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Chapter 18


Shaun Williams, PhD

Review of Terms


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.


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

  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 $$ \chem{Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O} $$


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

The Galvantic Cell

Oxidation is occuring at the anode in the presence of the reducing agent. Reduction is occuring at the cathode in the presence of the oxidizing agent. A wire connects the anode and cathode. Additionally the two chambers of the galvanic cell as connected through a porous disk.

Cell Potential

More on Galvanic Cells

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

Overall Balanced Cell Reaction

Line Notation

Description of a Galvanic Cell


Maximum Cell Potential

A Concentration Cell

A galvanic cell in which both solutions and both metals are the same. The anode cell contains a \(0.1\,\mathrm{M}\) solution while the cathode cell contains a \(1.0\,\mathrm{M}\) solution.

Nernst Equation


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 \(25^\circ\mathrm{C}\).


Concept Check

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

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

\( 1.03\,\mathrm{V} \)

One of the Six Cells in a 12–V Lead Storage Battery

A diagram of a lead acid battery showing alternating plates of anodes (made of lead) and cathodes (made of lead(IV) oxide). The plates are in a solution of sulfuric acid.

A Common Dry Cell Battery

Diagram showing a central graphite rod (the cathode) surrounded by a paste of manganese(IV) oxide, ammonium chloride, and carbon. Everything is encased in a zinc casing which acts as the anode.

A Mercury Battery

Diagram showing the top raised area of the battery acting as the cathode (made of steel) insulated from an outer casing of zinc which acts as the anode. Inside the casing is a paste of mercury(II) oxide in a basic medium of potassium hydroxide and zinc(II) hydroxide.

Schematic of the Hydrogen-Oxygen Fuel Cell

Hydrogen gas flows in through porous carbon from the left and oxygen gas flows in through porous carbon from the right. The gases react in the center forming water. A Wire connects the left and right porous carbon.


The corrosion of iron in a water droplet. The iron in contact with the water is acting as the anode while the iron around the water droplet is acting as the cathode.

Corrosion Prevention

Cathodic Protection

Buried iron pipes are protected from corrsion by connecting the pipe to a piece of magnesion. In the moist surrounding soil, the magnesium acts as an anode and the iron pipe acts as a cathode.


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

Producing Aluminum by the Hall-Heroult Process

Graphite rodes are inserted through the top of a container into a molten mixture of aluminum oxide and sodium aluminum fluoride. Electricity applied between the graphite rodes and the container causes the aluminum to be reduced to liquid aluminum which sinks to the bottom. A plug in the botton of the container can be opened to extract the liquid aluminum metal.

Electroplating a Spoon

A piece of silver and a spoon are in a solution of silver ions. They are connected to a power supply with the spoon connected to the cathode. Turning on the power supply plates solid silver onto the cathode (spoon).

The Downs Cell for the Electrolysis of Molten Sodium Chloride

A diagram showing an anode and a cathode in moten sodium chloride. Chlorine gas is produced at the anode and molten sodium is produced at the cathode from which is can be extracted.