Problem 70 - Chemistry The Central Science In Si Units

Question

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl^{-} ] will increase, decrease, or stay the same as the cell operates.

Step-by-Step Solution

Answer

(a) The electrode with 0.0150 M Cl⁻ is the cathode. (b) Standard EMF is 0 V. (c) Cell EMF is 0.264 V. (d) [Cl⁻] increases at the cathode and decreases at the anode.

Step 1: Identify the Cathode and Anode

In a galvanic cell, the electrode with the higher reduction potential under the given concentration conditions will act as the cathode. Compare the chloride ion concentrations: 0.0150 M versus 2.55 M. The lower concentration corresponds to a larger driving force for the reduction reaction according to Le Chatelier's principle. Hence, the half-cell with \( \left[ \mathrm{Cl}^{-} \right] = 0.0150 \text{ M} \) is the cathode.

Step 2: Calculate the Standard EMF

The standard reduction potential for the silver-silver chloride electrode is typically 0.222 V. Since both electrodes are made of the same material and standard conditions refer to a concentration of 1 M for all participants, when both electrodes are considered in their standard state, the standard EMF (\( E^\circ_{\text{cell}} \)) is zero because \( E^\circ_{\text{cathode}} = E^\circ_{\text{anode}} \). Therefore, \( E^\circ_{\text{cell}} = 0 \text{ V} \).

Step 3: Calculate the Cell EMF with Given Concentrations

Use the Nernst equation to calculate the cell EMF for the specific chloride ion concentrations: \[ E = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \left( \frac{ [\text{Cl}^-]_{\text{anion}} }{ [\text{Cl}^-]_{\text{cathode}} } \right) \]Where \( n = 1 \), \( R \) is the gas constant (8.314 J mol⁻¹ K⁻¹), \( T \) is temperature in Kelvins (assume 298 K), and \( F \) is the Faraday constant (96485 C mol⁻¹):\[ E = 0 - \frac{8.314 \times 298}{1 \times 96485} \ln \left( \frac{2.55}{0.0150} \right) = -0.0592 \times \log \left( \frac{2.55}{0.0150} \right) \]Calculating, \( E \approx 0.264 \text{ V} \). This positive value confirms the direction of electron flow.

Step 4: Predict the Change in \( [Cl^{-}] \) at Each Electrode

At the cathode (0.0150 M \( [\text{Cl}^-] \)), \( \text{Cl}^- \) is produced, thus concentration increases. At the anode (2.55 M \( [\text{Cl}^-] \)), chloride ions are consumed in the reverse reaction, so the concentration decreases.

Key Concepts

Electrodes

In a voltaic or galvanic cell, electrodes play a crucial role as they facilitate the flow of electrons through the cell. Understanding what happens at each electrode is essential for grasping how the entire cell operates.

An electrode is a conductor through which electricity enters or leaves an object, substance, or region in an electrochemical cell. In our given scenario with silver-silver chloride electrodes, each electrode is involved in specific half-reactions. In every cell, there are two types of electrodes: the anode and the cathode.

• The anode is where oxidation occurs. Electrons are released here, flowing out into the external circuit.


• The cathode is where reduction takes place. Electrons enter here, coming from the circuit.

In the given reaction, differences in chloride ion concentrations between the two half-cells dictate which electrode will act as the anode or cathode. The electrode with lower chloride ion concentration acts as the cathode since it's more favorable for reduction, while the electrode with higher concentration acts as the anode. This distinction is vital for the flow and harnessing of electrical energy in the cell.

Reduction Potential

Reduction potential is a measure of the tendency of a chemical species to be reduced by gaining electrons. It is essential for understanding the direction of electron flow in voltaic cells.

Each half-reaction or electrode has a standard reduction potential associated with it. The standard reduction potential ( E° ) is expressed in volts (V) and is compared to a standard hydrogen electrode. In our silver-silver chloride system, the standard reduction potential is 0.222 V under standard conditions.

So how does this potential help us identify the cathode and anode?

• Electrode with higher reduction potential tends to be the cathode, as these reactions gain electrons easily.


• In contrast, the electrode with lower potential usually acts as the anode.

For different chloride concentrations, the effective reduction potential changes. Le Chatelier's principle suggests that for the silver-silver chloride system, a lower concentration of chloride ions results in a higher driving force for reduction, making that electrode the cathode.

Nernst Equation

The Nernst equation provides a way to calculate the cell EMF under non-standard conditions by incorporating concentrations of the reacting species. It is an extension of the concept of standard reduction potential to real-world scenarios.

The Nernst equation is given by: \[ E = E^°_{\text{cell}} - \frac{RT}{nF} \ln \left( \frac{[\text{Cl}^-]_{\text{anion}}}{[\text{Cl}^-]_{\text{cathode}}} \right) \]

Let's break down the components of this equation:

• \(E\) is the cell potential at the given concentrations.
• \(E^°_{\text{cell}}\) is the standard cell potential, which in our exercise is 0.
• \(R\) is the universal gas constant (8.314 J/mol·K).
• \(T\) is the temperature in Kelvin; assuming a common condition of 298 K.
• \(n\) represents the number of moles of electrons exchanged, 1 in this case.
• \(F\) is the Faraday constant (96485 C/mol).

Using the provided chloride ion concentrations, the Nernst equation helps to compute a cell potential (EMF) of approximately 0.264 V. This output tells us that the system is actively converting chemical energy into electrical energy.

Cell EMF

The electromotive force (EMF) of the cell, also referred to as cell potential, is the capacity of the cell to drive an electric current through an external circuit. It's crucial for assessing the work that the cell can perform.
In a voltaic cell such as the one in our problem, cell EMF is the difference in potential between the two electrodes. How can we determine this value?

Under standard conditions, where concentrations of ions are 1 M, the

• Standard EMF ( E° ) is zero for identical electrodes, since their reduction potentials cancel each other out.

However, under non-standard conditions with given concentrations, the Nernst equation reveals a

• Cell EMF of approximately 0.264 V in our scenario.

This positive EMF indicates that the cell can indeed do work by driving electrons from the anode to the cathode.