Galvanic Cell with Inert Electrodes
HSC Chemistry Syllabus
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Conduct investigations to measure and compare the reaction potential of galvanic half-cells
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Predict the spontaneity of redox reactions using the value of cell potentials (ACSCH079, ACSCH080)
Inert Electrodes in Galvanic Cells Explained
What is an Inert Electrode?
Inert electrodes are substances that do not participate directly in the redox (reduction-oxidation) reactions within a galvanic cell. Unlike active electrodes, which are involved in the chemical changes of the cell, inert electrodes simply provide a surface for the transfer of electrons. The most common inert electrodes are platinum and graphite because of their excellent conductivity and chemical stability.
In a typical galvanic cell, you might see metals like zinc and copper acting as active electrodes. These metals participate in the redox reactions by either losing or gaining electrons. However, there are cases where the redox reaction involves species that are not solid metals—such as ions in solution or gases like hydrogen and chlorine. In these situations, an inert electrode is necessary because there is no solid material to serve as the electrode.
Example of an Inert Electrode in a Galvanic Cell
Consider the standard chlorine electrode, which consists of a platinum electrode in contact with `Cl_2` gas and a solution containing `Cl^-` ions. Neither chlorine gas and chloride ions are solids under stand conditions and thus cannot act as electrodes to facilitate the transfer of electrons.
The reaction occurring at this electrode can either be oxidation or reduction:
$$\text{Oxidation: } 2Cl^- \rightarrow Cl_2 + 2e^-$$
$$\text{Reduction: } Cl_2 + 2e^- \rightarrow 2Cl^-$$
Here, platinum acts as the inert electrode. It does not participate in the chemical reaction; instead, it provides a surface for chlorine gas molecules to gain electrons and form chloride ions in the reduction reaction.
Consider a galvanic cell which consists of nickel/nickel ion half-cell and chlorine gas / chloride ion half-cell. The chlorine gas half-cell requires an inert electrode e.g. platinum to conduct electrodes.
The reduction potential of chlorine gas is greater than that of nickel ions, therefore nickel metal undergoes oxidation.
Oxidation half-reaction:
$$Ni(s) \rightarrow Ni^2+(aq) + 2e^–$$
Reduction half-reaction:
$$Cl_2(g) + 2e^– \rightarrow 2Cl^–(aq)$$
In this example, the platinum electrode is the cathode as it is in the reduction half-cell. The platinum allows for the electrons released by the oxidation of nickel to travel to the cathode half-cell and reduce the `Cl_2` into `Cl^–` which otherwise would not be possible.
As the reaction proceeds, the nickel nitrate solution will become more green due to an increase in nickel ion concentration.
If an inert electrode is present in a galvanic cell, it is always included in the short-hand notation to represent the cell.
E.g. the notation for the galvanic cell above would be:
$$Ni(s) | Ni^{2+}(aq) || Cl^–(aq) | Cl_2(g) | Pt(s)$$
Example of Two Inert Electrodes in Galvanic Cells
In galvanic cells whereby neither oxidation nor reduction reactions contain chemical species suitable for conducting electricity, more than one inert electrode may be required.
Consider the following redox reactions:
$$2I^-(aq) \rightarrow I_2(s) + 2e^-$$
$$MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l)$$
In the oxidation reaction, iodine solid is a covalent substance with poor electrical conductivity. The remaining species in the redox reactions are aqueous and therefore cannot offer a suitable surface to conduct flow of electrons.
This is why two inert electrodes e.g. graphite rods need to be submerged in the electrolyte solutions shown above. The oxidation of iodide ions forms iodine solid on the surface of the graphite anode (left), and the associated electrons are then transferred to the graphite cathode (right) where they are then gained by permanganate ions in reduction.
Practice Examples
Example 1: write half-cell equations for this galvanic cell and calculate its standard potential
Standard potential:
$$2H^+(aq) + 2e^– \rightarrow H_2(g) \hspace{1 cm} E^{\circ} = 0.00 \text{ V}$$
$$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^– \hspace{1 cm} E^{\circ} = 0.76 \text{ V}$$
$$\text{Total potential: } 0.00 + 0.76 = +0.76 \text{ V}$$
Zinc metal electrode is the anode as its oxidation to form zinc ions will provide a positive potential for the galvanic cell to be spontaneous.
Notation:
$$Zn(s) | Zn^{2+} (1 M) || H^+ (1 M) | H_2(g) | Pt(s)$$
Example 2: Write half-cell equations for this galvanic cell and calculate its standard potential
Standard potential:
$$H_2(g) \rightarrow 2H^+(aq) + 2e^– \text{ E = 0.00 V }$$
$$Fe^{3+}(aq) + e^– \rightarrow Fe^{2+}(aq) \text{ E = 0.77 V}$$
$$\text{Total potential: } 0.00 + 0.77 = 0.77 \text{ V}$$
In this example, the hydrogen gas half cell is the oxidation half-cell as the reduction of `Fe^{3+}` provides a positive potential.
Notation:
$$Pt(s) | H^+ (1M) | H_2(g) || Fe^{3+} (1 M) , Fe^{2+} (1 M) | Pt(s)$$
Example 3: Write the half-cell equations for this galvanic cell and calculate its standard potential
Standard Potential:
$$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^– \hspace{1 cm} E^{\circ} = 0.76 \text{ V}$$
$$Fe^{3+}(aq) + e^– \rightarrow Fe^{2+}(aq) \hspace{1 cm} E^{\circ} = 0.77 \text{ V}$$
$$\text{Total potential: } 0.76 + 0.77 = 1.53 \text{ V}$$
Notation:
$$Zn(s) | Zn^{2+} (1 M) || Fe^{3+} (1 M) , Fe^{2+} (1 M) | Pt(s)$$
Example 4: The following reaction takes place in an electrochemical cell.
$$Cu(s) | Cu^{2+} (1 M) || Fe^{3+} (1 M), Fe^{2+} (1 M) | Pt(s)$$
Draw a diagram to represent this galvanic cell.
(Refer to Video)