Question

Consider the following galvanic cell: a. Label the reducing agent and the oxidizing agent, and describe the direction of the electron flow. b. Determine the standard cell potential. c. Which electrode increases in mass as the reaction proceeds, and which electrode decreases in mass?

Short answer

The galvanic cell is: Zn(s) | Zn^(2+)(aq) || Cu^(2+)(aq) | Cu(s). The reducing agent is Zn and the oxidizing agent is Cu^(2+). Electrons flow from Zn to Cu^(2+), with Zn being the anode and Cu^(2+) the cathode. The standard cell potential is 1.10 V. As the reaction proceeds, the mass of the Zn electrode decreases, while the mass of the Cu electrodes increases.

Step-by-step solution

1. Identify the redox reactions

To identify the reducing agent and the oxidizing agent, we must first be able to recognize the half-reactions occurring in the galvanic cell. The half-reactions can be found in a table of standard reduction potentials. Suppose the given galvanic cell is: Zn(s) | Zn^(2+)(aq) || Cu^(2+)(aq) | Cu(s) Zn -> Zn^(2+) + 2e^(-) (oxidation half-reaction) Cu^(2+) + 2e^(-) -> Cu(s) (reduction half-reaction)

2. Identify the reducing agent and the oxidizing agent

The reducing agent is the species that loses electrons (gets oxidized) and the oxidizing agent is the species that gains electrons (gets reduced). Therefore, for this galvanic cell: Reducing Agent: Zn Oxidizing Agent: Cu^(2+)

3. Determine the direction of electron flow

In a galvanic cell, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). Anode: Zn Cathode: Cu^(2+) Thus, the direction of electron flow is from Zn to Cu^(2+).

4. Determine the standard cell potential

To determine the standard cell potential, we use the standard reduction potentials of the half-reactions. The standard cell potential (E° cell) is found by subtracting the standard potential of the anode (E° oxidation) from the standard potential of the cathode (E° reduction): E° cell = E° reduction - E° oxidation Suppose the standard reduction potentials are: E° (Zn^(2+)/Zn) = -0.76 V E° (Cu^(2+)/Cu) = 0.34 V E° cell = E° (Cu^(2+)/Cu) - E° (Zn^(2+)/Zn) = 0.34 V - (-0.76 V) = 1.10 V

5. Determine the change in mass of the electrodes

As the reaction proceeds, the anode loses mass due to the loss of Zn atoms, which are transformed into Zn^(2+) ions. Conversely, the cathode gains mass as Cu^(2+) ions gain electrons and deposit as Cu atoms: Anode (Zn) - Decreases in mass Cathode (Cu^(2+)) - Increases in mass

Key concepts

Standard Cell Potential

The standard cell potential, commonly denoted as \( E^\circ_{\text{cell}} \), is a crucial concept in electrochemistry, which reflects the voltage difference between two electrodes when no current is flowing and conditions are at standard state (1 M concentration and 25°C). In essence, it measures the driving force behind an electrochemical reaction.

To compute the standard cell potential, we take the standard reduction potential of the cathode, which is where reduction or the gain of electrons occurs, and subtract the standard reduction potential of the anode, where oxidation takes place. If we take our earlier example with zinc and copper, where the potentials are \( E^\circ (\text{Cu}^{2+}/\text{Cu}) = 0.34 \) V and \( E^\circ (\text{Zn}^{2+}/\text{Zn}) = -0.76 \) V, their difference, \( 1.10 \) V, is the standard cell potential. This value indicates that the cell can perform work and that the reaction is spontaneous under standard conditions.

Reducing and Oxidizing Agents

A reducing agent is the chemical that donates electrons in a reaction and gets oxidized. An oxidizing agent, on the other hand, accepts electrons and is itself reduced. It's important to identify these agents because they dictate the flow of electrons in the cell.

In our example galvanic cell, zinc undergoes oxidation, meaning it loses electrons, and hence, it is the reducing agent. Conversely, copper(II) ions gain those electrons during reduction, making them the oxidizing agent. Simplifying these concepts, the reducing agent 'reduces' the other substance by giving away electrons, while the oxidizing agent 'oxidizes' the other by taking electrons. Recognizing these agents allows for a clearer understanding of redox chemistry and the roles each species plays in an electrochemical cell.

Electron Flow Direction

The electron flow direction is fundamental in understanding how galvanic cells work. Electrons move from the anode to the cathode, driving the cell's electrical current. The anode is where oxidation occurs, and the cathode is the site of reduction.

With the zinc and copper cell example, zinc is the anode; so, electrons flow from the zinc to the copper(II) ions at the cathode. This movement is driven by the cell potential, which results from a difference in energy levels between electrons in the anode and cathode materials. The electrons travel through an external circuit, providing electrical energy that can be harnessed to do work.

Change in Electrode Mass

During a galvanic cell's operation, one observes changes in electrode mass due to the redox reactions. The anode, which loses electrons through oxidation, decreases in mass because atoms leave the electrode and enter the solution as ions. Simultaneously, the cathode gains mass because ions in solution gain electrons and are deposited as solid metal atoms onto the electrode.

So, in our zinc-copper cell, zinc atoms at the anode release electrons and turn into \( \text{Zn}^{2+} \) ions, causing the anode to lose mass. In contrast, at the copper cathode, \( \text{Cu}^{2+} \) ions in solution accept electrons, become copper atoms, and increase the mass of the cathode. These processes of mass change are a physical manifestation of the chemical reactions taking place in the cell.