Question

A \(1 \mathrm{M}\) solution of \(\mathrm{AgNO}_{3}\) is placed in a beaker with a strip of Ag metal. A \(1 M\) 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?

Short answer

The anode is the copper electrode and the cathode is the silver electrode. The copper electrode loses mass and the silver electrode gains mass as the reaction proceeds. The overall cell reaction is \(Cu(s) + 2 Ag^{+}(aq) -> Cu^{2+}(aq) + 2 Ag(s)\). The emf generated by the cell under standard conditions is +0.46 V.

Step-by-step solution

Determine the half-reactions for the reduction and oxidation processes

The half-reactions for the reduction and oxidation processes are represented as follows: \(Ag^{+}(aq) + e^{-} -> Ag(s)\) \(Cu^{2+}(aq) + 2 e^{-} -> Cu(s)\)

Identify the anode and cathode by comparing the standard reduction potentials

To determine which electrode serves as the anode and which serves as the cathode, we need to compare the standard reduction potentials. Consulting a standard reduction potential table, we find the following values: For silver: \(E^{0}_{Ag+ -> Ag} = +0.80V\) For copper: \(E^{0}_{Cu^{2+} -> Cu} = +0.34V\) Since silver has a higher standard reduction potential, it will serve as the cathode, while copper will serve as the anode.

Determine which electrode gains or loses mass

At the anode, oxidation occurs, and atoms of the metal will lose electrons and enter the solution as ions. This means that the anode (copper) will lose mass as the reaction proceeds. At the cathode, reduction occurs, and metal ions in the solution will gain electrons and become solid metal atoms. This means that the cathode (silver) will gain mass as the reaction proceeds.

Write the overall balanced cell reaction

Now we need to balance the half-reactions and write the overall cell reaction. The balanced half-reactions are: Anode (oxidation): \(Cu(s) -> Cu^{2+}(aq) + 2 e^{-}\) Cathode (reduction): \(2 Ag^{+}(aq) + 2 e^{-} -> 2 Ag(s)\) Combining the two half-reactions, we get the balanced overall cell reaction: \[ Cu(s) + 2 Ag^{+}(aq) -> Cu^{2+}(aq) + 2 Ag(s) \]

Calculate the emf generated by the cell under standard conditions

To calculate the emf generated by the cell under standard conditions, we can use the Nernst equation: \[E_{cell} = E^{0}_{cathode} - E^{0}_{anode}\] Using the standard reduction potentials for silver (\(+0.80 V\)) and copper (\(+0.34 V\)), we get: \[E_{cell} = (+0.80 V) - (+0.34 V) = +0.46 V\] Therefore, the emf generated by the cell under standard conditions is +0.46 V.

Key concepts

Standard Reduction Potential

In electrochemical cells, understanding the concept of standard reduction potential is crucial. It refers to the tendency of a chemical species to gain electrons and be reduced. These potentials are measured under standard conditions: 1 molar concentration, 1 atm pressure, and 25°C. It's a way to compare how readily a substance will undergo reduction compared to the hydrogen ion's standard reference.For instance, in this exercise, we examine the standard reduction potential for silver and copper. Silver has a standard reduction potential of \(+0.80 \, V\), while copper's is \(+0.34 \, V\). This indicates that silver ions are more likely to gain electrons (be reduced) than copper ions under identical conditions. Knowing these potentials helps us predict which electrode will act as the anode or cathode.

Anode and Cathode Identification

The identification of the anode and cathode in an electrochemical cell is pivotal for determining the flow of electrons. The basic rule is: the anode is where oxidation occurs, and the cathode is where reduction takes place.

• The anode, often referred to as the negative electrode, is the site where electrons are lost (oxidation). In this exercise, copper acts as the anode because it has a lower standard reduction potential (\(+0.34 \, V\)) than silver.
• The cathode, the positive electrode, is where the gain of electrons (reduction) occurs. Silver plays the role of the cathode due to its higher reduction potential (\(+0.80 \, V\)).

The movement of electrons from the anode to the cathode generates current flow, reflecting the potential difference between these electrodes.

Cell Reaction Balancing

Balancing the cell reaction involves ensuring that the number of electrons lost in oxidation equals the number gained in reduction. This balance maintains charge neutrality and accounts for both mass and charge.

• At the anode, copper oxidizes: \(Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{-}\).
• At the cathode, silver ions reduce: \(2 Ag^{+}(aq) + 2 e^{-} \rightarrow 2 Ag(s)\).

Both half-reactions must be combined to depict the overall cell reaction:\[ Cu(s) + 2 Ag^{+}(aq) \rightarrow Cu^{2+}(aq) + 2 Ag(s) \]This reaction shows that copper metal loses electrons, while silver ions gain them, illustrating the entire redox process in the cell.

Electromotive Force (EMF) Calculation

The electromotive force (EMF) of a cell is a measure of the cell's potential to drive electric current. It's derived from the difference in standard reduction potentials of the cathode and the anode, calculated using the Nernst equation under standard conditions:\[ E_{cell} = E^{0}_{\text{cathode}} - E^{0}_{\text{anode}} \]Using our specific cell, with silver as the cathode \((+0.80 \, V)\) and copper as the anode \((+0.34 \, V)\), the EMF becomes:\[ E_{cell} = 0.80 \, V - 0.34 \, V = 0.46 \, V \]Thus, the cell generates an EMF of \(+0.46 \, V\) under standard conditions. This EMF represent the potential energy difference between the cathode and anode, causing the electrons to flow through the circuit and perform electrical work.