Problem 54 - Chemistry Matter and Change

Question

You design a battery that uses a half-cell containing Sn and \(\mathrm{Sn}^{2+}\) and another half-cell containing Cu and \(\mathrm{Cu}^{2+} .\) The copper electrode is the cathode, and the tin electrode is the anode. Draw the battery and write the half-reactions that occur in each half-cell. What is the maximum voltage this battery can produce?

Step-by-Step Solution

Answer

The battery can be represented as: Cu | Cu^2+(aq) || Sn^2+(aq) | Sn. The half-reactions for each half-cell are anode (Sn half-cell): \(Sn \rightarrow Sn^{2+} + 2e^-\) and cathode (Cu half-cell): \(Cu^{2+} + 2e^- \rightarrow Cu\). The maximum voltage this battery can produce, using the standard reduction potentials and the Nernst equation, is 0.48 V.

Step 1: Draw the battery

We can draw the battery using the following diagram: ``` Cathode (Cu half-cell) Anode (Sn half-cell) Cu | Cu^2+(aq) || Sn^2+(aq) | Sn ``` In the diagram, Cu and Cu^2+ represent the cathode half-cell, and Sn and Sn^2+ represent the anode half-cell. The vertical lines indicate the salt bridge connecting the two half-cells.

Step 2: Write the half-reactions

For each half-cell, we write the corresponding half-reactions: Anode (Sn half-cell): \(Sn \rightarrow Sn^{2+} + 2e^-\) Cathode (Cu half-cell): \(Cu^{2+} + 2e^- \rightarrow Cu\)

Step 3: Calculate the maximum voltage

The maximum voltage can be calculated using the Nernst equation and the standard reduction potentials of the half-reactions: \(E_{cell} = E_{cathode} - E_{anode}\) Using the following standard reduction potentials: Standard reduction potential for \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s): E^0_{Cu^{2+}/Cu} = 0.34 V\) Standard reduction potential for \(Sn^{2+}(aq) + 2e^- \rightarrow Sn(s): E^0_{Sn^{2+}/Sn} = -0.14 V\) We can calculate the cell potential by plugging in the values to the Nernst equation: \(E_{cell} = 0.34 V - (-0.14 V) = 0.34 V + 0.14 V = 0.48 V\) The maximum voltage this battery can produce is 0.48 V.

Key Concepts

Half-Cell Reactions

In the world of electrochemistry, a half-cell reaction is a fundamental concept that makes electrochemical cells work. Each half of an electrochemical cell is called a half-cell. It consists of an electrode and its surrounding solution, where a particular ion is involved in oxidation or reduction.
In our battery example, we have two half-cells. One contains tin (Sn) and its ion \(\mathrm{Sn}^{2+}\), and the other contains copper (Cu) and its ion \(\mathrm{Cu}^{2+}\).

• Anode (Sn half-cell): The reaction at the anode involves the oxidation of tin. Oxidation means losing electrons, so the reaction for this half-cell is: \(Sn \rightarrow Sn^{2+} + 2e^-\).
• Cathode (Cu half-cell): At the cathode, reduction takes place, which means gaining electrons. The copper ion gains electrons to form solid copper: \(Cu^{2+} + 2e^- \rightarrow Cu\).

These half-reactions are crucial because they allow us to understand how the flow of electrons generates electrical energy in the battery.

Standard Reduction Potentials

Standard reduction potentials are like a measure of how easily a species gains electrons, essentially how 'hungry' they are for electrons. This is crucial when predicting which half-reaction in an electrochemical cell will occur at the cathode and which at the anode. Standard reduction potentials are typically measured under standard conditions of 1 M concentration, 25°C, and 1 atm pressure.
Each half-reaction has its own potential, called the standard electrode potential \(E^0\). Here are the standard reduction potentials for our example:

• For the copper reaction: \[Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\], \(E^0_{Cu^{2+}/Cu} = 0.34\, V\).
• For the tin reaction: \[Sn^{2+}(aq) + 2e^- \rightarrow Sn(s)\], \(E^0_{Sn^{2+}/Sn} = -0.14\, V\).

The more positive \(E^0\) value indicates a stronger tendency to gain electrons and be reduced. In our battery, the copper half-cell has a higher \(E^0\) value, meaning it will be the half-cell that gets reduced, acting as the cathode.
By comparing these values, we can determine the directions of the electron flow and calculate the overall cell potential for the battery.

Nernst Equation

The Nernst Equation is a powerful tool in electrochemistry, enabling us to calculate the cell potential of an electrochemical cell under different conditions. Although standard reduction potentials assume ideal conditions, real-life scenarios rarely fit these parameters. This is where the Nernst Equation comes in handy.
The formula for the Nernst Equation is:

• \(E_{cell}=E^0_{cell} - \frac{RT}{nF} \ln Q\)

Where:

• \(E_{cell}\) is the cell potential at given conditions.
• \(E^0_{cell}\) is the standard cell potential.
• \(R\) represents the gas constant \(8.314 \text{ J mol}^{-1} \text{K}^{-1}\).
• \(T\) stands for the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred in the reaction.
• \(F\) is the Faraday constant \(96485 \text{ C mol}^{-1}\).
• \(Q\) is the reaction quotient, which becomes \(\frac{[Sn^{2+}]}{[Cu^{2+}]}\) for our battery.

This equation helps us adjust the predicted cell voltage, taking into account deviations from standard conditions like ion concentrations or temperatures.
In essence, while standard reduction potentials provide a snapshot under ideal conditions, the Nernst Equation paints a realistic picture of the electrochemical cell in action.