Question

Write the oxidation and reduction half-reactions for a silver-chromium voltaic cell. Identify the anode, cathode, and the direction of electron flow

Short answer

In the silver-chromium voltaic cell, the reduction half-reaction at the cathode is: \(Ag^+ + e^- \rightarrow Ag\), and the oxidation half-reaction at the anode is: \(Cr \rightarrow Cr^{3+} + 3e^-\). The cathode is the silver ion (Ag+), the anode is chromium (Cr), and the direction of electron flow is from the anode (chromium) to the cathode (silver).

Step-by-step solution

Write the half-reactions in their reduction form

Write the reduction half-reactions for silver ions (Ag+) and chromium ions (Cr3+) incorporating their respective ions and solid-state element: For silver: \(Ag^+ + e^- \rightarrow Ag\) For chromium: \(Cr^{3+} + 3e^- \rightarrow Cr\)

Determine the anode and cathode using standard reduction potentials

The standard reduction potential for the silver ion half-reaction is +0.80 V, while that of the chromium ion half-reaction is -0.74 V. In a voltaic cell, the half-reaction with the higher standard reduction potential occurs at the cathode (reduction) and the half-reaction with the lower standard reduction potential occurs at the anode (oxidation). Therefore, Cathode: \(Ag^+ + e^- \rightarrow Ag\) Anode: \(Cr^{3+} + 3e^- \rightarrow Cr\)

Write the oxidation half-reaction for the anode

Convert the anode half-reaction from its reduction form to its oxidation form by reversing the reaction: Anode oxidation half-reaction: \(Cr \rightarrow Cr^{3+} + 3e^-\)

Summarize the results

In the silver-chromium voltaic cell, the cathode (reduction) half-reaction is: \(Ag^+ + e^- \rightarrow Ag\) The anode (oxidation) half-reaction is: \(Cr \rightarrow Cr^{3+} + 3e^-\) The direction of electron flow is from the anode (chromium) to the cathode (silver).

Key concepts

Oxidation Half-Reaction

In a voltaic cell, oxidation involves the loss of electrons from a substance, which occurs at the anode. For a silver-chromium voltaic cell, you must convert the chromium reduction half-reaction into an oxidation half-reaction. Initially, to identify the oxidation half-reaction, you reverse the reduction half-reaction of chromium. Originally, chromium is represented as:

• \(Cr^{3+} + 3e^- \rightarrow Cr\)

This shows chromium ions gaining electrons. To express it in the context of oxidation, reverse it to become:

• \(Cr \rightarrow Cr^{3+} + 3e^-\).

Here, solid chromium loses three electrons to become chromium ions, effectively undergoing oxidation.Oxidation involves both gaining a positive charge and losing electrons because when electrons are removed, the atom’s positive charge increases due to the presence of more protons than electrons.

Reduction Half-Reaction

Reduction is the process where a substance gains electrons, occurring at the cathode in a voltaic cell. Within the silver-chromium cell, the reduction process hinges on silver ions gaining electrons to become silver metal. It’s represented by the half-reaction:

• \(Ag^+ + e^- \rightarrow Ag\)

Here, a silver ion accepts one electron to transform into metallic silver.Remember that the core of reduction is acquiring electrons, resulting in a decrease in oxidation state. This process aligns with an increase in the number of electrons a substance contains.

Electron Flow

Electrons naturally flow from a region of higher potential energy to lower potential energy. In a voltaic cell, electrons move from the anode to the cathode. For a voltaic cell with silver and chromium, the electron flow initiates at the chromium (anode). The conversion of chromium metal to chromium ions incites the release of electrons. These electrons travel through the circuit to reach silver, enabling the transformation of silver ions into solid silver at the cathode. Thus, the journey of electrons spans from oxidation to reduction sites, specifically from chromium to silver in this voltaic setup. The clear path of electrons ensures that the energy within the cell is harnessed effectively to perform electrical work.

Cathode

In electrochemical cells, the cathode is the site of reduction where electron gain transpires. For the silver-chromium voltaic cell, silver serves as the cathode. This choice is guided by the higher reduction potential of the silver reaction:

• \(Ag^+ + e^- \rightarrow Ag\).

The role of the cathode is essential for completing the circuit, enabling the cell to generate a sustained electrical current. Typically, the cathode is linked to the positive terminal of the energy source. It draws electrons from the external circuit,completing electron movement and continuing the silver ions' conversion into silver metal.

Anode

The anode in a voltaic cell is where oxidation occurs, which means electron loss takes place. For the silver-chromium cell, chromium is the anode. This determination is based on having a lower standard reduction potential for chromium compared to silver:

• \(Cr \rightarrow Cr^{3+} + 3e^-\).

In this reaction, chromium metal loses electrons to form chromium ions. The anode's function is crucial because it serves as the electron source for the cell, releasing electrons that travel through the circuit.In voltaic systems, the anode is often connected to the negative terminal, depicted as the beginning point for electron flow. It maintains the cycle of electron exchange, fueling much of the reactions within the cell.