
• Production of chlorine and sodium hydroxide

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