Problem 41 - Chemistry The Central Science In Si Units

Question

A \(1 M\) solution of \(\mathrm{AgNO}_{3}\) is placed in a beaker with a strip of Ag metal. A \(1 \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a second beaker with a strip of Cu metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass, and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

Step-by-Step Solution

Answer

The anode is Cu and the cathode is Ag. The Ag electrode gains mass, and the Cu electrode loses mass. The overall reaction is \( \mathrm{Cu}(s) + 2 \mathrm{Ag}^+ (aq) \rightarrow \mathrm{Cu}^{2+} (aq) + 2\mathrm{Ag}(s) \), and the cell emf is +0.46 V.

Step 1: Identify the Half-Reactions

To identify the anode and cathode, start by writing the reduction half-reactions for each electrode. For Ag: \( \mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag} \), and for Cu: \( \mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu} \). The electrode with the less positive standard reduction potential becomes the anode.

Step 2: Determine Anode and Cathode

The standard reduction potential for \( \mathrm{Ag}^+ \) is \(+0.80\,\text{V}\) and for \( \mathrm{Cu}^{2+} \), it is \(+0.34\,\text{V}\). Cu has a less positive reduction potential, so it serves as the anode where oxidation occurs, while Ag serves as the cathode where reduction takes place.

Step 3: Determine Mass Changes at Electrodes

At the cathode (Ag), the \( \mathrm{Ag}^+ \) ions in solution gain electrons and deposit as silver metal, increasing mass. At the anode (Cu), copper metal is oxidized to \( \mathrm{Cu}^{2+} \) ions, decreasing mass.

Step 4: Write the Overall Cell Reaction

Combine the half-reactions for the full cell equation: \( \mathrm{Cu}(s) + 2 \mathrm{Ag}^+ (aq) \rightarrow \mathrm{Cu}^{2+} (aq) + 2\mathrm{Ag}(s) \). The electrons balance out between the oxidation and reduction half-reactions.

Step 5: Calculate the Cell's EMF

To find the electromotive force (emf), use the equation: \( E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \). Substituting the known values, \( E_{\text{cell}} = +0.80\,\text{V} - (+0.34\,\text{V}) = +0.46\,\text{V} \). This is the standard emf for the cell.

Key Concepts

Redox Reactions

Electrochemical cells function based on redox reactions, which involve the transfer of electrons from one species to another. These reactions are broken down into two separate half-reactions, each occurring at one electrode.
The two types of half-reactions are oxidation, which occurs at the anode, and reduction, which takes place at the cathode. Oxidation is the loss of electrons, while reduction is the gain of electrons. In our example with the silver and copper electrodes, copper loses electrons (oxidation at the anode), and silver ions gain those electrons (reduction at the cathode).
We can split the overall reaction into:

• Reduction (at the cathode): \( \mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag} \)
• Oxidation (at the anode): \( \mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(aq) + 2\mathrm{e}^- \)

These half-reactions together form the overall redox reaction in the cell: \( \mathrm{Cu}(s) + 2 \mathrm{Ag}^+ (aq) \rightarrow \mathrm{Cu}^{2+} (aq) + 2\mathrm{Ag}(s) \).
Here, silver ions are reduced, and copper is oxidized, showcasing the foundational principle of redox reactions driving the electrochemical process.

Standard Reduction Potential

The concept of standard reduction potential is crucial for identifying the anode and cathode in electrochemical cells. It refers to the tendency of a chemical species to be reduced, measured under standard conditions.
Each half-reaction has a standard reduction potential measured in volts (V). These values allow us to predict which species will undergo reduction or oxidation.
In our cell example, the standard reduction potential for silver ions (\(\mathrm{Ag}^+\)) is \(+0.80\,\text{V}\), while for copper ions (\(\mathrm{Cu}^{2+}\)), it is \(+0.34\,\text{V}\).
With these values,

• The species with the higher (more positive) value is reduced. Silver, with \(+0.80\,\text{V}\), is reduced at the cathode.
• The species with the lower value, copper with \(+0.34\,\text{V}\), is oxidized at the anode.

This hierarchy allows chemists to deduce the flow of electrons in an electrochemical cell and therefore design the cell effectively.

Electromotive Force

Electromotive Force (EMF) is the driving force of electron flow in an electrochemical cell. Also known as cell potential, it represents the voltage between two half-cells under standard conditions.
Calculating the EMF involves subtracting the standard reduction potential of the anode from that of the cathode. This gives us the net potential difference that drives the redox reaction.
For our electrochemical cell:

• Cathode (silver): \(+0.80\,\text{V}\)
• Anode (copper): \(+0.34\,\text{V}\)

The EMF is calculated using the formula: \[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]Plugging in the numbers, \[ E_{\text{cell}} = +0.80\,\text{V} - (+0.34\,\text{V}) = +0.46\,\text{V} \]This means the cell can produce a potential difference of \(+0.46\,\text{V}\) under standard conditions.
Understanding EMF helps gauge how efficiently an electrochemical cell can do work, essentially determining its usefulness in applications such as batteries and galvanic cells.