A voltaic cell consists of two \(\mathrm{H}_{2} / \mathrm{H}^{+}\) half-cells. Half-cell \(\mathrm{A}\) has \(\mathrm{H}_{2}\) at 0.95 atm bubbling into \(0.10 \mathrm{M} \mathrm{HCl}\). Half-cell \(\mathrm{B}\) has \(\mathrm{H}_{2}\) at 0.60 atm bubbling into \(2.0 \mathrm{M} \mathrm{HCl}\). Which half-cell houses the anode? What is the voltage of the cell?

Short Answer

Expert verified
Half-cell B houses the anode. The voltage of the cell is approximately 0.1617 V.

Step by step solution

01

- Write the Nernst Equation

The Nernst equation is given by: \[ E = E^0 - \frac{0.0592}{n} \log Q \] where \( E^0 \) is the standard electrode potential, \( n \) is the number of moles of electrons transferred in the half-reaction, and \( Q \) is the reaction quotient.
02

- Identify Standard Electrode Potential

For the hydrogen electrode, the standard electrode potential \( E^0 \) is 0 V for both half-cells.
03

- Determine the Reaction Quotient (Q)

The reaction quotient, \( Q \), for the half-cell reaction is given by: \[ Q = \frac{[H^+]^2}{P(H_2)} \] For half-cell A, where \([H^+] = 0.10 \, M\) and \( P(H_2) = 0.95 \, atm\): \[ Q_A = \frac{(0.10)^2}{0.95} \approx 0.01053 \] For half-cell B, where \([H^+] = 2.0 \, M\) and \( P(H_2) = 0.60 \, atm\): \[ Q_B = \frac{(2.0)^2}{0.60} \approx 6.6667 \]
04

- Calculate Electrode Potentials for Both Half-Cells

Using the Nernst equation, calculate \( E \) for each half-cell. For half-cell A: \[ E_A = 0 - \frac{0.0592}{2} \log 0.01053 \] Solving this, \[ E_A \approx 0.0888 \, V \] For half-cell B: \[ E_B = 0 - \frac{0.0592}{2} \log 6.6667 \] Solving this, \[ E_B \approx -0.0729 \, V \]
05

- Identify the Anode and Cathode

The anode is the half-cell with the lower electrode potential. Comparing the two potentials: \[ E_A = 0.0888 \, V \quad (Higher) \] \[ E_B = -0.0729 \, V \quad (Lower) \] Therefore, half-cell B (with \( E_B = -0.0729 \, V \)) is the anode.
06

- Calculate the Voltage of the Cell

The cell voltage, \( E_{cell} \), is given by: \[ E_{cell} = E_{cathode} - E_{anode} \] Plug in the electrode potentials: \[ E_{cell} = 0.0888 - (-0.0729) \] \[ E_{cell} \approx 0.1617 \, V \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst Equation
The Nernst Equation is a key tool in electrochemistry for determining the cell potential under non-standard conditions. It is formulated as follows:
\[ E = E^0 - \frac{0.0592}{n} \,\log \, Q \]
Here, \( E \) represents the electrode potential, \( E^0 \) is the standard electrode potential, \( n \) is the moles of electrons transferred in the half-reaction, and \( Q \) is the reaction quotient.
The reaction quotient \( Q \) involves the concentrations of the reactants and products. For a hydrogen electrode, \( Q \) is calculated as follows:
\[ Q = \frac{[H^+]^2}{P(H_2)} \]
This relationship allows us to see how the cell potential changes with different hydrogen ion concentrations and gas pressures. Whether you're dealing with a voltaic or electrolytic cell, the Nernst Equation is instrumental in predicting cell behavior under various conditions.
Electrode Potential
Electrode potential refers to the voltage (potential) difference between an electrode and its surrounding solution. The standard electrode potential \( E^0 \) is measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C).
In the given exercise, both half-cells use a standard hydrogen electrode, for which \( E^0 \) is universally 0 V. Real-world cells rarely operate under these standard conditions, which is why the Nernst Equation is so essential.
Changes in hydrogen ion concentration and pressure lead to different reaction quotients \( Q \), and we use these to calculate the electrode potential for each half-cell. These calculations provide insights into which half-cell will act as the anode and which will act as the cathode.
Anode and Cathode Identification
In any electrochemical cell, identifying the anode and cathode is critical. The anode is where oxidation occurs, and for voltaic cells, it is the electrode with the lower (more negative) electrode potential. Conversely, the cathode is where reduction occurs and has the higher (more positive) electrode potential.
For example, in the given problem, half-cell A has an electrode potential of 0.0888 V, while half-cell B has an electrode potential of -0.0729 V. Therefore, half-cell B, with the lower potential, is the anode, and half-cell A is the cathode.
The flow of electrons occurs from the anode to the cathode, and knowing the correct identification of these electrodes is essential for properly calculating the cell voltage.
Cell Voltage Calculation
Calculating the cell voltage is a straightforward but crucial part of understanding voltaic cells. The cell voltage (also known as electromotive force or EMF) is given by the difference in electrode potentials between the cathode and the anode:
\[ E_{cell} = E_{cathode} - E_{anode} \]
In the given exercise, we have already identified the potentials:
\[ E_A = 0.0888 \, V \ (cathode)\]
\[ E_B = -0.0729 \, V \ (anode)\]
We plug these values into our equation:
\[ E_{cell} = 0.0888 - (-0.0729) \, V = 0.1617 \, V \]
This calculation tells us that the voltaic cell, as assembled, will have a cell voltage of approximately 0.1617 V. This voltage is essentially the driving force that pushes electrons through the circuit from the anode to the cathode.
Hydrogen Electrode
The standard hydrogen electrode (SHE) is a reference electrode that is universally recognized due to its stable and well-defined electrode potential of 0 V under standard conditions. It involves a half-cell where hydrogen gas is bubbled through a solution of hydrogen ions at 1 M concentration and 1 atm of pressure.
In practice settings, deviations from these conditions are common. As seen in the exercise, variations in hydrogen ion concentration and gas pressure impact the reaction quotient \( Q \), thus altering the electrode potential. This is why calculations with the Nernst Equation are necessary.
The reliance on the hydrogen electrode stems from its repeatability and predictability. For many cells, it serves as a baseline against which other electrodes are measured. Understanding the hydrogen electrode's role helps you better grasp why certain reactions occur and how those reactions drive a voltaic cell.

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