Question

Magnesium metal is oxidized and silver ions are reduced in a voltaic cell using \(\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) | \mathrm{Mg}\) and \(\mathrm{Ag}^{+}(\mathrm{aq}, 1 \mathrm{M}) |\) Ag half-cells. (IMAGE NOT COPY) (a) Label each part of the cell. (b) Write equations for the half-reactions occurring at the anode and the cathode, and write an equation for the net reaction in the cell. (c) Trace the movement of electrons in the external circuit. Assuming the salt bridge contains \(\mathrm{NaNO}_{3},\) trace the movement of the \(\mathrm{Na}^{+}\) and \(\mathrm{NO}_{3}^{-}\) ions in the salt bridge that occurs when a voltaic cell produces current. Why is a salt bridge required in a cell?

Short answer

Anode: Mg oxidizes, Cathode: Ag reduces, Net: \(\mathrm{Mg} + 2\mathrm{Ag}^+ \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{Ag}\), Electrons flow from Mg to Ag, Na+ to cathode, NO3- to anode for neutrality.

Step-by-step solution

Label Each Part of the Cell

In a voltaic cell, there are two half-cells. Here, we have the \(\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) | \mathrm{Mg}\) half-cell and the \(\mathrm{Ag}^{+}(\mathrm{aq}, 1 \mathrm{M}) | \mathrm{Ag}\) half-cell. The magnesium half-cell acts as the anode where oxidation occurs. The silver half-cell acts as the cathode where reduction occurs. Electrons flow from the anode to the cathode through the external circuit.

Write Half-Reactions and Net Equation

The oxidation half-reaction at the anode is \(\mathrm{Mg}(s) \rightarrow \mathrm{Mg}^{2+}(aq) + 2e^-\). The reduction half-reaction at the cathode is \(\mathrm{Ag}^+(aq) + e^- \rightarrow \mathrm{Ag}(s)\). Since two electrons are produced in magnesium oxidation and only one is consumed in silver reduction, the reduction equation is multiplied by two: \(2\mathrm{Ag}^+(aq) + 2e^- \rightarrow 2\mathrm{Ag}(s)\). The net reaction is \(\mathrm{Mg}(s) + 2\mathrm{Ag}^+(aq) \rightarrow \mathrm{Mg}^{2+}(aq) + 2\mathrm{Ag}(s)\).

Trace Electron Movement

Electrons produced at the anode move through the external circuit to the cathode. This movement from magnesium to silver ions facilitates the redox process by allowing electrons to perform work in an external device, like lighting a bulb.

Movement of Ions in Salt Bridge

In the salt bridge filled with \(\mathrm{NaNO}_3\), \(\mathrm{Na}^+\) ions move to the cathode compartment to balance the charge as silver cations are reduced and plated out as solid silver. \(\mathrm{NO}_3^-\) ions move to the anode compartment to balance the charge as magnesium atoms lose electrons to form \(\mathrm{Mg}^{2+}\) ions. The salt bridge maintains the cell's electrical neutrality and allows ions to flow, preventing charge buildup that would halt the reaction.

Role of the Salt Bridge

A salt bridge is necessary to maintain electrical neutrality in the half-cells. As electrons move from the anode to the cathode, the salt bridge compensates for ion imbalance by allowing ions to flow between the two half-cells. This keeps the reaction running by preventing charge buildup.

Key concepts

Electrode Reactions

In a voltaic cell, a chemical reaction takes place involving the transfer of electrons. This process is divided into two parts called half-reactions. These occur at the two electrodes, the anode and the cathode.

- **Anode Reaction**: At the anode, oxidation occurs. This means that a substance loses electrons. In our example, this is magnesium (Mg), which transforms into magnesium ions (\(\text{Mg}^{2+}\)), releasing two electrons in the process: \(\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-\).
- **Cathode Reaction**: Conversely, at the cathode, reduction occurs. A substance gains electrons. In the voltaic cell we're considering, silver ions (\(\text{Ag}^+\)) gain electrons to become solid silver: \(\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\). Since magnesium releases two electrons, this reaction happens twice to balance the electrons: \(2\text{Ag}^+(aq) + 2e^- \rightarrow 2\text{Ag}(s)\).

These reactions contribute to the electrical potential that enables the cell to conduct electricity. The combined equation for the full process is \(\text{Mg}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Ag}(s)\).

Electron Flow

The flow of electrons is fundamental to the operation of a voltaic cell. Electrons are transferred due to the events occurring at the electrodes.

- **Direction of Electron Flow**: In our cell, electrons flow from the magnesium electrode, the anode, through the external circuit, to the silver electrode, which is the cathode.
- **External Circuit**: The flow through this circuit can be harnessed to do work, such as lighting up a bulb. This is what powers devices in practical applications.

The movement of electrons completes the circuit between the anode and cathode, repeatedly running the reactions. This flow is vital as it allows the voltaic cell to perform its desired function.

Ion Movement in Salt Bridge

A salt bridge is a crucial component in maintaining the operation of a voltaic cell by enabling ion flow between the two half-cells. It's typically filled with a salt solution like sodium nitrate (\(\text{NaNO}_3\)).

- **Balancing Charges**: As electrons travel from the anode to the cathode, ions in the salt bridge equalize the charge build-up that occurs in the half-cells.
- **Movement of Ions**: In our voltaic cell: - \(\text{Na}^+\) ions move toward the cathode, countering the reduction of \(\text{Ag}^+\) ions to solid silver. - \(\text{NO}_3^-\) ions shift towards the anode, compensating for the creation of \(\text{Mg}^{2+}\) ions due to magnesium losing electrons.

**Role of the Salt Bridge**: The salt bridge prevents the build-up of excess positive or negative charges, which would otherwise slow and eventually stop the cell’s operation. It ensures the reaction continues by maintaining electrical neutrality, effectively allowing the cell to function without interruption.