Question

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array} $$

Short answer

The overall cell potential for the galvanic cell based on the given half-reactions is \(E_{cell}^° = 1.84\,\text{V}\), with gold (Au) acting as the cathode and thallium (Tl) as the anode.

Step-by-step solution

Identify the cathode and the anode

First, determine which half-reaction is taking place at the cathode (reduction) and which is taking place at the anode (oxidation). The substance with the higher standard reduction potential value will act as the cathode, as it has a stronger tendency to gain electrons and undergo reduction. In this case, gold (Au) has a higher standard reduction potential (\(\mathscr{E}^\circ = 1.50 V\)) than thallium (Tl), which has a standard reduction potential of \(\mathscr{E}^\circ = -0.34 V\). Thus, the cathode half-reaction is: $$ \mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}; \quad \mathscr{E}^{\circ} = 1.50 \mathrm{~V} $$ And the anode half-reaction is: $$ \mathrm{Tl}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{Tl}; \quad \mathscr{E}^{\circ} = -0.34 \mathrm{~V} $$

Calculate the overall cell potential

Now that we have identified the cathode and the anode, we can use their respective standard reduction potentials to calculate the overall cell potential: $$ E_{cell}^° = E_{cathode}^° - E_{anode}^° = E_{Au}^° - E_{Tl}^° $$ Plug in the given values for the standard reduction potentials and calculate the cell potential: $$ E_{cell}^° = 1.50\,\text{V} - (-0.34\,\text{V}) = 1.50\,\text{V} + 0.34\,\text{V} = 1.84\,\text{V} $$ Therefore, the overall cell potential for this galvanic cell is \(1.84\,\text{V}\).

Key concepts

Standard Reduction Potential

In the world of chemistry, the standard reduction potential is a critical concept, especially when dealing with galvanic cells. It is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. The standard reduction potential is usually expressed in volts (V) and is measured under standard conditions: 1 M concentration for each ion, a partial pressure of 1 atm for gases, and a temperature of 25 °C.

These values are crucial for predicting the outcome of electrochemical processes:

• Higher standard reduction potential: Indicates a greater likelihood of the species to be reduced. This is because it has a stronger attraction for electrons.
• Lower standard reduction potential: Suggests the species is less likely to be reduced and more likely to donate electrons during the reaction.

In the provided galvanic cell example, gold (\(\mathrm{Au}^{3+}\)) has a standard reduction potential of 1.50 V, which means it has a strong tendency to gain electrons compared to thallium (\(\mathrm{Tl}^+\)), which has a standard reduction potential of -0.34 V.
Identifying the cathode and anode rely heavily on comparing these potential values.

Cathode and Anode

In a galvanic cell, understanding the role of the cathode and anode is vital. These are the two electrodes where different half-reactions occur. The electrode where reduction takes place is called the cathode, while the site of oxidation is called the anode:

• Cathode: The electrode with the higher standard reduction potential. It gains electrons, as it is the site for the reduction process.
• Anode: The electrode with the lower standard reduction potential. It loses electrons, as oxidation occurs here.

For the example given, with the reactions:\[\mathrm{Au}^{3+} + 3 \, \mathrm{e}^{-} \longrightarrow \mathrm{Au}\]\[\mathrm{Tl}^{+} + \mathrm{e}^{-} \longrightarrow \mathrm{Tl}\]Gold serves as the cathode as it has a higher standard reduction potential (\(1.50 \, \mathrm{V}\)), making it more favorable for reduction. Conversely, thallium acts as the anode because of its lower reduction potential value (\(-0.34 \, \mathrm{V}\)).
Identifying which is the cathode and which is the anode is the first step in understanding how the galvanic cell functions as a whole.

Overall Cell Potential

The overall cell potential is a crucial indicator of the voltage that a galvanic cell can provide under standard conditions. It indicates the energy available to drive the cell's electron flow. To calculate it, we use the standard reduction potentials of both the cathode and the anode:\[E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ\]This equation shows that the overall cell potential is essentially the difference between the cathode's and anode's potentials. If it is positive, the reaction is spontaneous and the cell can do the work.

For the example in the exercise:

• The cathode reaction is for gold, with an \(E^\circ = 1.50 \, \mathrm{V}\) .
• The anode reaction is for thallium, with an \(E^\circ = -0.34 \, \mathrm{V}\).
• Plug these values into the formula:

\[E_{cell}^\circ = 1.50 \, \mathrm{V} - (-0.34 \, \mathrm{V}) = 1.84 \, \mathrm{V}\]Therefore, the overall cell potential for this galvanic cell is \(1.84 \, \mathrm{V}\), signifying a spontaneous process.