Question

Consider the standard galvanic cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{aligned} $$

Short answer

The standard galvanic cell is based on the given copper and silver half-reactions with their respective standard reduction potentials \(E_{\textnormal{Cu}^{2+}/\textnormal{Cu}}^{\circ}=+0.34 \, \textnormal{V}\) and \(E_{\textnormal{Ag}^{+}/\textnormal{Ag}}^{\circ}=+0.80 \, \textnormal{V}\). The copper half-reaction acts as the anode (oxidation), while the silver half-reaction acts as the cathode (reduction). The cell potential is given by \(E_{\textnormal{cell}}^{\circ}=(+0.80 \, \textnormal{V})-(+0.34 \,\textnormal{V})=+0.46 \,\textnormal{V}\).

Step-by-step solution

Identify the half-reactions and their standard reduction potentials

First, we need to find the standard reduction potentials for the given half-reactions. You can look up these values in a standard reduction potentials table. The standard reduction potentials (E°) for the half-reactions are: $$ \begin{aligned} \ \\ \textnormal{Cu}^{2+}+2 e^- &\longrightarrow \textnormal{Cu} & E_{\textnormal{Cu}^{2+}/\textnormal{Cu}}^{\circ}&=+0.34 \,\textnormal{V} \\ \textnormal{Ag}^{+}+e^- &\longrightarrow \textnormal{Ag} & E_{\textnormal{Ag}^{+}/\textnormal{Ag}}^{\circ}&=+0.80 \,\textnormal{V} \end{aligned} $$

Determine the anode and cathode reactions

In a galvanic cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. The half-reaction with the lower standard reduction potential will be the anode (oxidation). In this case, the copper half-reaction has a lower reduction potential, so it will act as the anode. The silver half-reaction, with the higher reduction potential, will act as the cathode. Thus, we rewrite the anode half-reaction as: $$ \textnormal{Cu} \longrightarrow \textnormal{Cu}^{2+}+2 \textnormal{e}^{-} $$

Calculate the cell potential

The cell potential E°(cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode. We can use the following formula to calculate it: $$ E_{\textnormal{cell}}^{\circ}=E_{\textnormal{cathode}}^{\circ}-E_{\textnormal{anode}}^{\circ} $$ By plugging in the known values, we get: $$ E_{\textnormal{cell}}^{\circ}=(+0.80 \,\textnormal{V})-(+0.34 \,\textnormal{V})=+0.46 \,\textnormal{V} $$ The cell potential for this standard galvanic cell is +0.46 V.

Key concepts

Understanding Standard Reduction Potentials

When delving into the world of electrochemistry, the concept of standard reduction potentials is critical. These potentials, often denoted as E°, are used to measure the tendency of a species to gain electrons, or reduce, under standard conditions, which are 25°C, 1 M concentration for each ion participating in the reaction, and a pressure of 1 atm for gases.

Standard reduction potentials are listed in tables for half-reactions, which show a reagent gaining electrons. A higher E° value indicates a greater likelihood of the species being reduced. In turn, these potentials are instrumental in predicting the direction of electron flow in a galvanic cell. To understand and predict galvanic cell outcomes, students should familiarize themselves with this concept and how to use standard reduction potential tables.

Anode and Cathode Reactions in Galvanic Cells

The heart of a galvanic cell lies in the anode and cathode reactions. In such cells, we have two separate half-reactions – oxidation at the anode and reduction at the cathode. The distinction between the two is straightforward: oxidation corresponds to the loss of electrons, while reduction is the gain of electrons.

Determining which reaction occurs at which electrode requires a look at the standard reduction potentials. The substance with the lower reduction potential will undergo oxidation at the anode; conversely, the substance with the higher potential will undergo reduction at the cathode. This concept is paramount when constructing a galvanic cell or trying to understand its operation. It helps predict the flow of electrons and the direction of chemical reactions.

Calculating Cell Potential

To harness the power of a galvanic cell, we must know how to perform a cell potential calculation. The cell potential, E°(cell), signifies the electromotive force of the cell — essentially, it measures how much voltage the cell can produce. It is calculated by subtracting the anode's standard reduction potential from the cathode's:
\( E_{cell}^{\boldsymbol{\theta}} = E_{cathode}^{\boldsymbol{\theta}} - E_{anode}^{\boldsymbol{\theta}} \)

For a positive cell potential, the reaction is spontaneous, indicating that electrons will spontaneously flow from anode to cathode. Remember, a negative cell potential implies a non-spontaneous reaction under standard conditions. While this formula seems straightforward, it's important for learners to practice this calculation regularly using various half-reactions to become adept at predicting cell behavior.