Problem 51 - Chemistry and Chemical Reactivity

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.(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?

Step-by-Step Solution

Answer

Anode: Mg, Cathode: Ag. Net reaction: \( \mathrm{Mg} + 2\mathrm{Ag}^{+} \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{Ag} \). Electrons flow anode to cathode; Na\(^+\) to cathode, NO\(_3^-\) to anode via salt bridge.

Step 1: Identify Cell Components

In the voltaic cell, - Anode is where oxidation occurs. For this cell, the magnesium metal (\( \mathrm{Mg} \)) is oxidized to magnesium ions (\( \mathrm{Mg}^{2+} \)). So, the anode is the magnesium electrode.- Cathode is where reduction occurs. Here, silver ions (\( \mathrm{Ag}^{+} \)) are reduced to silver metal. Thus, the cathode is the silver electrode.

Step 2: Write Half-Reaction Equations

Anode (oxidation): \[\mathrm{Mg} (s) \rightarrow \mathrm{Mg}^{2+} (aq) + 2 \mathrm{e}^{-} \]Cathode (reduction): \[\mathrm{Ag}^{+} (aq) + \mathrm{e}^{-} \rightarrow \mathrm{Ag} (s) \]Balance the reduction half-reaction by multiplying by 2:\[2 \mathrm{Ag}^{+} (aq) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag} (s) \]

Step 3: Write Net Cell Reaction Equation

Combine the balanced half-reactions to get the net reaction:\[\mathrm{Mg} (s) + 2 \mathrm{Ag}^{+} (aq) \rightarrow \mathrm{Mg}^{2+} (aq) + 2\mathrm{Ag} (s) \]

Step 4: Trace Electrons in External Circuit

Electrons flow from the anode (where they're released during oxidation of Mg) to the cathode (where they are used in the reduction of Ag\(^+\) ions) through the external circuit.

Step 5: Describe Ion Movement in Salt Bridge

In the salt bridge containing NaNO\(_3\), Na\(^+\) ions move towards the cathode to balance the negative charge, while NO\(_3^-\) ions move towards the anode to balance the positive charge.

Step 6: Explain the Purpose of Salt Bridge

The salt bridge maintains electrical neutrality by allowing ions to flow between the two half-cells, preventing the cell from rapidly reaching equilibrium and stopping the electron flow.

Key Concepts

Anode and Cathode Reactions

In a voltaic cell, the anode and cathode are crucial components where specific chemical reactions occur. The electrode where oxidation happens is called the anode. The magnesium (\( \mathrm{Mg} \)) serves as the anode in this particular cell, as it loses electrons and transforms into magnesium ions (\( \mathrm{Mg}^{2+} \)).The reaction at the anode can be represented as:

• \( \mathrm{Mg} (s) \rightarrow \mathrm{Mg}^{2+} (aq) + 2 \mathrm{e}^{-} \)

Conversely, the cathode is where the reduction process takes place. Here, the silver ions (\( \mathrm{Ag}^{+} \)) gain electrons to form silver metal.This reaction is depicted as:

• \( 2 \mathrm{Ag}^{+} (aq) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag} (s) \)

These half-reactions are essential as they demonstrate the electron exchange and the chemical transformations occurring in the voltaic cell.

Electron Flow

Electrons play a vital role in the function of a voltaic cell by moving through the external circuit. As the oxidation process happens at the anode, electrons are released. These electrons then travel through the conductive path to the cathode. This flow of electrons is what constitutes electric current, and it occurs continuously as long as the cell operates. The pathway can be traced as:

• From the magnesium anode: electrons are released during oxidation.
• Through the external circuit: electrons travel providing electric current.
• To the silver cathode: electrons are consumed in the reduction of silver ions.

The consistent flow of electrons is what powers the external circuit and is crucial for the cell's function.

Salt Bridge

In the design of a voltaic cell, the salt bridge serves a fundamental purpose. It helps maintain electrical neutrality by allowing ions to circulate between the two half-cells. This bridge typically contains an inert electrolyte like \( \mathrm{NaNO}_3 \).Here’s how it works:

• \( \mathrm{Na}^{+} \) ions move towards the cathode, balancing the negative charges as electrons arrive there.
• \( \mathrm{NO}_{3}^{-} \) ions move towards the anode to balance the positive charges from the departure of electrons.

Without the salt bridge:- The cell would quickly reach a state of equilibrium.- The flow of electrons would be hindered, stopping the cell's power generation.Therefore, the salt bridge is essential in maintaining the voltaic cell's continuous operation.

Oxidation and Reduction Reactions

The principle of oxidation and reduction reactions is at the heart of a voltaic cell’s operation. These reactions are collectively called redox reactions, which involve the transfer of electrons between substances.Oxidation reaction occurs when a substance loses electrons:

• In this case, magnesium (\( \mathrm{Mg} \)) loses electrons to form magnesium ions (\( \mathrm{Mg}^{2+} \)).

Reduction reaction is when a substance gains electrons:

• Silver ions (\( \mathrm{Ag}^{+} \)) gain electrons to become silver (\( \mathrm{Ag} \)) metal.

The net reaction combines both half-reactions:

• \( \mathrm{Mg} (s) + 2 \mathrm{Ag}^{+} (aq) \rightarrow \mathrm{Mg}^{2+} (aq) + 2 \mathrm{Ag} (s) \)

Understanding these processes helps in grasping how energy is transported and transformed in electrochemical cells.